Fluid Pressure, Density, Archimede & Pascal's Principle, Buoyant Force, Bernoulli's Equation Physics

in this video we're going to go over fluids we're going to talk about density buoyant force archimedes principle pascals law hydraulic lifts bernoulli's equation and things like that so the first thing that we need to talk about is density the density of an object is the mass divided by the volume this symbol is also known as rho now don't confuse it with pressure which i'm going to use capital p now if you rearrange the equation mass is density times volume the density of water is about one gram per cubic centimeter this is also equal to a thousand kilograms per cubic meter the specific gravity of a substance is the density of that substance let's say substance a divided by the density of water so for example if the density of aluminum is 2.7 grams per cubic centimeter what is the specific gravity of aluminum is the density of aluminum divided by the density of water and it's simply 2.7 so the specific gravity compares the density of a substance with that of water let's try this problem what is the weight of air inside a spherical balloon of radius 2 meters and you're given the density of air ignore the mass of the balloon feel free to pause the video and try this example so let's say if we have a balloon that has a spherical shape and there's gas particles inside the balloon and the gas even though it's light it does have mass so it exerts a downward weight force what is the weight force of those gas particles we know that weight is equal to mass times gravity and mass is density times volume so the weight of the air inside the balloon is the density times the volume times gravitational acceleration so it's 1.29 kilograms per cubic meter that's the density but we need to find the volume the volume of a sphere is four thirds pi r cube the radius is two meters so two to the third power that's eight eight times four is thirty two thirty two divided by 3 is 10.67 times pi this is about 33.51 cubic meters so that's the volume times the gravitational acceleration of 9.8 meters per second squared so if we multiply these three numbers this will give us a weight force of 423.6 newtons now let's talk more about density so let's say if you have a beaker containing water and a density of water is about a thousand kilograms per cubic meter now let's say if you place a chunk of aluminum what's going to happen will it sink or float the density of aluminum is about 2.7 grams per cubic meter i mean cubic centimeter which is 2700 kilograms per cubic meter so because aluminum has a higher density than water it will sink but let's say if you place an ice cube in water ice floats on top of water the reason why ice floats is because the density of ice is less than water the density of ice is about 917 kilograms per cubic meter so about 91.7 percent of the volume of ice will be below the surface of water so as you can see heavy objects sink and light objects they tend to float now the same behavior is true for gases let's say if you have a balloon that contains helium helium is lighter than air and so helium is going to rise now if you have a balloon containing carbon dioxide it's going to sink carbon dioxide is heavier than air air molecules are mostly composed of nitrogen and oxygen gas if you look at the periodic table oxygen has an atomic mass of 16 times the two oxygen atoms that you see here so the molar mass is about 32 grams per mole for n2 it's 14 times 2 is 28. now air is about 78 to 79 percent of nitrogen and approximately 21 oxygen there's some other gases too but it's around this number so we'll say that nitrogen is 78 and then the other one percent is the other gases so the average molar mass of air is somewhere between 28 and 32 the molar mass of helium is four the molar mass of co2 is 12 plus 16 times two that's 44. so because carbon dioxide is heavier than air it's going to sink but because helium is lighter than air it's going to rise another fun fact about air is temperature hot air rises and cold air sinks in hot air the gas particles are further apart and so hot air is less dense than cold air in cold air the gas particles are closer together and so cold air has a higher density than hot air so hot air tends to rise cold air tends to sink if you live in a three-story apartment or house if you turn on the ac you realize that the first floor even the basement will be the coldest area in the house and the attic is usually the hottest part of the house that's because all the hot air revises towards the attic and the cold air sinks towards the basement so this principle is useful for hot air balloons for example let's say if you have a a very massive hot air balloon let's say there's a torch so as you heat up the air inside the balloon the balloon expands and the hot air is less dense or lighter than the surrounding cooler air so let's say the temperature inside this balloon is maybe 50 degrees celsius and outside let's say it's 20. so as you heat up the air on the inside the hot air will rise now if you want the hot air balloon to descend you basically turn this off and the temperature will decrease and eventually the weight of the balloon and this package or the people inside it will eventually cause the hot air balloon to descent so by adjusting the temperature you can control if it's going to ascend or descend now here's a question for you is it easier to float in pure water or in salt water what would you say so is it easier to float let's say in a pool versus the ocean in a pool it's mostly pure water it might be some chlorine inside of it too but for the most part the density of the water that you'll find in a typical pool is about a thousand kilograms per cubic meter now the density of seawater which is basically salt water is a little bit higher it's about 1025 kilograms per cubic meter so therefore it's easier to float in the ocean than in a pool because the density of seawater is higher than that of a typical pool now let's say if you were to go to a very salty lake water let's say like the dead sea or something the density of the dead sea is very very high because it's so salty if you were to jump in and try to swim to the bottom you won't be able to the buoyant force there is so strong that it's going to lift you back to the top and you're going to float right on top of it the density of the dead sea water is about 1240 kilograms per cubic meter and the higher the density of a fluid the harder it is to sink it's easier to float so it's easier to float in sea water then there is the flow in the pool by the way a fluid is anything that could flow so liquids and gases because they can both flow are considered to be fluids let's try problem an empty bottle has a mass of 25 grams if you fill it with water the combined mass is 95 grams when filled with another fluid the combined mass is 75 grams what is the density of this unknown fluid so what would you do in this problem so we need to find the volume of the empty bottle of water because if we have that then using the mass of the fluid we can calculate the density of the fluid now we know the density of water it's about one gram per cubic centimeter so what is the mass of the water the combined mass of the bottle and the water is 95 grams and the mass of the bottle is 25 so the mass of water must be the difference between these two values it's 95 minus 25 which is 70 grams so now let's calculate the volume density is equal to mass over volume and volume is mass of a density so 70 grams of water divided by a density of one gram per cubic centimeter tells us that the volume of the bottle is 70 cubic centimeters now that we have the volume of the bottle let's calculate the density of the fluid so first we got to find the mass of the fluid the combined mass of the fluid and the bottle is 75 grams and the mass of the bottle is 25 so the mass of the fluid is 75 minus 25 so it's 50 grams therefore the density of the fluid is the mass divided by the volume so it's 50 divided by 70. 50 divided by 70 is about 5 over 7 or 0.714 grams per cubic centimeter now that we have the density if you need to change your units from grams per cubic centimeter to kilograms per cubic meter multiply by a thousand so this is equal to 714 kilograms per cubic meter so that's the answer for this problem 50 liters of water is mixed with 100 liters of alcohol what is the density of this mixture to calculate the density of the mixture all we need to do is take the total mass that is the mass of the water plus the mass of the alcohol and divided by the total volume because density is the total mass of the solution divided by the volume or the total volume of the solution now the mass of water is the density of water times the volume of water and the mass of the alcohol is the density of the alcohol divided by the volume of the alcohol and we're going to divide it by the total volume so let's separate this fraction into two smaller fractions so it's going to be to keep it simple p1 v1 divided by v total plus p2 v2 divided by the total volume now the reason why i did it that way is so that you could see that it really doesn't matter what the units of volume are because it's going to cancel if it's in cubic meters or liters it doesn't matter it's the ratio that's important so i don't have to convert liters into cubic meters to get the right answer so therefore the density of the mixture is going to be the density of water which is a thousand times the volume which is 50 liters divided by the total volume which is 150 liters plus the density of the alcohol which is 790 times the volume of the alcohol which is 100 liters divided by the total volume of 150 liters so we can cancel these two zeros so the first part a thousand times five divided by 15 is about 333 so that's the contribution from the water and for the other part 790 times 10 over 15 it's about 526.6 if we add the two it's 859.6 so that is the density in kilograms per cubic meter for the mixture and it makes sense though notice that the density of the mixture is between the density of water and the density of aluminum now it's not exactly in the middle if you average a thousand and 790 if you add the two numbers and then divide by two the average between those two numbers is 895. now if we were to mix 50 liters of water with 50 liters of alcohol then the density of the mixture should be 895 but notice that the density of the mixture is closer to the density of the alcohol it's between 790 and 895 as opposed to 8.95 in the thousand why is the density of the mixture closer to that of alcohol the reason for that is simple we make some more of the alcohol than the water so the density of the mixture should be closer to the density of the pure alcohol than pure water now let's talk about pressure so pressure is the ratio between force and area the unit for force is uh newtons and the unit for area is square meters the unit for pressure is pascals one pascal is one newton per square meter and one atm or one atmospheric pressure is 101.3 kilopascals kpa or 101 300 pascals now you might see units of tor or millimeters mercury one atm is equal to 760 millimeters of mercury and this is also equal to 760 units of tor now these units are usually more associated with chemistry but in physics the standard uniform pressure is pascals now let's go over the concept of pressure so let's say if we have a square that is one meter by 1 meter and if we have a larger square that is 10 meters by 10 meters now if we apply a force of a hundred newtons across the area of this square and the same force of 100 newtons across throughout this entire square which one has a larger pressure value is it the one on the left or the one on the right so pressure is force over area the area for the square on the left is one times one it's one square meter but for the figure on the right it's 10 meters times 10 meters so it's 100 square meters so we have a force of 100 newtons being applied over an area of one square meter so the pressure is force over area it's a hundred pascals now for the figure on the right a hundred newtons is being applied over an area of 100 square meters so 100 over 100 is one pascal so the pressure is significantly higher on the figure on the right so if you apply a force over a very small area the pressure is going to be very very high to illustrate this concept let's say if you push someone with your hand and you apply a gentle force it's not going to cause much damage but let's say if you apply the same force but with a pen and you poke someone with the pen which one is going to cause more damage pushing someone with your hand or applying the same force with a pen because the area of a pen is so small and if you apply the same force over such a small area the pressure is significantly higher for example if you push someone let's say you apply a force of 100 and let's say the area of this surface is one square meter the pressure is only 100 pascals but now let's say if you apply a force of 100 and let's say this area is significantly smaller let's say it's a point zero one square meters a hundred divided by point zero one is ten thousand so the pressure is significantly larger because the area is so much smaller and that's why poking someone with a pen causes much more damage than pushing someone gently with the palm of your hand it's because you're applying a significant force over a very very small area so the pressure is very large now let's say if we have a rectangular box and let's say the dimensions of the box are 5 meters by 4 meters and it has a height of 3 meters and let's say the mass of the box is 100 kilograms now what is the pressure exerted by the box at the ground below so let's say the box is on this horizontal surface what is the pressure exerted in this shader region so we know that pressure is force divided by area and the force exerted on the table in the purple shaded region is basically the weight of the box which is mg so the mass of the box is 100 the gravitational acceleration is 9.8 and the area of the bottom surface is basically the length times the width we don't need the height so it's 5 times 4. so the weight force is 100 times 9.8 which is 9 800 newtons applied over an area of 20 square meters so 980 divided by 2 is 490 so the pressure is 490 pascals now let's say if we have a very tall container that is filled with water and let's say the container is sealed so it's not open to the atmosphere how can we calculate the pressure at this point in this container so it's filled with water can we find a pressure let's say it's a very large tank and we want to find a pressure 5 meters below the surface what can we do so we know that pressure is force divided by area so the force is going to be the weight of the water mg that is currently above this surface where the point of interest is located so the mass of water exerts a downward force on this cross-sectional area and that wave forces the mass of the water above it times the gravitational acceleration divided by the area now because we have a fluid it's best to think of the mass of a fluid as being the density multiplied by the volume remember density is mass over volume so mass is density times volume so let's replace m with pv so the pressure at that surface is the density of the fluid times the volume of the fluid above it multiplied by gravitational acceleration times the area now the volume of the water above it is basically the volume of a rectangle which is the length times the width times the height so the volume as you mentioned before is l times h times w now left times width will give us the area of this surface which is basically the area of this surface here so length times width is area so volume equals area times height so let's replace v with area times height so the pressure is going to be the density times the area times the height times gravitational acceleration divided by the area so we can cancel a therefore the pressure due to the water above this cross-sectional area is equal to the density times the gravity times the height so now we can calculate it the density of water is about a thousand kilograms per cubic meter times g which is 9.8 times 5 meters so it's basically 5000 times 9.8 so the pressure due to the water above that point of interest is 49 000 pascals now what would be the pressure if this container was not closed if it was open to the environment in that case the pressure at this point would be higher because you have the weight of the atmosphere pushing down on the fluid so in this case the pressure at the point of interest is going to be called the absolute pressure that's p and that's equal to the pressure of the atmosphere plus the gage pressure the gauge pressure is simply the pressure that is above the atmosphere the atmospheric pressure is 101 300 pascals and the gauge pressure is the pressure due to the weight of the water above that cross-sectional area so that's the 49 000 pascals so the absolute pressure at that point is going to be a hundred fifty thousand and three hundred pascals let's try another problem let's say if we have a container that's open to the atmosphere and we have water and we also have let's say oil the density of the water we know it to be a thousand kilograms per cubic meter and let's say the density