A 450 kg block is placed on top of a 120 kg block that rests...

A 4.50-kg block is placed on top of a 12.0-kg block that rests on a frictionless table. The coefficient of static friction between the two blocks is 0.650. What is the maximum horizontal force that can be applied before the 4.50-kg block begins to slip relative to the 12.0-kg block, if the force is applied to each of the following? (a) the more massive block N (b) the less massive block N

Transcript

00:01 This problem may have two blocks stacked on top of each other.

00:04 The mass of the top block is 5 kilograms, and the mass of the bottom block is 12 kilograms.

00:17 We are given the coefficient of static friction between the top and the bottom block at 0 .6.

00:25 And the coefficient of correction between the bottom block and the surface that's on is 0.

00:35 We're asked to find the max horizontal force applied before slipping between the two blocks, it's supplied to both bottom and the top blocks.

00:43 First we'll draw a free body diagram showing the top block and the bottom block as separate systems.

01:00 Then we will apply the force and the friction of the top block on the bottom block.

01:08 And because friction applies to both in opposite directions, it also applies the top block.

01:14 And we have the normal force for the top and the gravitational force for the top block.

01:19 We also have the normal force for the bottom block.

01:21 Block, gravitational force for the bottom block, and because the top block is sitting on top of it, we also have the normal force for the top block.

01:35 If the top block were to slip, it would be in that direction.

01:40 Now we draw a free by diagram for both blocks as one system.

01:45 So we have the normal, at the top and the bottom block, we have the force, and we have the gravitational force for the top and the bottom block.

01:59 This is for no slipping only, because if they slip, then they are no longer one whole system.

02:11 Using newton's laws of sum of force equals mass times acceleration, we will start with the bottom block, the force applied to the bottom block.

02:24 So in the x direction is force minus friction equals the mass of the bottom block times the acceleration of the bottom block.

02:38 And in the y direction, we have the normal of the bottom block minus the normal of the top minus the gravitational force of the bottom equals zero.

02:52 And for the top block in the x direction, we have friction equals the mass of the top times the acceleration of the top, which using this we can then rearrange to find acceleration equals friction over a mass of the top.

03:10 Block.

03:13 In the y direction we have the normal of the top minus the gravitational of the top equals zero.

03:23 Given that there is no slip, the acceleration of the top block equals the acceleration of the bottom block which we will sum up to b a.

03:35 The formula for the max friction equals the normal of the top block times the coefficient of static friction.

03:44 Because we know the normal of the top equals gravitational force of the top which equals mass times gravity...

Question

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