Physics - Mechanics: Applications of Newton's Second Law (3 of 20) incline with 2 blocks

and here's our third example of how to use Newton's second law as you can see these problems are going to get a little bit more complicated as we go along as the examples continue here you can see we have a witch shaped object with a mass on top of the incline and then we have another mass hanging from a string attached to the first mass over pulley and you can see that in from intuitive intuition if this is a 10 kilogram mass and there's an 8 kilogram ask that the this big mass will probably pulled a smaller mass of the incline there's going to be an acceleration in this direction so the assumption is that the acceleration will be like this the question is what is that acceleration and what is the tension of the string over here and what is the tension in the string over there all right now we can solve this in various ways we can use what we call Freebody diagrams on each component here so we can take a look at the free body diagram here we take a look at the free body diagram here and then we could solve them simultaneously and we can get the answer that way although I'd like to reserve that kind of methodology to more complicated problems where that's really the only way we can do it there's actually an easier way in which we can solve the problem we can look at the eight kilogram mass the 10 kilogram mass and the string in between as a single system and we can look at all the forces acting on that system so if we use that taking that technique we'll go ahead and look at all the forces that act on the system and so we do not at this moment need to worry about the tension on the string because that's part of the system and so we can ignore that for now so we're just simply going to find out what are all the forces acting on this whole system also notice that I didn't write it down but also I'll just say it that the friction between incline and the mass here is equal to zero so we'll look at this as a whole system what are all the forces acting on it so first of all we have the force of gravity pulling down here so this is m1g and this is on an incline we're going to draw the perpendicular component and the parallel component to the incline so this would be m1 g times a cosine of theta and this is m1 G times the sine of theta all right now of course we still have the reactionary force of the force pushing down perpendicular an incline so we'll have the normal force pushing back this way that's a normal force and that's equal magnitude opposite direction to m1 G cosine of theta so they will cancel each other out notice I can now ignore this because I've divided this into the two components right there then we still have another force acting on the system over here we have the weight of this mass pulling down this is m2 G and notice there's no reactionary force because it's not resting on anything so there's nothing pushing back now what are the two or D well I can't gave it away didn't it what are the two forces that are now acting on the system the two net forces or the two forces that make up the net force on the system it would be this force right here and it would be this force right here now you say well how can you work with that because this this force is pushing this way or pulling this way this was pushed pushing this way and they're not pushing in the same direction well in a way they are because here you have a pulley and the string that connects the eight kilogram mass and 10 kilogram mass goes around this pulley it just kind of redirects the tension or the force around that pulley so in a way this force right here is taking the whole system and pulling it like this this force is taking the whole system and pulling it like this and if this is bigger than this the system will accelerate in this direction if this is bigger than this then the system will accelerate in this direction so it's okay to think of this as being two opposing forces even though they're not pointing in a linear direction all right if we then take that as the assumption we can say that F net is equal to mass total the whole system times the acceleration of the system or the acceleration is equal to F net divided by the mass total now I'm going to take the assumed direction as the positive direction with other words this acceleration from there up and this way is going to be considered the positive direction for the acceleration so any force pull are pushing in that direction I'm going to call a positive force any force opposing that in the opposite direction I'm going to call a negative force so we have this force right here which will aid the acceleration so this would be equal to m2 G and I'll call that positive minus this force right here because it opposes acceleration so it would be equal to m1 G sine of theta and the whole thing divided by the sum of the two masses m1 plus m2 and that's how I can find the acceleration of the whole system all right plugging in the numbers so this is equal to m2 which is 10 kilograms times 9.8 meters per second squared minus m1 which is eight kilograms times 9.8 meters per second squared let me move all the way so you can see that I still have to multiply times the sine of theta and that would be the sine of 30 degrees and the whole thing divided by the total mass which will be m1 which is 8 kilograms I believe yup and 1 8 kilograms plus 10 kilograms ok simplifying that a little bit that would be 98 Newtons - that would be half of this would be 4 times that