The two blocks (m = 20 kg and M = 99 kg) in the figure are...

Name: The two blocks (m = 20 kg and M = 99 kg) in the figure are not attached to each other. The coefficient of static friction between the blocks is ?s = 0.33, but the surface beneath the larger block is frictionless. What is the minimum magnitude of the horizontal force F? required to keep the smaller block from slipping down the larger block?
Uploaded: 2022-07-09T19:22:35-08:00
Duration: 5 min 54 s
Channel: Tarandeep Singh
Description: The two blocks (m = 20 kg and M = 99 kg) in the figure are not attached to each other. The coefficient of static friction between the blocks is ?s = 0.33, but the surface beneath the larger block is frictionless. What is the minimum magnitude of the horizontal force F? required to keep the smaller block from slipping down the larger block?

The two blocks (m = 20 kg and M = 99 kg) in the figure are not attached to each other. The coefficient of static friction between the blocks is ?s = 0.33, but the surface beneath the larger block is frictionless. What is the minimum magnitude of the horizontal force F? required to keep the smaller block from slipping down the larger block?

Transcript

00:02 So in this problem, we are given that a big block of mass m here has a small mass m right next to it.

00:11 They are not attached.

00:12 And there is a friction that's a force f being applied to the small mass m.

00:16 And we are given that the coefficient of static friction is 0 .33 between these two blocks.

00:22 And there is no friction between the larger block and the floor.

00:26 We are given the masses to be 20 kages and 99 kg.

00:31 And we are asked to find the minimum force f such that this small block of mass m stays stationary with respect to the big mass m.

00:46 It doesn't go down.

00:48 So what should be that minimum force given that this whole system can actually frictionlessly move or slip on the floor.

00:56 So this is the problem.

00:57 So what we will do is we will draw the free body diagrams of both these bodies.

01:01 And then we'll solve for this case.

01:05 So let's assume that because of this force, this whole system is accelerating in this direction with an acceleration a.

01:11 So we'll just name it a for now so we can mark it in the free body diagram.

01:16 So the small mass, if you draw, you will have a force f in this direction.

01:23 You will have a normal force from the bigger block, fn, let's say.

01:31 And you will have small m times g.

01:35 That's the weight of this block.

01:36 And there'll be a friction force because this block wants to go down there'll be friction force on the upside which will be equal to muus times this normal force and since this is accelerating at an acceleration a there'll be a suitor force small m times a so these are the forces on these bodies on the small body and so let's write down the equation so we can say that in the horizontal direction the force force f towards the right is equal to some of the forces towards the left because we have taken the pseudo force as a compensation for the accelerating frame of reference and created a inert flow of reference like this.

02:23 So in that case all the forces should balance each other out in all directions.

02:27 So that means that in the horizontal direction the right facing forces which is f should be equal to m times a plus fn.

02:38 So let's see this is equation one...

Question

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