static friction and pulley question

eagleye7

Hey all,
I was given this problem today by a lecturer as a way of helping us understand the section we just covered and i was wondering if anyone could help me with it.
we wont go over it again in a lecture as its a diy style thing but i thought maybe someone on here might find it interesting or may have seen it before and could give mae a hand ive pondered for a while myself and think i have an answer. any help is greatly appreciated hope it puzzles you as much as it does me!


Test question:
Block B, with mass mB, rests on blockA with mass mA, which in turn is on a horizontal tabletop.
The coefficient of kinetic friction between block A and the tabletop is μk, and the coefficient of static friction between block A and block B is μs. 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 mc that block C can have so that blocks A and Bstill slide together when the system is released from rest?

the diagram is attached.

eagleye7

The answer that i came up with is

mc= us.mb + uk.ma

for the max mc

im not 100% confident this is right as it was acquired through a bit of guess work.

Delphi91

Not entirely sure myself either but I "think" that:

m_c = mu_s * m_b

Sorry about the lack of scientfic notation - haven't figured out how to do it on here yet

Professor_Fink

Not sure how the others managed to get their answers, but they both seem to be wrong. Consider what happens when the force on B is small, so that B remains static on top of A. The acceleration of the combined mass A+B is given by [a_AB = m_C*g - u_k*(m_A + m_B)*g]/(m_A + m_B)

As long as this is smaller than u_s*g, mass B will remain static. Thus we have
u_s*g < [a_AB = m_C*g - u_k*(m_A + m_B)*g]/(m_A + m_B)

Rearranging this we get
m_C < (u_s + u_k)*(m_A + m_B)

eagleye7

Thats a brilliant way to think of it it makes a lot more sense to me now thanks Frinky :D