Tension: Example

by Jared Rovny

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    00:01 Here's a good example of a kind of pulley problem where you can see ropes and boxes together.

    00:07 Suppose you have a 10 kg mass on a slope, this time connected to a rope which itself goes over a pulley at the top and connects to a 15 kg a heavier mass that's just hanging by the rope.

    00:18 The question here is if the slope has an angle of 30 degrees with the horizontal and the masses start from rest, how long will it take for the 15 kg mass to fall, or sorry how far for the 15 kg mass fall in a time of 1 second, if it's connected to this other mass on the slope? So again, you should first give this a shot and see if you can use a picture like this which starts getting you started with the forces already, you can see some of the forces listed here you should try to remember, and see if you can find how far the 15 kg mass, M2, will fall in a time of 1 second.

    00:51 What we have listed here is a force of tension acting upwards on the second mass.

    00:56 It's also acting up the slope on the first mass, and as I just mentioned, this tension forces will be the same as each other.

    01:02 They'll be equal in magnitude because they're referring to the same rope.

    01:05 Also we have the second mass being pulled downwards by gravity and we have the mass on the slope having a normal force outward from the slope which again tries to keep the box from going into or out of the slope and we also have the gravitational force directly downwards, which I shaded because in this picture, you can already see that we've broken up that gravitational force into its two components, one in the x direction and one in the y direction.

    01:27 They're listed here one more time, make sure that you get those angles right.

    01:31 So this is Fg of the first box times the cosine of theta and then down the slope we have Fg for the first box times the sine of theta.

    01:40 And then lastly, we want to make sure we introduce a coordinate system, and you can introduce any coordinate system you want as long as you stick with that coordinate system throughout the problem.

    01:49 So, let's see what this problem looks like.

    01:52 We wanna find out how far this box falls in a time of 1 second.

    01:57 So I'm gonna call this d. Suppose we want to find d for this object falling a certain distance.

    02:03 The first thing we need to do is since we have a coordinate system for this first box over here, we should have a coordinate system also for the second box because we're going to write Newton's first, sorry second law for the first box and we're going to write Newton's second law for the second box, so we need a coordinate system for the second box as well.

    02:20 So what I'm going to do is say that the upward direction is the positive x direction for this box.

    02:26 Now the reason I'm doing that is just to be consistent with the coordinate system that we see for the first box.

    02:32 It's not necessary to do this as long as you're careful, but it's always a good idea if you see that the motion of the first box and the second box are going to go together.

    02:40 In other words, if the first box goes in the positive x direction for the first box, it's pretty clear from my diagram that the second box, the mass that's hanging is going to go up as well.

    02:51 Meaning that that motion is a corresponding motion.

    02:54 So we can say that they're both going in a positive x direction, and doing this can clarify some things for you and avoid some confusion with your signs later on.

    03:02 So using this convention, let's write Newton's second law for each box.

    03:05 So first, for the first box, let's write the Newton's second law equation for all the boxes in the x direction.

    03:13 So, first box in the x direction we have Fx equals the mass of the first box times it's acceleration.

    03:21 The forces are Fg times sine of theta in the positive x direction and then we have the tension force which is acting in what is the minus x direction for this first box.

    03:38 This is equal to the mass of the first box times its acceleration.

    03:43 And what is something that might be confusing here which is that when we're drawing out the forces on these diagrams, we draw the magnitude, or write rather the magnitude of the force like Fg1 sine of theta but we indicate the direction with the arrow, so I don't actually put minus signs or plus signs on the forces on the diagram.

    04:00 I just indicate directions with arrows.

    04:02 So I have a tension force which is in the negative x direction for this box, but I did not write minus the tension force, I just wrote the tension force itself.

    04:10 So now we have the tension force and we have Fg times the sine of theta acting in the positive direction and that's what this equation is telling us.

    04:17 So, now let's see what this looks like for the second box instead.

    04:21 And for the second box what we have is a force in the positive x direction, a force of tension and then minus, because this force is acting in the negative direction, Fg2.

    04:33 This equals the mass of the second box times the acceleration of the second box.

    04:39 Now one simplication I'm going to make here, and this is a very important physical insight to have about a problem like this which is that these two boxes are a coupled system and what I mean by that is if one moves, the other moves.

    04:50 Anytime you have two objects that are coupled like this that have to move together, they're going to have the same acceleration.

    04:57 So, the acceleration of the first box and the acceleration of the second box have got to be the same because again they move together.

    05:03 So, what I'm going to do is just introduce a letter a which is going to be the same thing as a1 and a2.

    05:09 They're the same acceleration. So with that being true, we have two equations and we'd like to solve for the acceleration. The reason is if we know the acceleration, we can figure out how far this box moves in a certain time, in this case 1 second.

    05:23 So let's see if we can find that acceleration with two equations.

    05:26 We have two equations and two unknowns.

    05:29 The unknowns that we have are the force of tension which we do not know and the acceleration which we also do not know.

    05:34 But any time you have the same number of unknowns as you have equations, you can always be confident that you're ready to solve and you'll always be able to solve these equations.

    05:42 There are a number of ways to solve two equations like this.