of the oil above it is 700 kilograms per cubic meter and let's say that the height of the oil is about 4 meters and the height of the water is about 5 meters and let's not forget that we have the weight of the atmosphere basically the weight of all the air molecules above this fluid pressing down on it exerting its own pressure so what is the pressure at the bottom of this container so that pressure the absolute pressure or the total pressure at that point is the sum of the pressures caused by the weight of the water above that point plus the weight of the oil and the weight of the atmosphere so the pressure or the gauge pressure due to water alone is going to be the density of water times the gravitational acceleration times the height and the pressure caused by the oil is the density of the oil times g times h plus the atmospheric pressure the density of water is about a thousand g is 9.8 and the height of the water is about five plus the density of the oil which is 700 times g times the height of the oil which is four and then plus the pressure due to the atmosphere which is a hundred and one thousand three hundred pascals so 5000 times 9.8 that's 49 000 that's the pressure due to h2o alone and then 700 times 9.8 times 4. that's 27440 plus the weight of the or the pressure due to the atmosphere so if we add the three numbers 27440 plus 49 000 plus 101 300 the absolute pressure at this point or the total pressure is 177 740 pascals so basically you need to add the gauge pressure of the water and the gauge pressure of the oil plus the atmospheric pressure of the air and that'll give you the total pressure at that point now here's a question for you let's say if we have a rectangular surface and it's eight meters by nine meters and we want to calculate the total force exerted by the atmosphere on this surface how would you do it so we know that pressure is force divided by area so the force exerted by the atmosphere on the surface is pressure times area and the area of the surface is the length times the width so the force exerted by the atmosphere is the pressure of the atmosphere which is 101 300 pascals times a length of 9 and a width of 8. so over this large area the force exerted by the atmosphere is 7 million 293 thousand and 600 newtons of force so that's the entire weight of the air above this particular surface so all of the molecules imagine if this was one big vertical column that extended from the ground to outer space the weight of all of the molecules in this vertical column is equal to this force right here so that's the weight of the atmosphere above this surface from the earth to outer space now the next topic of interest is the hydraulic lift now this device is filled with a fluid according to pascal's principle the pressure applied to a confined fluid increases the pressure throughout by the same amount so the pressure that we exert on one side let's call it the input pressure is equal to the pressure exerted on the other side the output pressure so if we apply a force over this circular area we're going to exert a pressure and that input pressure is the input force divided by the area and an output force will be exerted and that output force divided by a2 is equal to the output pressure let's call it f2 so if we cross multiply we're going to get the equation f1 a2 is equal to f2 a1 and if we solve for f2 by dividing both sides by a1 we get this equation f2 equals f1 times the ratio of the areas a2 over a1 so let's make some more space now let's apply an input force of 100 newtons and let's say that the area or the radius of that cylinder is 10 centimeters and let's say the radius of this cylinder is 30 centimeters what is the output force that's going to be exerted by this cylinder if we apply an input force of 100 newtons so using this equation f2 is equal to f1 which is 100 newtons times a2 which is basically pi times the radius squared over a1 which is pi r1 squared now we can cancel the pi values so simply the ratio of the radius of those two circles so r2 is 30 centimeters but we need to square it r1 is 10 centimeters and we need to square it as well so the unit centimeters will cancel we don't really need to change it to meters so it's 100 times 30 squared which is 900 divided by 10 squared which is 100. so f2 is equal to 900 newtons now let's think about what this means the output force was increased by a factor of nine it went from a hundred to nine hundred so that means that the mechanical advantage of this machine is equal to nine because it was able to multiply the force the input force by a factor of nine mechanical advantage is the ratio of the output force divided by the input force now even though the force was multiplied by a factor of nine it wasn't a freebie so to speak there was a cost for it it wasn't free so what was the cost that we had to pay to increase the force by a factor of nine it turns out that the input work that you apply to this device is equal to the output work so f1 times d1 is equal to f2 times d2 now let's say if the hydraulic lift went up by one meter that means that we have to apply a force of 100 newtons for a distance of 9 meters because the work can't be increased now we can lose energy by means of friction but let's say if we have a perfect ideal situation the input work equals the output work so the input work is basically a force of 100 newtons times the distance of 9 meters which is 900 joules the output work is the same the machine applies a force of 900 newtons but it only moves a distance of 1 meter so it exerts 900 joules of energy so there's no gain in energy in a system so we basically apply a small force over a longer distance and we get a larger force applied over a shorter distance the concept is similar to that of a lever a lever allows you to leverage your force to multiply so you can lift a larger object for example let's say if the lever arm on the left side is 8 meters and on the right side it's 2 meters if you apply an input force of 100 newtons the output force is going to be 400 newtons the torque created by f1 is equal to the force times the moment arm or the left arm which is perpendicular to the axis of rotation it's actually perpendicular to the line of action and it's between the axis of rotation or the pivot point so that's the moment arm so t1 is a hundred times eight so the torque on the left side is 800 newtons times meters now this creates another torque that has the same magnitude 2 times 400 is also 800 but as you can see with the lever the smaller force is associated with the longer side and the larger force is associated with the shorter side and the same is true for the hydraulic lift the larger force is associated with the shorter side and the smaller force is associated with the shorter distance so now let's work on another problem with the hydraulic lift so let's say if we have if we want to lift a crate that has a mass of 800 kilograms and we want to lift it by a distance of let's say 50 centimeters and let's say that the radius of this portion of the hydroglyph let's say it's 60 centimeters and the radius of this circle let's say it's 10 centimeters so what is the input force that we need to apply to lift the 800 kilogram crate by a distance of 50 centimeters and also what distance do we need to push down on the left side and also what is the pressure exerted on this surface and what is the work that we need to apply to move this object 50 centimeters high calculate the work used in two formulas the first thing that we should do is find the output force the minimum output force that we need to lift the 800 kilogram block or crate is the weight force so f2 has to at least be equal to the weight force in order to lift this object up 50 centimeters so f2 is equal to mg which is 800 times 9.8 so the output force is about 78 40 newtons so now that we have f2 we can calculate f1 f1 divided by a1 is equal to f2 divided by a2 the input pressure is equal to the output pressure according to pascal's principle so looking for f1 a1 is pi r squared but like in the last example we can cancel the pi constants so it's just going to be divided by r1 squared which is 10 centimeters squared and we could leave it in units of centimeters so a2 the pi is cancelled so it's just going to be r2 which is 60 centimeters squared and f2 is 78 40. so let's cross multiply so f1 times 60 squared let's get rid of this is equal to 10 squared times 78 40. 10 squared is 100 times 7840 that's 784 000 divided by 60 squared and so f1 or the input force is 217.8 newtons now how can we calculate the input distance how far do we need to push this particular cylinder down in order to lift the 800 kilogram block 50 centimeters up what do we need to do to calculate the answer the input work must be equal to the output work and we could find the output work which is the work required to lift an object against gravity and whenever you lift an object above ground level you're increasing the potential energy so the work required is basically the change in potential energy which is mg delta h and the weight force mg is basically f2 and the change in height is basically the vertical displacement so it's going to be force which is 780 or 7840 times the vertical displacement in meters which is 0.5 meters 50 centimeters is half of a meter and 100 centimeters is one meter so half of 7840 is 3920 joules so that's how much energy is required to lift the 800 kilogram crate by 50 centimeters which means that we need to apply the same amount of energy to uh push this cylinder down so that this cylinder can go up so the input work and the output work are both 3920 joules so now that we know the value of the input work we can calculate how far we need to push the cylinder down it's going to be the input force times the input distance so work is 3920 the input force is 217.8 and let's solve for g so it's 3920 divided by 217.8 so d is about 18 meters which is pretty high so we need to apply a force of 218 newtons and we need to push this cylinder down by a distance of 80 meters in order to lift the 800 kilogram block by only half a meter so there's no gain in energy however the force does increase so what is the mechanical advantage of this hydrauloglyph the mechanical advantage is the ratio of the output force divided by the input force so that's 7800 divided by 217.8 so the mechanical advantage is about 36 so this particular hydraulic lift multiplies the force by a factor of 36 so if you apply a force of 100 newtons the output force will be 100 times 36 or 3600 newtons now it turns out that there's another way to calculate the mechanical advantage the mechanical advantage of this hydraulic lift is also equal to the ratio of the input distance divided by the output distance so we had to push the cylinder on the left by a distance of 18 meters so that the 800 kilogram crate can go up by 0.5 meters so 18 divided by 0.5 is 36 so there's a trade-off if you want to increase the force you have to apply a smaller force for a longer distance so the work done is the same in both cases now the last thing we need to do is calculate the pressure exerted on the cylinder on the left side so pressure is force divided by area the force exerted on the left is 217.8 newtons and the area is the area of a circle which is pi r squared 10 centimeters is 0.1 meters so 217.8 divided by 0.1 squared and let's take that result divided by pi so this is equal to 6933 newtons well not newtons but pascals since this is now the pressure so that is it for the hydraulic lift so now the next topic of discussion is the mercury barometer so let's say if we have a beaker and if we have a tube and it's filled with mercury now this barometer is open to the atmospheric pressure so the weight of the air exerts a force and the pressure inside is zero so the difference in pressure is just the atmospheric pressure with this information how can you calculate the height of the mercury column how high is it above this level what's the difference in height now let's say that you know the density of mercury the density of liquid mercury is about 13 600 kilograms per cubic meter and you know the atmospheric pressure so what is the height of the mercury column now to figure this out we need to understand that the weight of the atmosphere which is associated with the atmospheric pressure and balances out the weight of this mercury column and the weight of the mercury column is associated with the gauge pressure so we could say that because the weight of the atmosphere must balance the weight of that mercury column the atmospheric pressure must equal the gauge pressure and the gauge pressure is simply the density of the fluid times the gravitational acceleration times the height now we know the atmospheric pressure it's a hundred one thousand and three hundred pascals the density of mercury is 13600 and g is 9.8 so we need to solve for the height so the height of the mercury column is about 0.76 meters which is approximately 76 centimeters now let's say if we replace mercury with water how high can the atmosphere support a column of water how high will the water column b so we can use the same equation but this time the density of water is significantly less than that of mercury so we should expect that we require a higher column to get the same mass of mercury mercury is 13.6 times more dense than water so to get the same mass we need a height that's 13.6 times greater than the height of mercury the density of water is a thousand g is 9.8 let's solve for h so the height of the water column that the atmosphere can support is about 10.3 meters which is very high now remember the height of the mercury column was 0.76 so if you take 0.76 and multiply by the ratio of the density of mercury by the density of water you will get this answer 0.76 times 13 600 divided by a thousand is equal to 10.3 so that's another way you can get the same answer if you know the density of one material and you want to find the height of the other material let's try this problem the height of mercury in an open barometer is 64 centimeters and the height of another fluid is 154 centimeters what is the density of the fluid and what is the pressure of the air so let's draw the open barometer and it has a mercury so let's calculate the pressure of the air first so the outside air is going to exert a pressure and that pressure pushes up the fluid so the weight of the air equals the weight of this column so therefore the pressure of the air is equal to the gauge pressure of that column which is density times gravity times height now for this equation the density of mercury has to be in the units kilograms per cubic meter so if it's 13.6 grams per cubic centimeter that's 13 600 kilograms per cubic meter g is 9.8 and the height of the mercury column in meters to convert centimeters to meters divide by 100 so this is going to be 0.64 so 13 600 times 9.8 times 0.64 that's about 85 299 pascals so that's the pressure of the air now what about the second part of the problem what is the density of the other fluid now the pressure of the air is equal to the gauge pressure of the other fluid and that's also equal to the gauge pressure of mercury so these two are equal to each other so we could divide by the gravitational acceleration those two will cancel so the density of the fluid times the height of the fluid is equal to the density of mercury times the height of mercury so the density of the fluid we're looking for that the height of the fluid is 154 centimeters the density of mercury we can use the units grams per cubic centimeter if we use those units then the density of the fluid will also be in grams per cubic centimeter and the height we can leave it in centimeters because the height of the fluid is in centimeters so the unit centimeters will cancel so the density of the fluid is simply the density of mercury which is 13.6 times the height of the mercury column divided by the height of the fluid column so the density of the unknown fluid is 5.65 grams per cubic centimeter so make sure you understand this because mercury has a higher density the height of the column is going to be less than the height of the fluid that has a lower density so water has a very low density but the height of the water column is very very high it was about like 10 meters which is a thousand centimeters but mercury because it has because it's very dense it has a lot of mass per unit volume the column for mercury is very low and so that's why mercury is very useful to use for barometers temperature gauges is because there's a lot of mass per unit volume now let's say if we have a column basically an open barometer with a column and it's going to be two fluids here's the first fluid and here is the second fluid and let's say on the inside the pressure is zero so let's call this fluid a and fluid b and the height of fluid b let's say it's 30 centimeters and the height of fluid a let's say it's 25 centimeters and let's say that the density of fluid a is 11 grams per cubic centimeter and a density of fluid b let's say it's 14 grams per cubic centimeter so with this information what is the atmospheric pressure how can you calculate it if you have two different fluids so the atmospheric pressure is the sum of the gage pressure for fluid b which represents or is associated with the