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

You are lowering two boxes, one on top of the other, down a ramp as shown, by pulling on a taut cable that is parallel to the surface of the ramp. The lower box has mass m1 = 20.0 kg, and the upper box has mass m2 = 10.0 kg. The two boxes move together at constant speed 0.125 m/s: the upper box does not move with respect to the lower box. The coefficient of kinetic friction between the ramp and the lower box is ???k = 0.250, and the coefficient of static friction between the two boxes is ???s = 0.900. (a) Find the magnitude of the force that the cable exerts on the lower box. m2 m1 3.00 m 4.00 m (b) Find the magnitude of the static frictional force that the lower box exerts on the upper box. Keep in mind that it is static friction that prevents the upper box from slipping off of the lower box. *Check* that your calculated value is smaller than the maximum possible value for static friction. (c) Draw a free-body diagram for the upper box, showing the direction and numerical magnitude of each force. Decompose the gravitational force into components, as a check that your calculated force values add up as expected. Please label your forces G=gravity, S=static, N=normal, E=Earth, L=lower, U=upper. (d) Draw a free-body diagram for the lower box, showing the direction and numerical magnitude of each force. Again decompose the gravitational force into components, as a check that your calculated force values add up as expected. Please label your forces G=gravity, S=static, K=kinetic, N=normal, E=Earth, L=lower, U=upper, R=ramp, T=tension (in Cable).

Transcript

00:01 In this problem, you're considering two boxes that are being let down a ramp.

00:08 So i've gone ahead and copied over the picture from the problem and written out all of the relevant information.

00:16 In part a, you're asked to figure out what is the tension in the rope.

00:23 What is the force on the rope on the bottom box? and so there are multiple ways that you can approach this.

00:36 I'm just going to be showing one of them.

00:40 And actually i'll be answering almost all of the other additional problems along the way, starting with drawing a free body diagram.

00:51 So i'm going to start with drawing a free body diagram for mass 1 and for mass 2.

00:58 If we're looking at mass one, there are a few different forces.

01:12 So we know that the tension force acts on mass one in a direction up the ramp.

01:21 And then we can actually look ahead to part d to think about some of the other forces.

01:30 So gravity is going to act on mass.

01:35 It's going to be straight down but we can decompose gravity into its x and y components and so i'll go ahead and draw those on here as well where this is the x component and this is our y component of gravity next we can think about the friction forces so mass one will experience kinetic friction between itself and the ramp, and because the mass is moving down the ramp, that force of kinetic friction will be in a direction up the ramp.

02:21 So i'm going to draw this next to the tension force.

02:27 It's also going to experience, and again, you can call this different things.

02:35 It's going to experience a force due to its interaction with mass one.

02:42 You could call this the force u for the upper mass.

02:50 I'm going to actually call it the static friction force because it's going to be due to static friction between mass 1 and mass 2.

03:01 For mass 2, that static friction keeps it from sliding down so it's directed up.

03:07 And so the static friction that it exerts on mass one will be pointed down the ramp in the opposite direction.

03:16 So our static friction force is going to be in this direction.

03:23 We can also think about the interaction between the block and the ramp.

03:28 Again, it looks like you could call this force, the force from the ramp, but i'm going to call it the normal force because it is a force that is normal.

03:39 To the direction of the surface.

03:52 And we also have to consider the weight of the second block.

04:01 So there's going to be an additional, almost normal force in the opposite direction, but that normal force will be due to the lower block, or sorry, the upper block.

04:20 So i'm going to instead label it as upper.

04:27 And specifically, it's going to be similar to the gravitational force.

04:36 It's going to be the y component of the gravitational.

04:41 Force from the upper block, which is acting as a normal force downward on block one.

04:49 And i think at that at this point, we've included all of the forces.

04:53 We have gravity on here.

04:56 We have a normal force from the ramp.

04:58 We have a normal force from block two, which i've called upper.

05:02 We have our two friction forces and the force of tension.

05:09 I'm going to go ahead and do the same thing for mass two.

05:13 It's a lot easier for mass two though because mass two because we went ahead and called it upper in our free body diagram for mass one i'm going to call its gravitational force upper so you again you could call this g2 or something like that but because i used that before that's what i'm going to use here.

05:50 But again, we would decompose it into its x and y components.

05:57 On our upper block, the only other things that we need to consider are the normal force and the force of friction.

06:08 So it will experience a normal force as well.

06:14 And because these normal forces aren't the same, i'm going to go ahead and label them.

06:20 So i'm going to label this one as n1 and this one as in 2.

06:25 Again, we could instead label the normal force on mass 2 as n and the normal force on mass 1 as r because it's coming from the ramp.

06:39 It is really up to you.

06:42 You want to choose how we label the forces.

06:46 Matter too much as long as we know what forces we're referring to you're consistent with your labeling and you include all of the relevant forces.

06:58 But with that being said, we have one more force here and that's the static friction force on m2.

07:04 It's keeping it from falling down the ramp.

07:07 So it's going to be directed up the ramp and has a magnitude equal to s.

07:14 And these static friction forces are going to be equal in magnitude, so i'm going to go ahead and leave it as s.

07:23 Okay, so this actually answers parts c and d, actually this i think is d, parts c and d for this problem.

07:38 And we haven't gotten to finishing up a yet.

07:41 The next step is to utilize newton's second law.

07:48 So i focus on mass one because that is the one that includes the tension force.

07:55 And the tension force is in the x direction if i consider positive x to be up the ramp and positive y to be out of the ramp.

08:07 Because we're told that the boxes move with a constant speed, we know the acceleration is zero, and thus the sum of the forces in the x direction for mass 1 must be equal to 0.

08:24 In the x direction, we have a positive t, a positive k, and we have a negative s and a negative.

08:39 With g sub x or x component of the gravitational force...

Question

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