32. Each of these boxes is pulled by the same force F to the...

32. Each of these boxes is pulled by the same force F to the left. All boxes have the same mass and slide on a friction-free surface. Rank the following from greatest to least: a. The accelerations of the boxes b. The tensions in the ropes connected to the single boxes to the right in B and in C 47. Why does a rope climber pull downward on the rope to move upward? 48. You push a heavy car by hand. The car, in turn, pushes back with an opposite but equal force on you. Doesn't this mean that the forces cancel each other, making acceleration impossible? Why or why not? 62. A stone is suspended at rest by a string. (a) Draw force vectors for all the forces that act on the stone. (b) Should your vectors have a zero resultant? (c) Why or why not?

Transcript

00:01 So in this problem, we have this first part where we have a section of boxes, one box with an applied force, two boxes with the same applied force, or three.

00:14 And if there's multiple boxes, they're attached by a rope.

00:20 So they're all friction -free, and all the boxes have the same mass.

00:23 And we want to rank the acceleration between them.

00:26 So they each have mass m.

00:31 So we can relate force, mass, and acceleration with newton's second law, sum of forces is equal to ma.

00:38 We only have one force in this case, though, force, and that's going to be equal to ma.

00:43 So this is the case for one box.

00:46 As we see, we've ended up with those same letters.

00:48 And so this first row will have an acceleration in the same direction as force, we would call it a.

00:55 Let's look at the second case, though, if this was the case one, two, and three.

01:01 In case 2, we have the same force, but now there's two boxes.

01:06 So the total mass of the situation is 2m times a.

01:13 And so in case 1, if acceleration is f over m, just solving it, acceleration case 2 is force over m and then one -half because we have that extra 2.

01:27 So now we have an acceleration here, 1 -half a.

01:31 We can see where this is going for case 3, where there is three boxes.

01:35 Now.

01:38 And so the acceleration here is equal to one -third f over m.

01:43 So the acceleration is equal to one -third a because we can always compare it to this first case, right? we have acceleration equals f -f -m.

01:53 So any time we have f -r -m in these other equations, that's also equal to a, but a of two should be equal to one -half and so forth because of the fraction.

02:06 So even though these are separate bodies because they're attached by a rigid string, we can kind of treat them as acting like individual bodies.

02:14 Now you can work this out, have a box and it's through by the diagram with the tension, and then the box in the middle with tensions and work through all the way, and then you have force on the other side, and you'd get the same answer.

02:34 So we can rank the accelerations and we see this.

02:37 The greatest acceleration is here, and then the least for boxes is three, and then the case b is in the middle.

02:51 In the next part, we want to look at the tension in the ropes.

02:55 So let's look at, and we can basically use the same working out.

03:00 To cause box m to have acceleration a, we need force f.

03:06 So let's imagine we just have this box with a rope.

03:10 We know the force here is tension, but what should it be equal to? this box is accelerating a mass a, and so this tension is equal to m a to accelerate that box with the appropriate acceleration.

03:25 And that's going to be the same thing here, where the tension here is going to be equal to m a.

03:33 But what about the two boxes here in the string here? well, we have two boxes and a force, which has some tension here.

03:41 That's going to cause a certain acceleration.

03:43 Well, what is that? well, it's like this case.

03:46 We need force to carry 2m through.

03:50 So the tension in the second string should be equal to 2m .a because it's two times the mass to cause the same acceleration.

04:00 And so we can rank the strings if we have our rope 1 here, rope 2 here, rope 3 here.

04:10 We see the tension in rope 1 is equal to the tension in rope 2 because that's the same situation.

04:16 But the tension in rope 3, now that it's carrying two boxes, has to be greater than the tension rope 1 or the tension in rope.

04:24 Two because of course they're equal so if t3 is greater than one it's equal if greater than the other and the next problem she's going to talk about some more of newton's second laws so here we have in the next part a rope climber that's hanging on the rope now they pull downwards on the rope and we want to see why well this is going to talk about newton's third law where every force has an equal and opposite reaction so if there is this force from the climber onto the rope then there's an equal and opposite force on the other body.

05:02 So this is the force from the climber onto the rope.

05:08 So there's an equal and opposite force on the person, the force of the rope on the climber...

Question

Please give Ace some feedback