What Is the Maximum Mass of Block C to Keep Blocks A and B Sliding Together?

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Homework Help Overview

The problem involves two blocks, A and B, resting on a frictionless surface, with a third block, C, suspended by a string over a pulley. The goal is to determine the maximum mass of block C that allows blocks A and B to slide together without separation, given the coefficient of static friction between A and B.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to calculate the forces acting on the blocks, including weight and static friction, to find the maximum mass of block C. Some participants question the logic of assuming a net acceleration and its implications on the sliding of block B over A. Others seek clarification on how to approach the problem without knowing the acceleration.

Discussion Status

The discussion is ongoing, with participants exploring different interpretations of the problem. Some guidance has been offered regarding the need to consider equilibrium rather than acceleration, but there is no explicit consensus on the correct approach yet.

Contextual Notes

Participants express confusion about the topic and the calculations involved, indicating a potential lack of understanding of the underlying principles of static friction and equilibrium in the context of the problem.

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Homework Statement


Block b with mass 5kg rests on block a with mass 8kg which in turn is on a horizontal tabletop. there is no friction between block a and the table top but the coefficient of static friction between block a and b is 0.75.A light string attached to block a passes over a frictionless pulley and c is suspended from the other end. what is the largest mass the block c can have so that a and b still slide together when the system is released from rest

Homework Equations



w=mg
fs=usmg
f=ma

The Attempt at a Solution


I figured i needed to work out what force is required to move a and b together

w=mg
w=(5+8)x9.8
=127.4N

I then worked out the static force
fs=usmg
= 0.75x5x9.8
=36.75

I then decided that this was the maximum force that the two blocks could be moved with

There for for c we know acceleration = 9.8
F=ma
36.75=mx9.8

m=3.75kg

im not sure if this is right as i don't think my logic makes much sense
any help would be much appreciated thank you very much
 
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I then decided that this was the maximum force that the two blocks could be moved with

There for for c we know acceleration = 9.8
F=ma

That is where you've gone wrong. If there is a net acceleration, surely the block B would slide over A, right?
 
sorry I am not sure then how to correct my error
any hints would be greatly appreciated thanks
i am so totally confused with this whole topic
 
What you've been asked to find out, is the max. value of m so that the system is still in equilibrium.
What is the max. static frictional force acting on B?
 
i thought the max static force was 36.75N as calculated in the above post
but then how do i calculate the mass of block c if you do not no the acceleration.
sorry i must seem so dumb
 

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