weight of that column plus the gauge pressure of fluid a so it's going to be pa plus pb so the gauge pressure for fluid a is the density of fluid a times the gravity times the height and for fluid b is simply pgh as well so the density for a now we need it in kilograms per cubic meter so that's going to be 11 000 times g which is 9.8 and the height of fluid a in meters 25 divided by 100 is 0.25 meters and then plus the density of b which is 14 000 times 9.8 times the height of b which is 0.3 meters and we just got to add the two gauge pressures so 11 000 times 9.8 times 0.25 so the gauge pressure for fluid a is 26 950 pascals now for fluid b it's 14 000 times 9.8 times 0.3 so it's about 41 160 pascals so now let's add these two values so the pressure of the atmosphere under these conditions is about 68 110 pascals so that's how you can calculate it with an open barometer that contains two different fluids now let's say if we have an open tube manometer we're going to have two of them so both manometers will contain mercury but the shapes are going to be a little bit different so on the outside it's open to the atmosphere so this is the atmospheric pressure and we also have the atmospheric pressure on this side as well and this is going to be the pressure of the gas which we can just call it p and then we have the height of the mercury column or the height difference which we'll call h so how can you write an equation between the pressure of the atmosphere and the pressure inside the gas for each of these situations so these two pressures the pressure of the gas and the pressure of the atmosphere they're going to be they're going to differ by the gauge pressure which is pgh that's the gauge pressure of the mercury column so how can we find out which equation will be which so there's a question for you which pressure is stronger on the left side is it the atmospheric pressure or the pressure of the gas notice that the gas pushes down more on the mercury than the atmospheric pressure for the picture on the left therefore this pressure is greater than the atmospheric pressure and it differs by the gauge pressure of the mercury column so p is equal to the atmospheric pressure plus the gauge pressure which is the density times gravity times the height of the mercury column now on the other side the air pushes down more on the mercury column than the gas so the atmospheric pressure is stronger so therefore the atmospheric pressure is equal to the pressure of the gas plus the gage pressure so let's work on some examples now let's say if we have a fluid let's say it's some type of oil and the height difference is about let's say 40 centimeters and the density of this oil let's say it's pretty heavy let's say it's three grams per cubic centimeter now let's say the pressure of the gas is 30 kilopascals or rather let's say the pressure of the atmosphere is simply 101.3 kilopascals what is the pressure of the gas so first we need to find out which pressure is greater the atmospheric pressure or the pressure the gas so the atmospheric pressure pushes down more on the oil than the gas so the atmospheric pressure is greater than the gas so the atmospheric pressure is equal to the pressure of the gas plus the gauge pressure of the oil so that's going to be p plus pgh so we already have the atmospheric pressure which we know it to be a hundred and one thousand three hundred pascals and this is equal to the pressure of the gas plus the gauge pressure of the oil which is the density of the oil in kilograms per cubic meter which is 3000 times g and times the height of the column which is 0.4 meters so three thousand times nine point eight times point four is about eleven thousand seven hundred sixty so the pressure of the gas is 101 300 minus 11 760. so it's going to be 89 540 pascals so that's how you can calculate the pressure of the gas inside a manometer you simply need to add or subtract the gauge pressure either from or to the atmospheric pressure and just ask yourself is the gas pressure is it less or greater than the atmospheric pressure and then you can either add or subtract pgh to it the pressure that's greater is the one that pushes the fluid further into the manometer because the fluid is at a higher level on the right side that means the pressure of the gas is less than the pressure of the atmosphere here it's lower which means that this pressure is greater it exerts a greater downward force now how is it that if we place an object that's that has a higher density than water it will sink but if the density is less than water it will float why do light objects float and heavy objects sink every object has a downward weight force so what causes the object to move up towards the surface whenever an object is placed in the fluid there's an upward buoyant force and according to archimedes principle the buoyant force on a body immersed in a fluid is equal to the weight of the fluid displaced by that object so what does that mean anytime you place an object in a fluid that's going to be an upward buoyant force now how can we calculate this upward buoyant force first we need to draw a better picture so let's draw a cube and let's say this cube is submerged in water now at the top surface there's going to be a gauge pressure and at the bottom surface is another gauge pressure and these two pressures because they're at different depths that a different height the pressures will be different the pressure at the bottom is higher than the pressure at the top so because you have a difference in pressure there's going to be a net upward force whenever you have a high pressure and a low pressure there's a force that is directed from high to low and so that that force is going to be the buoyant force that's going to lift the object in the upward direction now let's say that the top surface has an area called a and the same is true for the bottom surface so the force acting on the bottom surface we can call f2 and the force acting on the top surface is f1 so the net force which is the buoyant force is the difference between f2 and f1 f2 is the gauge pressure times the area and f1 is the gate pressure times the area as well but these are two different pressures but the area is the same so what we can do is factor out the area so the buoyant force is the area times p2 minus p1 which is basically the change in pressure now remember that the gage pressure is the density times the gravity times the height so the buoyant force is equal to the area times the density times the gravity times the change in height which is basically the height of the box so now area times height is volume so buoyant force is equal to the density of the fluid times gravity times the volume of the object that is submerged so this is the buoyant force that's how you can calculate it let's try this problem so we have a tank that contains water and there's a rectangular block of wood we want to know if it's going to sink or if it's going to float so a simple way to find out is to calculate the density of the block and see if it's greater than or less than the density of water the density of the block is simply the mass divided by the volume the mass is 175 kilograms the volume of a rectangle is the length times the width times the height but we need to convert the dimensions to meters and we simply need to divide it by 100 to convert centimeters to meters so 50 divided by 100 is 0.5 so the dimensions are 0.5 meters by 0.75 by 0.8 you have to divide each by a hundred so 0.5 times 0.75 times 0.8 gives us the volume of the block which is 0.3 cubic meters so now we can take the mass and divide it by the volume so 175 divided by 0.3 gives us a density of 583.3 kilograms per cubic meter so notice that the block has a density that's less than that of water so it's going to float it's going to rise to the top so what this means is that the upward buoyant force acting on the block exceeds the weight force that brings the block down so there's a net upward force that causes it to rise to the surface so let's calculate the upward buoyant force now according to archimedes principle the buoyant force is equal to the weight of the fluid displaced by the object and the weight of any object is mass times gravity so the weight of the fluid is the mass of the fluid times the gravitational acceleration and we know that mass is density times volume because density is mass divided by volume so we can replace the mass of the fluid with the density of the fluid times the volume of the fluid displaced which is equal to the volume of the object that is submerged so this is the buoyant force equation that we got earlier in the video so the density of the fluid is a thousand the volume of the fluid is uh or the volume that's displaced is point five times point seven five times 0.8 which is the volume of the block and that's 0.3 cubic meters times the gravitational acceleration of 9.8 so the upward buoyant force is 2940 newtons so now let's calculate the weight force so the weight force is simply the weight of the object it's mg so it's 175 kilograms times g which is 9.8 so the weight force is basically 1715 newtons now the net force is the difference between the upward buoyant force and a downward wave force so the net force is 2940 minus 1715. so the block experiences an upward force of 1225 newtons so now that we have the net force what is the upward acceleration of the block according to newton's second law net force is equal to mass times acceleration so if the net force is 1225 and if the mass of the block is 175 then the acceleration is 1225 divided by 175 so the block has an upward acceleration of 7 meters per second squared now acceleration is the rate of change of velocity that means that the upward velocity is increasing by 7 meters per second every second so after a time of 5 seconds the upward velocity will be it's going to be 35 meters per second and you could use the kinematics equation to find out what the final velocity will be at the top or you could find displacement using the kinematics equation so if you need to review that topic you can find my video on youtube so now that we have the net force and the upward buoyant force what volume of the block will remain above the surface of the water at equilibrium now the buoyant force is proportional to the volume of the object that is submerged so as the object rises above the surface of water the buoyant force decreases and it's going to rise above the water when it reaches equilibrium that is when the buoyant force is equal to the weight of the object at that point the net force is zero and the block of wood no longer continues to rise so when the net force is zero the upward boiling force is equal to the downward weight force so the buoyant force of the object which is the density of fluid times the volume submerged times g is equal to the weight of the object which is mg now the mass of the object is the density of the object times the volume of the entire object times g so at this point we could divide both sides by g so we'll get a very useful equation the density of the fluid times the volume of the fluid displaced or the volume of the object that is submerged that is under water is equal to the density of the object times the volume of that object that's the entire volume so here's our picture this portion here is the volume that is submerged the part in green whereas the entire block is the volume of the object so let's see the amount that is above the surface of the water and how much is below it so the density of the fluid is a thousand we don't know the volume submerged we know the density of the object it's uh 175 divided by the density which was uh actually the density we had it already was 0.3 i just forgot for a moment and the volume of the object is 0.5 times 0.75 times 0.8 well actually the volume is 0.3 i take that back the density of the object was 175 over 0.3 which is 583.3 so basically this whole part right here is just the mass of the wood if you multiply 583.3 times 0.3 you get 175. so this whole thing is just the mass of the object but i'm going to leave the equation in this form because it's useful if you don't know the mass of the object sometimes you'll have the density and the volume of the object so it's going to be 175 divided by a thousand so the volume submerged is the mass of the object divided by the density of the fluid which is 0.175 so if the volume submerged is 0.175 cubic meters the volume that's above the surface of the water is the total volume which is 0.3 minus the volume submerged so 0.3 minus 0.175 is 0.125 so that is the volume of the block that is above the surface of the water now we can turn it into a percentage 0.125 divided by 0.3 is 41.7 percent so 41.7 percent of the block is above the surface of the water and 0.175 divided by 0.3 gives you the other 58.3 percent which is below the surface of the water that is the amount that's submerged now in this problem we have an ice cube floating on top of liquid mercury so let's say this is liquid mercury and here is the ice cube so we want to find out what fraction of the ice cube will be submerged in liquid mercury now we know that the density of the fluid times the volume submerged is equal to the density of the object times the volume of the object now let's think about what this equation means because we know that mass is equal to density times volume the density of the object times the volume of the object is equal to the total mass of the object the density of the fluid times the volume of the object submerged is basically the density of the fluid times the volume of the fluid that's displaced vs is the same as the volume displaced so therefore the density of the fluid times the volume displaced is equal to the mass of the fluid that is displaced and this is the basic idea behind archimedes principle the mass of the fluid that's displaced is equal to the the mass of the object now how can we find the fraction of the ice cube that's going to be submerged we need to take the volume that is submerged and divided by the total volume so we need to rearrange this equation so first let's divide both sides by vo and then let's multiply both sides by one over pf so the fraction that we're looking for is equal to the density of the object divided by the density of the fluid so it's going to be nine hundred seventeen divided by thirteen thousand six hundred so it's about point zero six seven four which represents six point seven four percent so about six point seven percent of the ice cube will be below the surface of mercury the other 93.3 percent is above the surface of the fluid so the fraction of the ice cube that's submerged is 0.0674 or 6.7 percent so here we have a cube that is made up of wood and it's 15 centimeters in all three dimensions now we have oil on top represented in yellow and water beneath that and this block of wood sits it floats literally between the oil and water phases now how is that possible how can this block of wood just float at that point it doesn't rise all the way to the top it doesn't sink all the way to the bottom it just stays in the middle now we know that heavy objects they sink light objects flow so if the density of the oil is 800 and the density of the water is a thousand that means that the density of this block of wood must be between 800 and a thousand if it was more than a thousand it would sink to the bottom if it was less than 800 it would stay above the oil surface so because it's between 800 and 1000 it's somewhere in the middle now notice that the cube is 15 centimeters in length that means that five centimeters is below the interface between the oreo and the water and 10 centimeters is above it so knowing that would you say that the density of this block of wood is it between 800 and 900 or 900 and a thousand now let's see if the block of wood was right in the middle let's say if it was 7.5 centimeters above and it was 7.5 centimeters below the interface in that case the density would be exactly 900 it would be the average of 800 and a thousand but notice that the majority of the block of wood is above the interface so is it between 800 and 900 or 900 and a thousand well we know heavy objects sink light objects flow so because it's higher it's more into the oil phase but above the and less than the water phase so to speak that means it's less dense than 900 it should be between 800 and 900 if more of the block of wood was below the water oil surface let's say this was 10 and this was five then the density would be between 900 and a thousand so let's go ahead and calculate the density of this block of wood and all the other things that we need to get now the first thing we're going to do is calculate the gauge pressure at the lower and upper face of the block so let's calculate the gauge pressure at this point and at this point let's call the the pressure at the lower point p2 and the pressure at the upper point p1 so p2 is going to be the gauge pressure of the water and the oil combine so this is going to be pressure of h2o plus the gauge pressure of the oil now the gauge pressure caused by water is only the portion of the water that is above that point anytime you have an object above a certain point the weight of that object exerts a force on anything below it and whenever you have a force