You are lowering two boxes one on top of the other down the...

You are lowering two boxes, one on top of the other, down the ramp shown in the figure (Figure 1) by pulling on a rope parallel to the surface of the ramp. Both boxes move together at a constant speed of 19.0 cm/s . The coefficient of kinetic friction between the ramp and the lower box is 0.430, and the coefficient of static friction between the two boxes is 0.785. Part A What force do you need to exert to accomplish this? Part B What is the magnitude of the friction force on the upper box? Part C What is the direction of the friction force on the upper box?

Transcript

00:01 Okay, we have blocks being slowly let down this ramp and we aren't given any dimensions, so i used some ones i found, and some values for the mass, hopefully they're right.

00:15 But i'll do it symbolically too, so that's fine.

00:18 So we consult for that angle of inclination theta, and we're looking to find the force, what i've called, let me rename it t.

00:29 Trying to find this force t required to keep the, to let the box down at a constant speed, which means they're not accelerating.

00:42 So what we first want to do here is just look at, we want to draw a free body diagram.

00:51 There's our free body of m.

00:55 Let's do m1 first.

00:56 Here's m1.

00:57 So m1 we have the normal force.

01:00 We have the friction, static friction force from block 2, we have gravity.

01:14 On the m2 we have gravity, we have tension, and kinetic friction.

01:29 We have the normal force, and then we have the frictional force from the block sitting on top of the of it.

01:41 And then we also have the normal force from the top block pushing down on the first block.

01:53 Okay, so what we want to do here is look at the forces on m2 in the x direction.

02:09 So first off, draw a little triangle here, there's angle theta.

02:19 So for m2, some of the forces in the x equals 0.

02:30 Some of the forces in the x equals 0.

02:34 Okay, we have our tension plus static friction.

02:48 Minus the gravity force in the x direction, minus the frictional force from that top block.

03:02 That equals 0.

03:03 Now i want to sum the forces in the y.

03:08 So we have n2 minus n1.

03:16 Minus m2g cosine cosine theta equals 0 okay um actually looking at this now i don't think we actually need to do that but that's fine okay looking at m1 some of the forces in the x equals 0 so friction minus m1g sine theta equals zero.

03:54 So the frictional force is equal to m1g sine theta.

04:03 Now we have a third law pair here so f12 equals minus f21.

04:09 So that's just the friction in between those two blocks.

04:12 So the one block is pushing on the other one and the other block is pushing in the opposite direction.

04:19 So we don't, or that's to find the tension, we need f2 .1.

04:27 But we know what f21 is because we know what f12 is.

04:31 So going back to that tension equation, so we get tension plus friction minus m2g sine theta.

04:44 And then it's going to be minus m1g sine theta.

04:50 So that's what we just saw for using third law pairs...