    05:45 One easy one if you see a pattern like we see here is that you have a minus force of tension in the first equation and a plus force of tension in the second equation, so you can do something which is to add the two equations together.

    05:57 So, all I do is I add the left side of these equations minus Ft plus Ft is equal to zero plus Fg1 sine of theta and then we have a minus Fg2.

    06:13 So all I've done is I've added the left side of both of these equations for the simple reason that the forces of tension would clearly cancel out if I did so.

    06:19 This is equal to the sum of the right hand side which is m1a plus m2a.

    06:28 Just so I can make a quick point, I'm gonna rewrite the right-hand side of this equation very quickly in a way that might show us something about the physics of this problem, so let's do that.

    06:36 If I factor out the mass and the acceleration, what you can see is something that looks exactly like Newton's second law if I'd written Newton's second law for the entire problem all at once.

    06:52 In other words, what if I considered my system, the thing that I'm looking at, to be both boxes together.

    06:58 If I do that then I can say, what are the forces acting on the boxes from the outside' In other words, we don't look at internal forces of a system, parts of the system acting with each other like the tensions here with both boxes acting on each other.

    07:10 If I just think of the whole system as sort of a single thing and ask what forces are acting on my whole system, then those forces would be the gravitational forces tugging one box one way and one box the other way.

    07:21 In that case I would have F equals ma.

    07:24 In other words, the forces on my system equals the mass of my system m1 plus m2 times the acceleration of my system.

    07:32 And so you can see there are many ways to approach our problem.

    07:35 If we were being clever we could've tried to jumped straight to that one and considered this system as a whole right off the bat, but it wasn't necessary in solving this problem at all.

    07:42 Now, that we have an equation like this, we can simply find the acceleration by dividing both sides by the mass.

    07:47 So, we have Fg1 times the sine of theta minus Fg2 divided by the entire mass of the system m1 plus m2.

    07:57 This is our acceleration, now we have one last easy step which is to use our equations of motion one more time and say that the initial position of the second box, so this is the mass m2, is equal to or sorry the final position of mass m2, is equal to its initial position, plus its initial velocity times time plus 1/2 its acceleration which we just found times the time squared.

    08:23 If we want, we can call this initial position zero or anything we'd like.

    08:28 It doesn't really matter what we call it because we're just looking for the distance, the total distance that it travelled which will not matter what we call its original position.

    08:34 So we can say x minus x nought or final position minus initial position.

    08:39 In other words, the distance that it travelled or the displacement as we introduced it will be equal to and then we have these other terms initial velocity times time, we don't think it was over or the problem gives us that it was not moving initially, otherwise they would have given us a velocity, equals 1/2 the acceleration that we just found, so we put that whole thing in here, Fg1 sine of theta minus Fg2 divided by the total mass m1 plus m2 times the time squared.

    09:13 And of course the last thing we'll have to do in any problem including this one is to plug in our numbers so let's go ahead and do that now.

    09:19 We have 1/2 and now we have a 15 kg and a 10 kg mass.

    09:24 So, we have mass 1 times g1 sine of this angle which is a 30 degree angle still from the original setup.

    09:38 So sine of 30 degrees is 1/2 minus the force of gravity on mass 2 which is mass 2 times g divided by the total mass, the sum of the masses.

    09:51 We have t squared where the time is 1 second and so that won't change our calculation.

    09:58 So, let's write our final conclusion here.

    10:00 Let's keep this equation going in one last place, we're almost done.

    10:04 We have 1/2 times, let's factor out the g, and we have mass 1 divided by 2 which is 10 divided by 2 or 5 minus mass 2 which is 15 divided by the total mass which is 10 plus 15 or 25, and we factored out the g, so that is approximately 10 as we're assuming in these problems.

    10:30 Then we get our answer, which is 1/2 and the 10, we can cancel and get a 5.

    10:36 You know 5 minus 15 is minus 10 so we have minus 10 times 5 over 25 which will be minus 2.

    10:46 So, there's something interesting about this answer.

    10:49 First of all, it's negative. We got a minus 2 meters for the distance that this thing went.

    10:53 But that actually makes sense in our coordinate system.

    10:55 Look at our coordinate system. It started with this assumption that we're calling the upward direction, the positive x direction.

    11:02 So, I should expect when I find the displacement, that the displacement will be a negative number because the object is moving downward.

    11:08 So, be careful as you're going through problems to keep track of your minuses and your pluses and to make sure that it has some physical intuition behind it and makes sense at the end of the day.

    11:17 Because many times, just by looking at the plus and minus signs, we can figure out whether a given answer is right or wrong.

    About the Lecture

    The lecture Tension: Example by Jared Rovny is from the course Force.

    Included Quiz Questions

    1. 12 N
    2. 15 N
    3. 16 N
    4. 10 N
    5. 8 N
    1. –180 m
    2. 120 m
    3. 185 m
    4. -120 m
    5. 130 m
    1. 147 N
    2. 588 N
    3. 15(9.8)sinθ N
    4. 60(9.8)sinθ N
    5. 50(9.8)sinθ N

    Author of lecture Tension: Example

     Jared Rovny

    Jared Rovny

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