acting on a certain area there's a pressure at that point so the gauge pressure of water is going to be the density of water times g times the height of water above this point which is 5 centimeters plus the gauge pressure of oil which is the density of the oil times sheet times the height of the oil above this point so the height of oil above this point represents the 20 centimeters it's not 20 plus 5 because the 5 centimeters correlates to the water but the 20 centimeters it's only it's all the oil that's above that point so make sure you don't put 25 because we've incorporated the 5 centimeters from water so the density of water which is 1000 times g which is 9.8 plus the height of the water above p2 that's 5 centimeters which is 0.05 meters the density of the oil that's going to be 800 times g times the height of the oriole that's above the point which is 20 centimeters divide that by 100 you get point 20 meters 100 times nine point eight times point zero five that's about four hundred ninety and eight hundred times nine point eight times point twenty is about fifteen sixty eight so the total pressure p2 due to the weight of the oil and the water above that point is 2058 pascals now what about p1 how can we calculate the gage pressure at this point now notice that there's no water above p1 so the gauge pressure is therefore the pressure of oil above that point which is equal to pgh the density of the oil is 800 g is 9.8 but what is the height of the oil what should we plug in for h so the height that we need to use is the height of the oil above that point of interest the total height of the oil layer is 20 centimeters and the portion of the block that is above the oil water interface is 10 which means that the height of oil above p1 is also 10 because 10 and 10 is 20. so 10 centimeters is about 0.1 meters if you divide it by 100 therefore the gauge pressure at p1 is 800 times 9.8 times 0.1 which is 784 pascals so now that we have the gauge pressure at p1 and p2 let's calculate the forces that the fluid exert at these points so if there's a pressure p2 then that must mean that there's a force that the fluid exerts at p2 and there's another force that the fluid exerts at p1 let's calculate these two forces we know that pressure is force divided by area so if you multiply both sides by a pressure times area is equal to the force so let's calculate f2 first so at the lowest surface of this cube it has an area of 15 centimeters by 15 centimeters or 0.15 meters by 0.15 meters and the fluid below it exerts an upward force p2 or i mean f2 so f2 is going to be the pressure p2 times the area of the lower surface p2 is 2058 pascals or newtons per square meter and let's multiply by point 15 meters squared which is the area of the bottom face of the cube so f2 is 46.3 newtons so now we need to calculate f1 which is the force that's exerted on the top face of the cube so f1 is going to be p1 times a and p1 is 784 times the area of point 15 square meters or point 15 meters and then square so f1 is equal to 17.64 newtons and f2 is 46.3 newtons so now that we have the force that the fluid exerts at the bottom and top surfaces now we can calculate the buoyant force the buoyant force which is an upward force is the difference between f2 and f1 f2 will always be larger than f1 because the pressure at a greater depth is always higher than the pressure at a higher position within a fluid since there's more weight of the fluid at a lower position so the buoyant force is f2 minus f1 which is going to be uh 46.3 minus 17.64 so the upward buoyant force is 28.66 newtons and now it turns out that there's another way to calculate the buoyant force and that's used in this equation the buoyant force is also equal to the density of the fluid times the volume of the fluid displaced which is the same as the volume of the object that is submerged in that fluid times the gravitational acceleration so in this problem there's two fluids so therefore the buoyant force is going to be pvg for the water plus pvg for the oil so you have to be careful in the way you uh use this equation so let's start with the water the density of the water is a thousand now what is the volume of the block that is submerged in the water so now think of a cube the volume of a cube is basically the length times the width times the height so if we draw a cube well first let me make some space so here's a rough sketch of a cube so this is the left of the cube this is the width and notice that the cube is not not all of it is in the water phase only a portion so the height of the cube that's in the water phase is about five centimeters so the length and the width of the cube that's point fifteen times point fifteen or point fifteen squared times the height of the cube that's in the water which is point zero five times g so that's the buoyant force due to the water being displaced that buoyant force is equal to the weight of the object up to this level now the rest of the object the weight of the upper part of the object is due to the oil so we need to use the height of the object that's in the oil phase for the second part of the equation so that's going to be the density of the oil plus the length and the width which is point fifteen squared times the height of the cube that's in the oil which is uh ten centimeters as opposed to five so this is gonna be point ten in meters times nine point eight so a thousand times point fifteen squared times point zero five times nine point eight that's about 11.025 so that's the buoyant force due to the the water alone and 800 times 0.15 squared times .10 times 9.8 is 17.64 so that's the buoyant force due to the the oil layer that's due to the oil being displaced around the upper part of the cube now if we add the two values we're going to get 28.66 which is the total upward boiling force acting on the cube so now you know how to calculate the buoyant force provided by the water and the oil and you know how to calculate the total upward boing force now how can we calculate the mass and the density of the block now that we know the upward buoyant force so because the block is in equilibrium it's staying it's basically floating between the two layers that means that the net force then that upward force on the block is equal to zero which means that the upward buoyant force has to balance the weight force in order for it to float at the interface so the buoyant force equals the weight force which means that the buoyant force of 28.66 is equal to mg the weight force of the block so 28.66 divided by the gravitational acceleration will give us the mass of the block which is 2.924 kilograms now that we have the mass we can calculate the density of the block density is mass divided by volume the mass of the block is 2.9 kilograms and the density of the cube is basically the side length raised to the third power is the length times the width times the height or 0.15 times point fifteen times point fifteen which is point fifteen meters cube and so the density of the block is 866 kilograms per cubic meter which makes sense for this block avoided to float between the oil and the water layer the density has to be between 800 which is the density of the oil and 1000 which is the density of water as you mentioned before if it's more than a thousand it's going to sink to the bottom if it's less than 800 it's going to float to the top but if it's between it's going to flow in between the two layers and since more of the block of wood is in the oil layer the density has to be between 800 and 900 if the majority of the block of void was in the water layer the density would have to be between 900 and a thousand consider this u-shaped tube so it's filled with water and then oil is added on top of it and when it comes to rest it forms this particular shape so based on the shape and the relative heights that we see here how can you calculate the density of the oil what would you do to find it now it's important to understand that the weight of the oil above the purple line is equal to the weight of the water above the purple line the masses have to be the same so that it's balanced once you understand that you'll realize that the gauge pressure due to the oil is equal to the gauge pressure due to the water above the purple line so because the volume of oil above the purple line is greater than the volume of water above the purple line that means that the oil is lighter in order to have the same mass the oil requires a larger volume which means that the density of the oil is less than the density of water so we should get an answer that's less than a thousand let's go ahead and calculate that answer since the weight of the fluids above the purple line must be the same the gauge pressure of the fluids at the purple line must be equal to each other so the gauge pressure of the oil is equal to the gauge pressure of water at that point at the purple line so pgh is equal to pgh we can cancel g so now we can calculate the density of the oil the height of the oil above the purple line is 40 centimeters the density of water is a thousand and the height of water above the purple line is the difference between 40 and 18 which is 22 centimeters now we don't need to convert centimeters to meters because the unit centimeters will cancel so it's going to be a thousand times 22 divided by 40. so the density of the oil is 550 kilograms per cubic meter which is less than a thousand since there's more oil above the purple line than water in terms of volume so if the mass is the same mass is density times volume if you increase the volume the density has to decrease so since the volume of oil above the purple line is greater than the volume of water the density of the oil has to be less than the density of the water if mass is the same density and volume are inversely related now let's say that we have a block of aluminum metal and there's a rope attached to it and it's suspended in the vacuum now the upward tension force that supports the weight of the aluminum block let's say it's 1960 newtons this is basically the force that a scale would read so that's the tension force how can you calculate the mass of the aluminum block now in order for it to remain suspended in a vacuum the tension force must be equal to the weight force such that the net force in the y direction is zero so if t equals w that means that the weight force is 1960 which is equal to mg so 1960 divided by 9.8 will give you the actual mass of the aluminum block as measured in a vacuum so the mass is 200 kilograms now what about if the block of aluminum was placed in a tank of water so in this case what is the tension force in the rope at this point and what is the apparent mass on the aluminum block if it's fully submerged in water so there's going to be a downward weight force but because it's displaced in water or it's submerged in water there's an upward buoyant force acting on the block now if it's balanced in this location then that force in the y direction is equal to zero now the sum of all forces in the y direction is going to be the upward tension force plus the upward buoyant force minus the downward weight force and it's equal to zero if it's at rest so if we solve for the tension force it's going to be the weight force minus the buoyant force now the tension force it measures the apparent mass in the last example in a vacuum the tension force measures the actual mass but if you submerge an object in a fluid when you lift it up in a fluid it's going to feel lighter because the buoyant force supports the weight of the object so as you lift it up the buoyant force helps you to lift up the object so it feels lighter when lifting an object in water but once you lift it out of water it's it feels heavier because you no longer have the support of the buoy force so the tension force in the rope it measures the apparent mass which is mg the weight force is based on the actual mass and the buoyant force is the density of the fluid times the volume that submerged times the gravitational acceleration so we can divide everything by g so the apparent mass is equal to the actual mass minus the mass of the fluid displace which is the density of the fluid times the volume displaced or the volume of the object that is submerged mass is equal to density times volume so if we solve for volume volume is equal to mass divided by density and if the object is fully submerged the volume that is displaced by the water is equal to the actual volume of the object when fully submerged which is the mass of the object divided by the density of the object so let's replace vs with m over po so the volume that is submerged is equal to the mass of the object divided by the density of that object so now let's factor out the gcf on the right side of the equation which is m so the apparent mass is equal to the actual mass times one minus the density of the fluid divided by the density of the object so this equation is true only when the object is fully submerged so now let's calculate the apparent mass for this particular problem so we know that the actual mass of the aluminum block is 200 kilograms and the density of the fluid which in this case is water is one thousand and the density of the object which is aluminum is twenty seven hundred so if you type this in the calculator exactly the way you see it you'll see that the apparent mass of aluminum when placed in water is 125.9 kilograms so the apparent mass is always less than the actual mass so remember this is the apparent mass when it's fully submerged in water now that we have the apparent mass we can calculate the tension force the tension force is the apparent mass times g so it's just 125.9 times 9.8 which is 1233.8 newtons so far we see that the mass measured by an object changes if you place that object in a fluid so in a vacuum an object weighs heavier than it would in a fluid if you place it in water it weighs lighter due to the upward buoyant force now the same is true in air air is a fluid it contains gases so if you want to measure the true mass of an object in air you need to take into consideration the density of the air because air also provides an upward buoyant force for any object displaced within it so let's try this problem what is the apparent mass of a 1000 kilogram block of ice in air so the apparent mass formula is m a is equal to the true mass of an object times one minus the density of the surrounding fluid divided by the density of the object so the mass of the block is 1000 the density of the air which is the fluid is 1.29 and the density of ice is 917. so the apparent mass of this block of ice is about 998.6 kilograms per cubic meter so because the density of air is very low the apparent mass is very close to the true mass it doesn't differ as much the difference is significant for objects that are light but objects that are very heavy with a very large density the difference is negligible but for light objects let's say like a piece of paper then the upward boing force provided by air becomes significant now let's answer the second part of the question so let's say if you have a scale and you place a block of ice on top of it and the scale says that the block of ice is 476 476.3 kilograms now because is measured in air this is the apparent mass it's not the true mass so how can you calculate the true mass knowing the density of air and the density of the block of ice so the apparent mass is 476.3 we need to calculate m and pf is 1.29 and po is 917. 1 minus 1.29 divided by 917 is about 0.998593 so 476.3 divided by that number gives you a true mass of 476.97 kilograms so as you can see the true mass of the object is always going to be higher than the apparent mass so whatever the scale reads know that the true mass is slightly higher than the apparent mass let's try this word problem a 15 kilogram block has an apparent mass of 12 kilograms when fully submerged in water what is the density of the block so we can use the apparent mass formula to calculate the density of an object submerge in a fluid of known density and we can also use it to calculate the density of an unknown fluid if we know the density of the object that is in the fluid so let's rearrange this formula to calculate the density of the fluid and the object so let's divide both sides by m so m a divided by m is equal to 1 minus pf over po now we're going to take this term which is negative on the right side we're going to move it to the left which is going to be positive on the left side and we're going to take this term which is positive on the left and move it to the right so it's going to be negative on the right side so p f divided by p o is equal to one minus m a divided by m o now let's multiply both sides by the density of the object so that these terms cancel so now we have this equation the density of the fluid is equal to the density of the object times 1 minus the apparent mass divided by the actual mass now let's plug in what we have so we know that the density of the fluid which in this case for the first part it's water we're looking for the density of the block the apparent mass is 12 kilograms and the actual mass is 15. 