Three-mass System with Friction

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The discussion revolves around calculating the maximum mass of block C that allows blocks A and B to slide together without block B moving independently. The coefficient of static friction between blocks A and B is 0.750, and the user attempts to find the acceleration and forces involved. Initial calculations led to confusion regarding the correct mass to use in the equations, resulting in incorrect values for block C's mass. It was clarified that the maximum force block A can exert on block B is 0.75 times the gravitational force, which affects the overall acceleration of the system. The correct approach involves keeping gravitational acceleration in the equations to derive accurate results, ultimately leading to the conclusion that the maximum mass for block C should be 39 kg.
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Homework Statement

Block B, with mass 5.00 kg, rests on block A, with mass 8.00 kg, which in turn is on a horizontal tabletop. There is no friction between block A and the tabletop, but the coefficient of static friction between block A and block B is 0.750. A light string attached to block A passes over a frictionless, massless pulley, and block C is suspended from the other end of the string.

What is the largest mass that block C can have so that blocks A and B still slide together when the system is released from rest?

Homework Equations



F=ma

The Attempt at a Solution



I can't figure this out for the life of me. I took blocks A and B as a system in themselves to find the acceleration for which the block B would not move.

I set Fapplied-.75(Mb*9.81)=36.79N.

I then took the force applied and set it as F=ma, 36.79=(Ma+Mb)a, a=2.83 m/s^2.

This should be the most acceleration that block A can experience without having block B move. I used both Ma+Mb in the equation as it appears that the hanging block would need to move the two as a system.

I then took the acceleration I had found and set a new force equation, utilizing the entire system:

Fnet=Mc*g, then substitute in F=ma, Msys*a=Mc*g, (Ma+Mb+Mc)a=Mc*g.

I solved for Mc and got a negative number, which obviously cannot be true.

I'd appreciate any help on how I should go about solving this...
 
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The 36.79 N is the maximum force block A can exercise on block B. So a is a little bigger. To accelerate block B with 2.83 m/s2, a force of 14.15 N is sufficient.
Better leave the g in there until you have a final expression. It cancels out.
 
Also want to buy new solving equipment, because (Ma+Mb+Mc) * a=Mc*g solves easily : Mc = (Ma + Mb)*a/(g-a). Positive for all a up to g (Mc can't get higher then)
 
BvU said:
The 36.79 N is the maximum force block A can exercise on block B. So a is a little bigger. To accelerate block B with 2.83 m/s2, a force of 14.15 N is sufficient.
Better leave the g in there until you have a final expression. It cancels out.

First off, thanks for the assistance.

So there is an error in the maximum acceleration I found? Should I have divided by only the mass of block A rather than the mass of A + B? If so, I get a=4.6m/s^2.

I found an error I made when inputting the equation for Mc in the calculator, thanks for pointing out my mistake. However, after correcting, with the new acceleration of 4.6 m/s^2 I still receive an incorrect answer of 10.31 kg. The correct answer should be 39 kg. Which equation is incorrect?
 
Eqns are OK. But filling in the mass of A is problematic. A can exercise a maximum force of 0.75 g on B, so B can be accelerated 0.75 g maximum by A. So A + B can be accelerated by 0.75 g maximum, etc.
Remember my tip on leaving g in ? That way you get the exact answer (which, granted, you would also have gotten if you consistently use the same value for g...)
 
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