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Force and static friction

  • Thread starter e_burke
  • Start date
1. 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


2. Homework Equations

w=mg
fs=usmg
f=ma

3. 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 dont think my logic makes much sense
any help would be much appreciated thank you very much
 
133
0
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 im not sure then how to correct my error
any hints would be greatly appreciated thanks
i am so totally confused with this whole topic
 
133
0
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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