1 minus 12 over 15 is about 0.2 and a thousand divided by 0.2 is 5 5000 so the density of the object is 5000 kilograms per cubic meter so now that we know the density of the object let's calculate the density of the fluid so if we place the same object in an unknown fluid the apparent mass is found to be 10 kilograms what is the density of the unknown fluid so the density of the object is 5000 the new apparent mass in this unknown fluid is 10 kilograms and the actual mass is 15. 1 minus 10 over 15 is about point 33 or 0.33333 repeating times 5 000 is equal to a density of 16 67 kilograms per cubic meter so that is the density of the unknown fluid let's work on this problem a 60 kilogram block is attached to a rope connected to a vertical spring in a vacuum if the spring stretches 12 centimeters to support the weight of the block what is the spring's constant so a vertical spring can be used as a scale if you want to measure the weight of an object but first we need to know what the spring constant is so what you need to do first to measure the spring constant is you need to apply a block which you know the mass of so here we have a block of known mass which is 60 kilograms and so the weight force is going to cause the spring to stretch whenever you apply a downward force to stretch a spring the spring wants to go back to its natural length there's an upward restoring force it turns out that the tension force that's in the rope is equal to the force required to stretch the spring so those two are the same so what is the tension force in this problem if the net force in the y direction is equal to zero the upward tension force that supports the weight of the block has to equal the weight of the block and the tension force is basically the force required to stretch the spring and according to hooke's law f is equal to kx and the weight force is mg now here's a question for you we know that x represents the amount that the spring is stretched from its equilibrium position and we have the value of x in centimeters so if we use 12 centimeters k is going to be the spring constant in newtons per centimeter should we keep x in centimeters or should we convert it to meters and this problem really doesn't matter but it's best if we keep it in centimeters because when we answer the second part of the problem it's in centimeters as well so keep track of the units so x is 12 centimeters the mass is 60 kilograms and g is 9.8 so the weight force is 60 times 9.8 which is 588 newtons and it stretches the spring by 12 centimeters so the spring constant is 49 newtons per centimeter that means that a force of 49 newtons will stretch the spring by one centimeter so therefore a force of 900 i mean 588 newtons will stretch the spring by 12 centimeters so now that we have the constant k we can use that to measure the mass of an unknown object now let's answer the second part of the problem so a cubicle block of aluminum of unknown mass is attached to a vertical spring which stretches 54 centimeters in air so here's the spring and here's the rope attached to the block so we have the downward weight force and the upward tension force that supports the weight of the block and the tension force is the same as the force required to stretch the spring and let's say this is the original natural length of the spring x is the amount that the spring is stretched by from its natural length so how can we calculate the mass of the block so once again once the block reaches equilibrium when it no longer stretches the spring the net force in the y direction is zero so the tension force equals the weight force and the tension force is basically the force required to stretch the spring which is kx and the weight force is mg so this time k is 49 newtons per centimeter and x is now 54 centimeters so the unit centimeters will cancel giving us the force in newtons which is equal to the weight force now keep in mind this is an air it's no longer in a vacuum so the mass that we're going to calculate is not the true mass it represents the apparent mass the reason why that's the case is due to the fact that air is a fluid whenever you have an object placed in a fluid there is an upward buoyant force in a vacuum there is no upward point force so the tension is equal to the true weight of the object but in the fluid the tension equals the apparent weight so the apparent mass times the gravitational acceleration is really the apparent weight so what is the apparent weight equal to well for this object to be in equilibrium the sum of the forces must be equal to zero meaning the upward tension force plus the upward buoyant force minus the downward weight force is equal to zero so if you solve for tension t is equal to the true weight minus the buoyant force so since t which is equal to the apparent weight that apparent weight is the true weight minus the boring force but for this particular problem we don't have to worry about all that what we need to understand is that the tangent force is equal to the apparent weight and not the true weight which means that the tension force is equal to the apparent mass times the gravitational acceleration once we understand that we can find the apparent mass and then using the other formula we can calculate the true mass so let's find the apparent mass so it's going to be the spring constant 49 times the amount that the spring is stretched by which is 54 centimeters that's equal to the apparent weight which is 26 46 newtons divided by 9.8 and so the apparent mass is equal to 270 kilograms so that's how you can calculate the apparent mass of an object using a vertical spring you need to know the spring constant and you need to measure with a ruler how much the spring stretches by and you can calculate the spring constant using a block of known mass and see how much it changes by and then you can find the mass of an unknown object so now that we have the apparent mass let's use this equation apparent mass is equal to the true mass times 1 minus pf divided by po so the apparent mass is 270. the mass of the the density of the fluid is uh 1.29 because it's in air and the density of the object is 2700 so 1 minus 1.29 divided by seven hundred it's roughly about one it's like point nine nine nine five two two two so 270 divided by that number is about 270.13 so in air the true mass is very close to the apparent mass they don't really differ by that much so now that we know the true mass and the apparent mass how far will the spring stretch if 40 of the aluminum block is submerged in water so let's draw a new picture so let's say if we have a tank contained in water and naturally if you place aluminum in water aluminum is more dense so it's going to sink to the bottom but if we attach a rope to it then it won't sink the upper tension force will support the weight of the object and it's still attached to a spring now only 40 percent of the aluminum block is submerged in water so we're still going to have a downward weight force but now we have an upward buoyant force provided by the water as well there's also an upward brain force provided by the air as well so what we're going to do is we're going to calculate the amount that this spring will stretch neglecting air and also using air so first let's neglect the effects of air let's just look at the contribution of the buoyant force just by water alone now we know that there's an upward tension force which is equal to the force required to stretch the spring so we got to find out what delta x is which we could just simply say x so feel free to pause the video and determine the distance that the spring will stretch if 40 of the aluminum block is submerged in water so to calculate the amount that the spring will stretch we need to calculate the tension force because that tension force is the force that's pulling on the spring which is equal to kx we know the value of k it's 49 newtons per centimeter so if we can calculate the tension force then we could find x so how can we calculate the tension force so we know that the tension force is equal to the weight force minus the buoyant force the weight force is mg and we know the true mass which is i'm going to write it on the side 270 point 13 kilograms and then minus the buoyant force which is the density of the fluid times the volume of the object that is submerged times the gravitational acceleration so we know the density of the fluid that's a thousand so we need to find the volume that is submerged so what is the volume of the cube how can we calculate that answer now we know that mass is equal to density times volume so if you want to solve for volume it's mass divided by density now the volume of the object that is submerged is 40 of the aluminum block so that's 0.4 times the volume of the entire object and we know that the volume of the entire object is basically the mass of that object divided by the density of the object so now we can calculate the volume that is submerged so vs is equal to 0.4 times the mass of the object which the true mass is 270.13 divided by the density of the aluminum block which is 2700 kilograms per cubic meter so 270.13 divided by 2700 so the volume of the aluminum block is about 0.1 kilograms per cubic meter multiplied by 0.4 so the volume that is submerged it's about point zero four zero zero two so now that we have the volume submerged we can now calculate the tension force so it's going to be mg which is 270.13 times 9.8 minus the buoyant force due to water which is the density of water times the volume submerged which is point zero four zero zero two times g which is nine point eight so the weight of the object 270.13 times nine point eight that's about two thousand six hundred forty seven point three minus the upward bound force caused by the displacement of water that's a thousand times point zero four zero zero two times nine point eight and that's about three hundred ninety two point two so if we subtract these two numbers the tension force is equal to 2255.1 units so the tension force is equal to kx where k is 49 newtons per centimeter so 22 55.1 divided by 49 is equal to 46 centimeters so that's how far the spring will stretch if the aluminum block is partially submerged so notice that it stretches 54 centimeters in air but in water it's only going to stretch 46 centimeters because water supports the weight of the aluminum block by means of the upward buoyant force so it's going to be less than 54. now if we take into account the upward buoyant force of air the amount that the spring stretches by should be even less right now it's 46.02 so that is the current answer if we don't take into account air but now let's take into account the effects of air so this is the buoyant force due to water now we must also subtract the buoyant force due to air let's call it fb2 so keep in mind w minus fb is 22 55.1 now we need to subtract it by the boiling force due to air to get the new tension force now the new buoyant force due to air is equal to the density of the air times the volume of the object submerged in air times g so let's find the volume of the object's emerging air so if 40 is below water 60 percent is above water so we need to find 60 percent of the volume of the object and the volume of the object we know is the mass of the object divided by the density of the object so it's 0.6 times the mass of the object which is 270.13 divided by the density of aluminum which is 2700 so the volume that is submerged is about point zero six zero zero three so now let's calculate the buoyant force due to air so the density of the air is 1.29 and the volume of the object submerged in air as we said it's 0.06003 and g is 9.8 so the buoyant force due to air is very very small even negligible it's about 0.76 newtons so the tension force is 2255.1 that's without air but due to the buoyant force of air it's now going to be 22 54.34 newtons so this tension force is equal to kx because it's used to stretch the spring so let's divide it by 49 so due to the effects of air the spring will stretch by a distance of 46.007 centimeters so the difference is minute but there's still a difference either case the spring will stretch by 46 centimeters whether you take into account air or if you choose not to take into account a block of iron is attached to a spherical balloon of radius 10 meters that is filled with helium so let's say this is the balloon and attached to it we have a block of iron the radius of the balloon is 10 meters and it contains helium we also have the mass of the skin of the balloon what is the upward buoyant force acting on a balloon the buoyant force is equal to the density of the fluid in this case the surrounding fluid is the air around the balloon times the volume of the balloon which is the volume of the object submerged in the fluid times g the density of the air is simply 1.29 the volume of a balloon is 4 over 3 pi r cubed or pi times 10 to the third power times g ten to the third is a thousand times four pi divided by three times one point two nine times nine point eight the upward buoyant force acting on the balloon is 52 955 newtons now that we have the upward buoyant force let's go ahead and calculate the mass of iron that the balloon can support so the upward buoyant force has to support the weight force of iron it must also support the weight of the helium that's inside the balloon and also it has to support the weight of the balloon itself because the structure of the balloon has mass so therefore the upward buoyant force is equal to the weight of the load or the iron block plus the weight of the helium particles inside the balloon plus the weight of the balloon structure itself so the buoyant force which is uh 52 955 which for now i'm going to write it as p of the fluid times the volume times g that's equal to the weight of the load which is the mass of the iron load times g plus the weight of helium which is the mass of helium times g plus the weight of the block so we could divide everything by g so we have the density of the air times the volume which equals the mass of the load now the mass of helium is equal to the density of the helium gas times the volume of the balloon plus the mass of the balloon so what can we do to solve for the missing variable now at this point let's plug in everything that we have but let's move this and this term to the other side so the density of the air times the volume of the balloon minus the density of helium times the volume minus the mass of the balloon equals the mass of the iron load so let's take out v so now we can solve for m so the volume of the balloon is four over three pi times the radius to the third power the density of the air is 1.29 minus the density of helium which is 0.179 minus the mass of the balloon which is 125 so the mass of the load once you type this information in it's 4529 kilograms now that we have that information what is the buoyant force acted on the block of iron because the block of iron is also displaced in air as well so the air is going to exert an upward bound force however the volume of the block of iron is insignificant so chances are we can neglect it but let's go ahead and calculate the buoyant force acting on the block of iron so that buoying force is going to be the density of the fluid which is the density of the air times the volume of the block of iron which displaces the air times g now we don't have the volume of iron but we do know the mass and we do know the density so if mass is density times volume the volume is mass divided by density so the buoyant force acting on the block of iron is the density of the air times the mass of the block of iron which is 45 29 divided by the density which is 7 800 times g so the buoyant force acting on the block of iron it's about 7.34 newtons now keep in mind the buoyant force that was acting on the balloon was 52 955 newtons so as you can see the buoyant force provided by the block of iron is insignificant compared to the buoyant force provided by the balloon and the reason for that is the volume of the balloon is so much more larger than the volume of the block of iron and so that's why we could ignore this buoyant force the next thing that we need to talk about is the mass flow rate the mass flow rate is the amount of mass that passes through a certain point per unit time so it's the change in mass divided by the change in time now we know that mass is density times volume so the change in mass is the density times the change in volume now let's say if we have a pipe which is in the shape of a cylinder we know that the volume of a cylinder is pi r squared times height so the volume of the cylinder is the area of the circle which is pi r squared times the height which is basically the length of the pipe so the change in volume is the area times the change in height the change in height is basically the displacement displacement divided by time is equal to velocity d over t is equal to v so therefore we can say that the mass flow rate is equal to the density times the area times the velocity so if the mass flow rate is equal to density times area times velocity how can we calculate the volume flow rate to find the volume flow rate multiply both sides by one over p mass divided by density is volume we know that m is equal to pv so if you divide by p mass over density is volume so delta m over p is delta v so the volume flow rate is basically the area times the velocity now let's work on some problems based on the