Maximum Pulling Force for Two Blocks and a Cord

AI Thread Summary
The discussion focuses on calculating the maximum pulling force that can be exerted on a 5.00 kg block resting on a 10.0 kg block without causing it to slip, given a coefficient of static friction of 0.350. The user outlines their approach by drawing free body diagrams (FBD) for both blocks, identifying forces such as gravitational, normal, pulling, and frictional forces. They express uncertainty about their diagrams and calculations, particularly in relating the forces to the system's acceleration. The equations derived include frictional forces based on the mass of the blocks and the coefficient of friction. The goal is to eliminate acceleration from the equations to solve for the maximum force.
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Homework Statement


A block of mass 5.00kg rides on top of a second block of mass 10.0kg. A person attaches a string to the top block and pulls the system horizontally. Friction between the two blocks keeps the 5.00kg block from slipping off. If the coefficient of static friction is 0.350, what is the maximum force that can be exerted by the string on the block without causing the top block to slip.


Homework Equations


F=ma

The Attempt at a Solution


First of all, I drew my FBD:

top block: gravitational force going down, normal force going up, Pulling Force going to the right and frictional force to the left

bottom block: gravitational force going down (mass of top and bottom block?), normal force going up. I have a feeling that I'm missing something in this diagram.

Can someone help me? thanks
 
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top block - mg downwards (m = 5), N2 upwards, F towards the right, Ff1 towards the left.

bottom block FBD - Mg downwards (M = 10), N1 upwards and friction (Ff1) towards the right on top, and Ff2 towards the left at the bottom.

And the whole system is accelerating towards the right with 'a', so you can the pseudo forces to the two blocks.

Note that Ff1 = \mumg ans Ff2 = \mu(m+M)g

You can now get two equations, eliminate 'a' and find F.
 
graphene said:
top block - mg downwards (m = 5), N2 upwards, F towards the right, Ff1 towards the left.

bottom block FBD - Mg downwards (M = 10), N1 upwards and friction (Ff1) towards the right on top, and Ff2 towards the left at the bottom.

And the whole system is accelerating towards the right with 'a', so you can the pseudo forces to the two blocks.

Note that Ff1 = \mumg ans Ff2 = \mu(m+M)g

You can now get two equations, eliminate 'a' and find F.



Thanks!
 
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