that would be hmm thirty-nine point two thirty nine point two Newtons all divided by 18 kilograms at this point I'm going to need my calculator never leave home without it 98 - thirty-nine point two divided by 18 and I get three point two seven so that's equal to three point two seven meters per second squared all right so now I have my acceleration of the system which means this will accelerate in this direction at three point two seven meters per second squared and this will accelerate up the incline at three point two seven meters per second squared now I need to find so I got the acceleration I need to find the tensions I'm going to now concentrate on this right here so take the mass m2 at ten kilograms so this was equal to m2 and I have the force of gravity pulling down on it I like to use blue for that side of the four is this way so it would be m2 G and then a half a force pulling upward and that would be tension two now which of those two is bigger well it appears to me that m2 G should be bigger than t2 because the acceleration in this direction so that makes sense what I could do here is I could use the second laughter I'm not settle Thurman I'm the second law of Newton that we say that F net equals M times acceleration and since the net force would be m2 G minus tension - because again I'm assuming the acceleration to be in this direction so I'll call this the positive acceleration so it's this minus this this is aiding this is opposing the acceleration and that is equal to m2 times a and so therefore if I solve this for T - I can say that if I move this over there and move this over here I get m2 G minus m2 a equals T 2 and verse in the equation T 2 is equal to m2 G minus m2 a all right now the plug in the numbers m2 is 10 kilograms that's 10 kilograms times 9.8 meters per second squared - 10 kilograms times the acceleration of 3.27 meters per second squared and what will they give us nine point eight minus 3.2 seven times ten or sixty five point three Newtons so that would be t2 now what about t1 well turns out that T 1 should be equal to t2 now why can I say that because it's a single string connected from one master the other mass and so whatever the tension is here must be the tension there assuming that there's no friction on this pulley and the pulley has no mass words the mass is so small we can ignore it and so we don't have to worry about the moment inertia that's something for a different topic so we can say therefore that T 1 must equal T 2 and therefore it is equal to 65 point 3 Newtons now what's but if you say well I don't necessarily believe that show me prove to me that that's the case another way in which we can look at it is that T 1 is the force required to hold up this mass against the m1 G sine theta component plus the acceleration of this mass of the incline at this acceleration so another way of looking at it say well I can say that T 1 must equal to m1 G sine theta plus the m1 a so it's the force required to hold up this mess against the gravity component along the incline plus the force required to pull this mass of the incline and if we plug in the numbers there we'll get the same answer well we can see that's true and 1 is 8 I'll leave out the unit's 9.8 sine of 30 degrees is 1/2 plus the mass which is 8 times acceleration of 3.27 and if you now plug it into your calculator so have 8 times 9.8 times 0.5 plus eight times three point two seven equals sure enough six point or times ten 65.4 so give me sixty five point four Newtons which of course is the same hope almost the same slight run off error because I didn't take it off - enough significant figures but you can see you get the same answer now one more thing coming back over here and looking at this situation right here an easy way to figure out what the tension and the string is that have something suspended from it which may or may not be accelerating is that the tension in this case t2 is always going to be equal to the weight of this object which is in this case m2g plus or minus the force required to accelerate it m2 a now I said plus or minus well it is plus if the object is being accelerated upward because done not only do you have to hold it up against gravity mg you also have to have additional force accelerated upwards against gravity so that game would be m2 G + m2 8 but if the object is allowed to accelerate downward then you don't need as much tension in the string you need enough to hold up against gravity - the force that will allow us to to accelerate downward with grabs in that case will be m2 G minus m2 a remember the equation that we ended up with when we use Newton's second law on this Freebody diagram indeed we did get m2 G minus m2 a so it always be mg while this case was M - of course we'll always be mg plus MA or mg - ma plus if its accelerated upward - when its accelerated downward and of course if you look at it over here it will be mg of course times the sine of theta because it was an incline plus MA because it was being accelerated up the incline so I have to add the plus ma to it and that's how you do problems like that

Channel: Michel van Biezen

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Keywords: ilectureonline, ilectureonline.com, Mike, Mike van Biezen, van Biezen, ilecture, ilecture online, Physics, Mechanics, Newton's First Law, Newton's 2nd Law, Applications of Newton's Second Law, Acceleration, Force, Wedge, Pulley, 2 Objects, Tension, yt:stretch=16:9

Id: wY1AkwzKpj4

Length: 12min 18sec (738 seconds)

Published: Tue Aug 27 2013