equations that we just came up with so water flows through a circular pipe that has a diameter of 20 centimeters and it flows through the pipe at a speed of three meters per second so what is the mass flow rate so the mass flow rate is equal to the density of the fluid times the cross-sectional area times the velocity so the cross-sectional area is basically pi r squared since it's the area of the circle so if the diameter is 20 centimeters the radius is 10 centimeters but in meters if you divide by 100 that's 0.1 meters so it's going to be the density which is a thousand times pi times r squared multiplied by the speed of three meters per second so the mass flow rate is about 94.25 kilograms per second so what this means is that every second 94.25 kilograms of water passes through this region now that we have the mass flow rate what mass of water will enter a storage tank in two hours so let's say this is the storage tank and water flows out of the pipe into the storage tank so after two hours how many kilograms of water will enter the storage tank so we know the mass flow rate is 94.25 kilograms per second so what we need to do is convert hours into seconds so we know that there's 60 minutes in an hour and also there's 60 seconds in a minute so two hours is about seventy two hundred seconds so if we take the time in seconds and multiply by the mass flow rate the unit seconds will cancel and we can get the total mass that's going to flow into the tank in two hours so it's 7 200 seconds multiplied by 94.25 kilograms per second so in two hours 678 thousand and six hundred kilograms of water is going to flow into the tank now what about the last part of the question how long will it take to add 25 000 kilograms of water to the tank so we're given mass and we need to find the time you can use an equation or you can simply use units to get the answer let's start with the mass in kilograms so we have 25 000 kilograms and we want to convert it to some unit of time so now we know the mass flow rate which we set it to be 94.25 kilograms per second so in one second 94.2 kilograms of water will enter into the tank so notice that the units kilograms cancel and now we have the units in time but let's go ahead and convert it to minutes so there's 60 seconds in a minute and let's see if minutes is an appropriate unit so if we want the answer in seconds 25 000 divided by 94.25 is about 265 seconds so let's convert that to minutes so let's divide that result by 60. so it's going to take about 4.42 minutes for twenty five thousand kilograms of water to enter into the tank at this rate let's try another problem so water flows through a circular pipe of radius eight centimeters so that's uh this distance here and it flows through this pipe at a speed of 2.5 meters per second what is the volume flow rate the volume flow rate change in v divided by change in t is equal to the cross-sectional area of the pipe times the velocity so the cross-sectional area is the area of a circle which is pi times the radius squared eight centimeters is basically point zero eight meters squared times the speed which is 2.5 meters per second so the volume flow rate is about point zero five zero three cubic meters per second so as you can see here we have square meters times meters to the first power two plus one is three now that we have the volume flow rate how long will it take to fill a pool if the dimensions are ten by four by fifteen so first we need to calculate the volume of the pool which is length times width times height 10 times 4 times 15 is about 600 cubic meters so now that we have the volume and the volume flow rate we can calculate the time so let's start with the volume which is 600 cubic meters and we know that in one second a volume of 0.0503 cubic meters of water is going to flow through the pipe into the pool so notice that the volume cancels the cubic meters so right now this is going to give us the time in seconds but it's going to be a pretty large number so let's convert it to minutes now we know that there's 60 seconds in a minute and also they're 60 minutes in a single hour so these units will cancel and those units will cancel 600 divided by 0.0503 is about 11 928 seconds divided by 60 that's 198.8 minutes divided by another 60. this is going to give us a time of 3.3 hours so that's how long it's going to take to fill a pool with water if the volume of the pool is 600 cubic meters now there's some other stuff that you need to know let's say if we have another circular pipe and let's call this position a and position b the rate at which water flows into the pipe at a equals the rate at which water leaves the pipe at b and this is associated with the equation of continuity so the mass flow rate at position a is equal to the mass flow rate at position b and the mass flow rate is basically the density times the area times the velocity so this equation is useful if a fluid can be compressed if you have a compressible fluid like a gas the density can change so let's say if the area is constant but if the density increases that means that the velocity has to decrease and let's say if the density is constant if density is constant you can cancel these two terms and then you get another equation which is a1 times v1 and that's equal to a2 times v2 the area times the velocity as you mentioned is the volume flow rate so if the density is the same that means that you have an incompressible fluid if the density is constant so for incompressible fluids the volume flow rate is constant so if you increase the area the velocity will decrease and if you decrease the area the velocity will increase so let's say if we increase the area so let's say a1 has an area of 2 centimeters squared and a2 has an area that's twice the value let's say it's four square centimeters so if the velocity that enters this cross-sectional area is five meters per second what is the velocity that leaves this pipe so what's v2 so if we use the equation a1 v1 is equal to a2 v2 a1 is 2 v1 is 5 a2 is 4 so let's find v2 so it's 2 times 5 which is 10 divided by 4 that's 2.5 so if you increase the area by a factor of 2 from 2 to four then the velocity will decrease by a factor of two so as you increase the area the velocity decreases proportionally now let's say if we decrease the area let's say if the radius is 10 centimeters and let's say over here the radius is now 5 centimeters if the velocity that enters the pipe let's say it's 10 meters per second what is the velocity that leaves this smaller section so what's v2 if we have v1 so using the same equation a1 v1 is equal to a2 v2 but this time we have the radius of the of the pipe at these two points so for circle area is pi r squared now we can cancel pi and we don't have to convert centimeters into meters because we have a ratio and as long as r1 and r2 are both in centimeters the units will cancel and it's going to work so r1 is 10 centimeters and v1 is 10 r2 is 5 and let's solve for v2 so 10 squared times 10 is a thousand divided by 5 squared which is 25 a thousand divided by 25 gives you a velocity or speed of 40 meters per second so if we reduce the radius by a factor of two the velocity increases by a factor of four because the velocity is inversely related to the square of the radius so if you decrease the radius by two two squared is four the speed will increase by factor four if you decrease the radius by three three squared is nine the speed will increase by a factor of nine so in this case let's say if r one was ten over three centimeters then the speed would be ten times nine it's gonna be ninety meters per second now what about the pressure what happens to the pressure as the speed increases from 10 to 40. now according to bernoulli's principle wherever the velocity of the fluid is high the pressure is low and where the velocity is low the pressure is high so on the left side the speed or the velocity is very low so the pressure is high and on the right side the speed is high so the pressure must be low now it makes sense whenever you have a pressure there is a force because pressure is forced over area so on the left side if we have a region of high pressure there must be a very large force that accelerates the fluid towards the right and on the right side we have a low pressure so the force that is created by that pressure is weak to notice that this force is greater than this one so therefore there's a net force that accelerates the fluid towards the right and so it speeds up from 10 meters per second to 40 meters per second so the potential energy that's stored in the form of this high pressure is converted to kinetic energy as the pressure reduces so as the pressure decreases from high to low it loses potential energy and the water molecules they gain kinetic energy and so they speed up and as you can see in order for the mass flow rate to be the same the water has to flow faster in its confined space so that the amount of water that enters the pipe equals the amount of water that leaves the pipe now if we wish to calculate how the pressure changes along the pipe we need to use bernoulli's equation and bernoulli's equation comes from the conservation of energy so let's say if we have the same situation and the pipe accelerates due to the cross-sectional area decreasing so there's a velocity v1 and an increases to a velocity v2 and we have a pressure p1 and a pressure p2 so if there's a change in pressure there's going to be a net force which accelerates the fluid so the work done by this net force should be equal to the change in kinetic energy and the change in kinetic energy is equal to the final kinetic energy minus the initial kinetic energy so work plus the initial kinetic energy is equal to the final kinetic energy and the work done by a force is that force which is basically the net force times the displacement and the initial kinetic energy is one-half mv1 squared and the final kinetic energy is one-half and v2 squared now we know that pressure is force divided by area so if you multiply both sides by a the pressure times the area is equal to the force so we can replace f with pressure times area times the displacement plus one-half mv one squared equals one-half mv two squared so at this point what we're going to do is divide everything by the volume area times displacement equals volume let's say if you have a cylinder and it has an area a and a displacement or height of d then the volume is equal to a d the volume of a cylinder is the area which is pi r squared that's the area of the circle and the vertical or the horizontal displacement is basically the height of the cylinder so volume equals a d so we have pressure we're going to replace a d with uh volume which is divided by volume plus one half m divided by v mass divided by volume is density so we get this equation and v will cancel so the pressure plus one half pv one squared is equal to one half pv two squared and i do have to make a correction because force equals pressure times area but the net force is really the difference in two forces it's the difference between f1 and f2 so then net force it's really f1 minus f2 which is p1a minus p2a so if you take out a then that forces a times p1 minus p2 which is basically a times the change in pressure so this should be delta p by the way so change in pressure plus one half pv one squared is equal to one half pv two squared and delta p we said it's uh p1 minus p2 so if we add p2 to both sides of the equation we're going to get this form of the equation now what if the height changes because sometimes the pipe needs to go to a different elevated position so let's say this is the pressure p1 this is the pressure p2 and we have a fluid moving at a speed v1 and here at a speed v2 so the pressure has to apply a force to lift up the fluid against gravity and anytime the height of the fluid is changing the gravitational potential energy is changing so over here the pressure has to be high and here the pressure is low assuming if the velocity is the same as well so the energy stored in the high pressure will be converted to gravitational potential energy to lift up the object or to lift up the fluid to a higher position so as energy so energy is going to be transferred from pressure to gravitational potential energy as the fluid goes up so the work done by the difference in pressure is equal to the change in mechanical energy the reason being is dependent on the difference in pressure the pressure could increase the speed as well which really depends on the relative difference between a1 and a2 the air the cross-sectional area of the pipe so if the area decreases the speed is going to increase so the pressure can be used to speed up the fluid or it can elevate the fluid to a different height so the kinetic energy of the fluid can change and also the gravitational potential energy can change the mechanical energy is basically the sum of the kinetic and the potential energy so if work equals change in mechanical energy in this particular situation then it equals the final mechanical energy minus the initial mechanical energy so work plus initial mechanical energy is equal to the final mechanical energy and the initial mechanical energy is basically the initial kinetic energy plus the initial gravitational potential energy and that's equal to the final kinetic energy plus the final potential energy so p1 is going to exert a force f1 and p2 will exert a backward force called f2 so the work is basically the net force times the displacement and the net force is the difference between these two forces so it's f1 minus f2 times d and we know that force is basically pressure times area so it's uh p1a minus p2a times d and if we factor out a it's going to be a times p1 minus p2 times d so let's replace w with the area times the change in pressure times d initial kinetic energy is one half v one squared and initial gravitational potential energy is mgh1 and that's equal to final kinetic energy which is one half mv two squared plus final potential energy which is mgh2 so now what we're going to do is replace area times displacement with volume so a d is basically volume times p1 minus p2 plus everything else which i guess i could just copy this now let's divide everything on both sides of the equation by the volume so these two cancel so it's p1 minus p2 plus mass over volume is density so it's one half density times v one squared plus density times g h one which equals one half p p2 v2 squared plus pg h2 squared or just pg h2 and now we're going to take this and add it to the other side so now we have bernoulli's equation p1 plus one half pv one squared plus pgh1 is equal to one half p2v2 squared or just pv2 squared plus pg h2 plus pressure 2 so that's bernoulli's equation it helps you to see how pressure changes when the height changes or when the speed of the fluid changes so let's review how this is going to work so let's say if as you mentioned before if the speed increases so let's say this is v1 and this is v2 whenever the speed increases the pressure is going to decrease so if it's moving slow at the left and if it's moving fast on the right side whenever the fluid is moving slow the pressure is high and whenever the fluid is moving fast the pressure is low so you need a high pressure to accelerate the fluid to increase the speed whenever the speed increases there's an acceleration which means that there's a net force and the net force created by fluid is usually due to pressure differences so as the pressure changes from high to low the speed is going to increase if of course the cross-sectional area decreases because as the area decreases the velocity will increase and the pressure will decrease now let's say if we have a pipe and the height decreases so let's say this is p1 p2 but the cross sectional area is the same so that means that v1 is going to equal v2 if a1 equals a2 so therefore the velocity is constant now whenever an object travels from a high position to a low position gravitational potential energy is being released so the question is what is absorbing that gravitational potential energy because typically when an object slides down an incline the gravitational potential energy is converted to kinetic energy and so the object speeds up so if the cross sectional area is the same according to the volume flow rate equation if a1 equals a2 v1 equals v2 if the fluid is incompressible so if the fluid density doesn't change the velocity has to be the same which means that the kinetic energy is the same so where does that gravitational potential energy go if kinetic energy is the same at this position the pressure is low but now the pressure is high so whenever a fluid falls from a high position to a low position the pressure increases the gravitational potential energy is being transferred to another form of potential energy which is stored in the form of high pressure so whenever you have a fluid at high pressure that fluid is capable of doing work because once that pressure changes from high to low there's going to be in that force and whenever there's a net force work can be done energy can be transferred because force times displacement is work so as the fluid travels from a high position to a low position the gravitational potential energy is being converted in the form of pressure so the pressure increases likewise if the fluid flows from a high position or rather from a low position to a high position the energy that's stored in the form of high pressure is going to be released to push up the fluid to a higher position so pressure energy is being converted to gravitational potential energy so let's see if we can put this information to good use so let's say this is point a point b point c and point d let's see how the pressure changes as the fluid moves through this uh pipe now let's review the basic rules we know that if the cross-sectional area decreases the fluid is going to speed up the velocity is going to increase and whenever the velocity increases if the speed is high and the pressure is low so the pressure decreases and this is at constant height now let's say if the area is the same that means velocity or the speed is going to be the same now if the height decreases the pressure will increase and if the height increases the pressure will decrease so now let's go through this problem as the fluid flows from position a to position b the cross-sectional area decreases which means that the velocity of the fluid increases so that means that the pressure decreases so if a is at high pressure b is at a relatively low pressure compared to a so now let's compare b and c so the cross-sectional area from b to c is the same so which means that the velocity is going to be the same but the height is changing so the position is decreasing so gravitational potential energy is being converted to pressure energy which means that the pressure is going to increase so c is at a higher pressure relative to position b now what about position d the cross sectional area increases so the cross sectional area increases that means that the velocity has to decrease and according to bernoulli's principle whenever the speed is low the the pressure is high so the pressure increases so d is at a higher pressure relative to c now let's work on some problems so let's say if we have a pipe and the cross-sectional area decreases let's say the pressure at position a is 150 kilo pascals or 150 000 pascals what is the pressure at b if the cross-sectional area at a let's say it's 30 square centimeters and a2 is 15 square centimeters and let's also say that the velocity at position a let's say it's four meters per second so with this information if you have v1 p1 a1 and a2 what is p2 how would you calculate the pressure at position b so feel free to pause the video so the first thing we need to do is calculate v2 using the volume flow rate equation a1 v1 is equal to a2 v2 so this equation is true if the fluid is incompressible where the density is the same so a1 is 30 square centimeters v1 is 4 a2 is 15 what's v2 so if the area decreases by a factor of 2 from 30 to 15 the speed should increase by a factor of 2 from 4 to 8. 30 divide well 30 times 4 divided by 15 gives you eight so that is the velocity at position b so now that we know the velocity we can calculate the pressure using bernoulli's equation so p1 plus pgh1 plus one half pv1 squared is equal to p2 plus pgh2 plus one half pv2 squared now the height of the center of the fluid is the same so the gravitational potential energy doesn't change so energy from pressure is being converted to kinetic energy so p2 has to be lower than p1 because the speed is increasing whenever the velocity is high the pressure is low so let's calculate p2 so p1 is 150 000 pascals now we're going to use water so the density of water is a thousand kilograms per cubic meter the velocity at position a is four we're looking for p2 and v2 is 8 4 squared is 16 times 0.5 that's eight times a thousand so this is a thousand pascals and eight squared is 64. half of that is 32 times a thousand so over here this is 32 000. so 150 000 plus 8 000 minus 32 000 gives you a pressure value of a hundred and twenty six thousand so as the speed increase from four meters per second to eight meters per second the pressure decreased from 150 000 pascals to 126 000 pascals so notice that the pressure doesn't change proportionally so to speak even though the speed doubled the pressure didn't it wasn't reduced by half it decreased but it wasn't exactly by half so anytime the speed increases the pressure decreases now let's try another problem so let's say the cross-sectional area is 10 square centimeters and at a2 the cross-sectional area is the same and let's say the fluid is moving at a speed of 5 meters per second and also since the cross-sectional area is the same the speed is the same now let's see the pressure at p1 or position a let's say it's 150 000 pascals and the height difference between these two positions let's say it's uh five meters what is the pressure p2 so as you mentioned before anytime the area is the same the speed will be the same so the pressure energy is being converted to gravitational potential energy anytime the height of the fluid increases the pressure is going to decrease assuming the speed remains constant so how can we calculate p2 so let's start with bernoulli's equation now since the velocity is constant if e1 equals v2 so we can eliminate this term p1 is 150 000 pascals times pgh the gauge pressure now we're going to say that this is ground level so position a is at a height of zero if that's at ground level so this disappears p2 are looking for that and then pgh the density of water times g which is 9.8 times the height position b is 5 meters above the reference level so 1 000 times 9.8 times 5. is about forty nine thousand so p two is a hundred fifty thousand minus forty nine thousand so p two is about a hundred and one thousand so as we can see because the height increased the pressure decreased from 150 000 pascals to 101 000 pascals now it turns out that there's two ways to increase the speed of the fluid one way is for the fluid to decrease in height if the object decreases in height it can speed up meaning the gravitational potential energy can convert to kinetic energy the other way to speed up the fluid is to decrease the pressure if you decrease the potential energy stored in the form of pressure it can be converted to kinetic energy and the fluid can speed up and as we know if the height decreases the pressure can increase sometimes now let's say if the gravitational potential energy decreases by 200 joules so basically the height is decreasing now sometimes this decrease in height can increase the pressure and it can increase the speed for example if the energy stored in pressure goes up by 120 joules that means that the kinetic energy which is quantified by the speed and the mass must go up by 80 joules so a decrease in height is capable of increasing the pressure and the kinetic energy but now let's say if the height decreases and the gravitational potential energy as a result decreases by 200 joules but sometimes the kinetic energy can increase more than the amount that the gravitational potential energy decreased by now let's say if the potential energy decreases by 200 joules but the kinetic energy increases by 300 joules how is that possible the only way that the kinetic energy can increase by a magnitude greater than the potential energy is that the pressure has to decrease so if the pressure loses 100 joules of energy that 100 joules is transferred to kinetic and the 200 joules lost by the gravitational potential energy is also transferred to kinetic energy so if the speed increases by a lot the height and the pressure could both decrease so you have multiple situations that you can deal with here but let's illustrate this example so let's say if we have another pipe and the height decreases but also the cross-sectional area increases so let's say the cross sectional area is 10 square centimeters and at this position the cross sectional area doubles to 20 square centimeters and let's say the pressure is 200 kilopascals or 200 000 pascals and the speed let's say it's currently 10 meters per second so that's v1 so we need to calculate v2 the new speed and we want to calculate p2 and let's say the height difference between these two positions let's say it changes by 10 meters so first let's find a new speed if the cross sectional area doubles well actually i don't want it to double i want it to decrease so the speed can increase so let's say the new cross sectional area is five so if the cross sectional area is reduced by a half that means the speed has to increase it has to double so the speed is 20 meters per second according to the equation a1 v1 equals a2 v2 so if the area decreases the speed increases so we know the kinetic energy is increasing now let's calculate the new pressure so p1 plus p1 or just pv squared one half pv one squared plus pgh1 is equal to p2 plus one half pv2 squared plus pgh2 so let's say this is the reference level so h2 is zero p1 is 200 000 plus one half the density of the fluid which will use water times v1 squared or 10 squared plus p times g times h1 which is 10 meters above the reference level equals p2 plus one half p times v two or twenty squared ten squared is a hundred times the thousand times point five so this is fifty thousand and a thousand times nine point eight times ten is ninety eight thousand and 20 squared times a thousand times 0.5 is 200 000 plus p2 so these two cancel which means p2 is basically 50 000 plus 98 000. so p2 is now 148 000. so now let's think about what happens here now what do you think happened so we know that the speed increased which means that the kinetic energy of the fluid went up but notice that the gravitational potential energy decreased because the fluid it decreased in height and also the pressure decreased from 200 000 to 148 000. now just because the height decreases it doesn't always mean that the pressure will increase if the speed was constant then the pressure has to increase if the height decreases but in this case the increase in kinetic energy exceeded the decrease in gravitational potential energy so in order to increase the kinetic energy further the pressure had to decrease as well so let's say if the change in height was in 10 meters let's say if it's 20. so let's say if the decrease in gravitational potential energy was significant what's going to happen now so p1 which is 200 000 pascals plus pgh that's a thousand times g times the new height difference of 20 meters not centimeters plus one half pv squared or 1000 times 10 meters per second squared this is equal to p2 pg h2 is zero because this is going to be at the reference level plus one half p v two squared so we know this is twenty thousand which is going to cancel with this number so p two is going to be a thousand times nine point eight times twenty which is a hundred ninety six thousand plus a thousand times ten squared times point five which is 50 000. so if we add those two numbers we could see that p2 is 240 thousand pascals now notice that the pressure is higher than what it was before in the last example it was lower than what it was before so in this case because we increased the change in height because it's 20 meters and not 10 meters the decrease in gravitational potential energy was more than enough to increase the kinetic energy and so that extra decrease in gravitational potential energy was used to increase the pressure so whenever the height decreases the pressure will increase if the speed is constant if the speed is not constant the pressure can increase or it can decrease depending on the relative changes in potential and kinetic energy so if the speed were to increase by let's say a small amount the pressure will increase as well however if the height decreases and if the speed is to increase by a large amount the pressure has to decrease to also contribute to the increase of that speed and therefore the kinetic energy so if the speed changes the pressure may or may not increase the pressure is most likely to increase if the speed increases by a small amount but if the speed increases by a large amount the pressure has to decrease so it helps to think of the energy changes so let's say if the gravitational potential energy decreases by 500 joules if the kinetic energy increases by an amount that's less than the gravitational potential energy let's say 400 then the pressure energy has to go up by a hundred so in this case the decrease in height caused the speed to increase by a small amount so the pressure is going to increase by a small amount now let's say if the potential energy decreases by 500 joules but this time the kinetic energy increases by a large amount an amount that's greater than the decrease of the potential energy so let's say the kinetic energy increases by 700 joules to support that increase the pressure has to go down by 200 joules so in this case since the height decreased and since the velocity increased by an amount that's greater than the height the pressure has to decrease so let's summarize what we've just considered and consider the relative sizes of the arrows so if the height decreases and if the speed increases by an amount that's less than a decrease of the height then the pressure has to increase as well so notice that the height of these two arrows is equal to the the height of that arrow let's make this a little bit bigger so now let's say if the height decreases but the speed increases by an amount that's greater than a decrease in the height in this case the pressure has to decrease as well to support the increase of the speed so that's how you can tell if pressure is going to go up or if it's going to go down when the height decreases and if the height decreases but if velocity doesn't change then the pressure has to go up by an amount that's equal to the decrease in the gravitational potential energy so now you know how to tell if the pressure is going to go up or if it's going to go down so anytime the speed is constant if the height decreases pressure goes up if the height decreases and the speed increases by a small amount the pressure will go up if the height decreases but if the speed increases by a large amount greater than a decrease of the height then the pressure is going to go down to support the increase in speed so hopefully that makes sense try this problem so let's say if we have a similar situation let's say that the area is 2 centimeters squared and over here the area is four centimeters squared and let's say the velocity v one is currently two meters per second at a pressure of 300 kilopascals now let's say the height difference is 5 meters so if the area let's reduce the area rather than increasing it let's say the area is one square centimeters so if the area reduces by a factor of two what is the new velocity so if we decrease the area by two we know the velocity is going to double p2 so we know it's going to be 300 000 pascals plus one half the density times the velocity squared plus pgh and that's equal to pgh2 which is zero plus p2 plus one half times a thousand times four squared if you solve for p2 you should get 343 000 so notice that the pressure increased now let's say if you decrease the area to an even lower amount let's say that the area is decreased to let's say it decreases by a factor of four so to 0.5 squared if it decreases by a factor of 4 that actually let's say it decreases by a factor of 8 let's say it's 0.25 what's the new velocity if it decreases by that amount so if it decreases by a factor of eight the velocity should increase by a factor of eight two times eight is sixteen so notice that the speed increases by a large amount calculate the new pressure so it's going to be p1 which is 300 000 pascals plus one half pv squared plus pgh and pgh2 is zero plus p2 plus one half pv squared so a thousand times nine point eight times five is forty nine thousand two squared is four times a thousand which is four thousand times half that's two thousand so two thousand plus forty nine thousand that's fifty one thousand plus three hundred thousand the entire left side is three hundred fifty one thousand sixteen squared times a thousand times point five is a hundred twenty eight thousand so p2 is 351 000 minus 128 000 which is 223 000 or 223 kpa so as you can see as the height decreased and when the velocity increased by a small amount the pressure went up from 300 kilopascals to 343 now when the height decreased by the same amount but when the velocity increased by a large amount notice that the pressure went down from 300 kpa to 223 kpa because the kinetic energy of the system increased by a lot the decrease in gravitational potential energy was not enough to support the increase in kinetic energy so the potential energy stored in the form of pressure had to go down now let's consider the reverse situation let's say that the pipe goes up this time and the cross-sectional area decreases how will the pressure change will the pressure increase or decrease so since the area is decreasing that means that the velocity has to increase which means that the kinetic energy of the system has to go up the velocity has to increase if the volume flow rate is to remain constant meaning that the fluid is incompressible the density has to remain the same so if the velocity goes up kinetic energy has to go up now the height is increasing which means that the gravitational potential energy is increasing as well so what has to happen to pressure in order for the kinetic energy to go up and the potential energy to go up so pressure has to decrease the potential energy stored in the form of pressure has to go down in order that the gravitational potential energy can go up and the kinetic energy can go up so here the fluid is moving at a high speed and here it's moving at a low speed so therefore the pressure has to be high and here the pressure is low so the pressure decreases so if the height increases the pressure has to decrease to support the increase in height and if the speed increases the pressure has to decrease even more to support the increase in speed so in order for these two forms of energy to go up pressure has to go down now let's say if the area increases rather than decreasing as the fluid travels upward so if the area increases that means that velocity has to decrease and if the speed is decreasing the kinetic energy has to decrease now we know that height is increasing which means that the gravitational potential energy has to be increasing so the pressure is it increasing or decreasing so it depends so here are two scenarios let's say if let's say if the change in kinetic energy is greater than the change in potential energy gravitational potential energy and in the other situation the change in kinetic is less than the change in potential so if the change in kinetic energy is greater than the change in potential energy will the pressure have to increase or decrease now let's use numbers so if the change in kinetic energy is greater let's say that the kinetic energy decreases by 300 joules that means the potential energy has to go up by something that's less than 300 so let's say the potential energy goes up by 200 that means that the pressure has to go up by a hundred negative 300 is opposite to positive 300. if negative 300 joules of energy was lost by means of kinetic energy then if 200 joules were gained by potential energy then the other 100 has to show up as an increase in pressure so if the decrease in kinetic energy exceeds the increase in potential energy the pressure has to go up now if the change in kinetic energy is less than the change in potential energy the pressure has to go down now it helps if you put numbers to it so let's say if the kinetic energy decreased by 400 joules and if the gravitational potential energy increase or change more than that let's say it changed by 700 joules then to support the increase in potential energy the pressure has to decrease by 300. so this is negative negative and positive so the change in energy is zero here if you add these three values energy is conserved the change in energy is zero now let's say if you have a tank filled with water and there is an opening let's say this is the height of the water and you want to find out the velocity of water that shoots out of this tank now let's say the height of water above that position let's say it's 20 meters how can you calculate the velocity of water that flows out using torricelli's term the equation that you need is velocity is equal to the square root of 2 g h there's many ways in which you can get this equation one way is to realize that because the height of the water is decreasing the potential energy is being converted to kinetic energy the pressure inside here is basically a vacuum and the pressure outside here is a vacuum tube is basically the atmospheric pressure so there's really no change in pressure the height is decreasing and that change in gravitational potential energy is converted to kinetic so pe is equal to ke so mgh is equal to one-half mv squared so you can cancel m and then if you multiply both sides by two two g h is equal to v squared and if you square root it the velocity is simply the square root of 2gh so it's going to be the square root of 2 times 9.8 times the difference in in height between these two points and this is going to give you a speed of 19.8 meters per second now you can use bernoulli's equation to get the same result so you want to calculate the velocity and let's draw the picture again so right now the rate at which the velocity or the height of this fluid is decreasing is very slow so we could say v1 is approximately zero we're looking for v2 now let's say this is the reference level let's say that's ground level the height difference is basically h1 h2 is basically going to be zero if we choose that to be the reference level so we can get rid of this portion of the equation and once the water exits the opening it's basically exposed to the atmosphere which has a pressure of 101.3 kpa and if this surface is open too the pressure there is equal to the atmospheric pressure which is 101 kpa so these are the same they cancel so now what we have left over is the gauge pressure pgh1 is equal to one-half p1 or just p and v1 squared which is basically the gauge pressure is associated with the gravitational potential energy mg mgh and that's this this pressure here is associated with the kinetic energy of the fluid which is one half mv squared so we can cancel the density the same way we can cancel the mass and then gh is equal to one half of v squared so if you multiply both sides by two two g h is equal to v squared and if you take the square root the velocity is equal to the square root of 2gh so using bernoulli's equation or simply using the conservation of energy you're going to get this equation so if you increase the height of the water tank the speed at which water flows out of that tank will increase let's try this problem water at a pressure of 9 atm flows into a building at ground level at 3.2 meters per second through a pipe eight centimeters in diameter the pipe rises 15 meters high and decreases to 3.5 centimeters in diameter what is the velocity and pressure of the water at this point so let's draw a picture so it rises and then the diameter decreases so the height difference is 15 meters feel free to pause the video and work on this example now if the height increases and if velocity was constant the pressure should decrease the increase in gravitational potential energy is provided by the decrease in the potential energy due to the decrease in pressure now because the area decreases the velocity has to increase if the velocity increases and if the height was the same the pressure would have to decrease so in this case because the gravitational potential energy is increasing and plus the kinetic energy is increasing the pressure has no choice but to decrease if the height goes up and the speed goes up the pressure has to go down there's no other way so let's calculate the new pressure so at ground level the pressure is 9 atm that's p1 we need to find p2 but before we can find p2 we got to find v2 so the velocity v1 is 3.2 meters per second so what is v2 so let's use the volume flow rate equation a1 v1 is equal to a2 v2 and a is pi r squared so these two cancel r one is basically half of eight centimeters if the diameter is eight centimeters the radius is 4 centimeters v1 is 3.2 and r2 is half of 3.5 so it's 1.75 so 4 squared times 3.2 divided by 1.75 squared is equal to a velocity of 16.7 meters per second so the radius went down from about four to less than two if the radius decreases by factor of two the speed will have to increase two squared or a factor of four so here this the radius decreased by more than two so the speed increase by more than four so now that we have the speed let's use bernoulli's equation to calculate the pressure so it's going to be p1 which is non-atm times 101 300 and one atm is 101 300 pascal so 9 times that value gives you p1 which is 911 700 pascals plus pgh so let's say this is at ground level so the height is zero so there's no pgh value and then plus one half p1 v1 squared so the density of water is 1000 times v1 squared which is 3.2 squared now that's equal to p2 plus pgh a thousand times 9.8 times the height of 15 plus one half pv squared now three point two squared times a thousand times point five that's fifty one twenty plus nine eleven seven hundred on the left side the total pressure value is nine hundred sixteen thousand eight hundred twenty and then a thousand times nine point eight times fifteen plus point five times a thousand times sixteen point seven squared that's uh two hundred eighty six thousand four hundred forty five and then plus p two so nine sixteen eight twenty minus 286 445 gives you a p2 value of 630 and 375 pascals so as you can see because the height increase and the speed increase the pressure had to decrease it decreased from 911 000 pascals to 630 000 pascals and if you wanna convert it back to atm divide six thirty three seven five by one three hundred so in atm this pressure is about six point twenty two atm so generally speaking if you want to raise a fluid to a high position above ground level you need to decrease the pressure if the speed is going to be the same and if you want to increase the speed where the height is the same you need to decrease the pressure now if water flows to a lower level and if the speed is the same the pressure is going to increase now if water's slowing down from high speed to low speed and if the height is the same the pressure has to increase so make sure you understand when the pressure is going to go up and when it's going to go down but generally speaking if the water if the speed decreases or if the height decreases where the other thing is constant so if the speed decreases and the height is constant pressure increases if the speed is constant and the height decreases pressure increases now let's consider this problem so let's draw a house let's say this is the window of the house now if wind blows over the house what is the net upward force acting on the roof so let's say this is the side view of the roof now there's air underneath the roof and there's air above it so there's no wind blowing the pressure above and the pressure below is the same so the pressure below exerts a force and the pressure above exerts the force so the pressure is the same these two forces cancel out and the net force is zero so remember the pressure of air at sea level is about 101.3 kpa now remember bernoulli's principle whenever the velocity is low the pressure is high and when the velocity is high the pressure is low so if wind passes over the top surface of the roof you have high speeds so the pressure is going to be low so if you have a low pressure system or low pressure region created by this wind basically the wind creates a vacuum it decreases the pressure above the roof and inside the house there's not much wind so the pressure inside is high therefore the upward force exerted by the high pressure is created by the downward force exerted by the lower pressure so there's the net upward force that's going to lift from the roof so if the speed of the one is very very high the pressure inside the house can literally lift up the house and that's how a hurricane can pull off roofs from homes the speed of the wind above the roof if it's fast enough can decrease the pressure above the roof of the house causing the pressure inside the house to lift up the house so if that roof is not attached strongly to the house it's flying off in a hurricane now how can we quantify the net force so we need to use bernoulli's equation so p2 plus pgh2 plus one half pv1 squared is equal to p1 plus pgh1 plus one half p v2 squared so we're dealing with air which means the density is pretty low and the height difference of the roof is negligible so we can cancel these two terms so basically what we can do is move p1 to the left side so it's going to be p2 minus p1 on the left side and let's take this term move it to the right side and that's equal to one half p v two squared minus one half pv one squared so the change in pressure is basically one half times p and then that's v2 squared minus v1 squared so the change in pressure is equal to this equation and the change in pressure which is p2 minus p1 if we multiply both sides by a p2a minus p1a is equal to one half p times a v2 squared minus v1 squared now the reason why you want to do that is because pressure is force divided by area so pressure times area is 4. so p2a is basically f2 and p1a is f1 which equals what we have here so let's make some space first so therefore f2 minus f1 is basically the net upward force which is one half the density of the air times the area of the roof times the difference in the squares of the speeds so now we can calculate the net force so it's one half times the density of the air times the area of the roof which is 300 square meters times the difference in the squares of the speeds now inside the house there's probably no wind so we could say it's zero so v1 is zero outside of the house above the roof we have a wind speed of 50 so it's going to be 50 squared so it's 0.5 times 1.29 times 300 times 50 squared so the net upward force is 483 thousand and 750 newtons so as you can see the force depends on the density of the air the area upon which the wind acts upon and the square of the speeds so if you double the density of air the force will double if you double the area upon which the wind act upon the force will double however if you double the speed of the wind the force will quadruple because the force is proportional to the square of the speed so as the wind speed increases the force increases exponentially so to speak now if you think about it let's say if you have a ball and it's windy the wind is going to exert a force on the ball but it's not going to be that much now imagine if it's windy and if you carry let's say a very large cardboard with a huge area so when the wind blows on the cardboard it's going to exert a large force the larger the area the greater the force exerted by the wind and if the cardboard is large enough the wind could probably literally carry you away now here's another question for you let's see if you have a block of water something and there's a wind speed traveling this way and this way as well but the wind speed on the right side let's say it's 40 meters per second and the wind speed on the left side is 90 meters per second what is the direction of the net force on this uh wooden block or this piece of cardboard isn't that force directed upward downward to the left or to the right what would you say now i do want to mention something in the last example the wind was moving in a direction perpendicular to the surface of the cardboard in a situation like this the wind will exert a force in a direction of motion but in this example the wind is traveling parallel to the surface of the block so according to bernoulli's principle whenever the wind speed is low the pressure is high and whenever the wind speed is high the pressure is low so this the wind speed is parallel to it so it's going to create a high pressure when the speed is low which means that that high pressure will exert a larger force towards the left and the low pressure will exert a small force towards the right so therefore the net force is going to be in the direction of the side with the higher speed so in the last example then that force was an upward force because the wind speed above the roof was larger than the speed underneath the roof and so that's why there was an upward force now the same concept applies to the wing of an airplane so as an airplane moves forward the wind that travels on the upper surface has to travel a longer distance based on the design of the wing the upper surface is longer than the lower surface so the wind that travels above it has to travel at a high speed and the wind that flows beneath it has to travel at a low speed so according to bernoulli's principle whenever the velocity of the fluid is low the pressure is high and whenever the velocity of the fluid is high the pressure is low so since we have a high pressure underneath the wing this creates an upward lift force so the faster the plane is moving forward the greater the upward lift force will be so let's say if you have an object and there's a speed above the object that is about 80 meters per second and the speed below it is 30 meters per second calculate the upward lift force exerted on this block if the area of the block where the wind passes parallel to it is about let's say 400 square meters so we already have the equation we know that the net force is basically one half the density of the air times the area times the difference in the squares of the speeds so it's one half the density of air which is 1.29 times the area of 400 square meters times v2 squared which is uh 80 squared minus v1 squared which is 30 squared so it's 0.5 times 1.29 times 400 that's 258 and then times 80 squared minus 30 squared which the total force is 1 million 419 000 newtons

Channel: The Organic Chemistry Tutor

Views: 1,065,508

Keywords: fluids, physics, mechanics, rest, motion, pressure, density, absolute pressure vs gauge pressure, pascal's principle, hydraulic lift, open tube manometer, mercury barometer, buoyant force, mass flow rate, volume flow rate, archimede's principle, buoyancy, toricelli's theorem, bernoulli's principle, apparent mass, apparent weight, bernoulli's equation, pascal's law, continuity equation, atmospheric pressure, specific gravity, gauge pressure, absolute pressure

Id: nUKUzpRgdbY

Length: 242min 34sec (14554 seconds)

Published: Fri Oct 14 2016