Block B of mass `m_(B) = 0.5 kg` rests on block A, with mass `m_(A) = 1.5kg` which in turn is on a horizontal tabletop (as shown in figure) .The coefficient of kinetic friction between block A and the tabletop is `mu_(k) = 0.4` and the coefficient of static friction between block A and blockB is `mu_(s) = 0.6` A light string attached 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 `m_(c)` (in kg) that block C can have so that block A and B still slide together when the system is replaced from rest?

Answer

5

Explanation

Denote the common magnitude of the maximum acceleration as a For block A as remain at rest with repect to block B , `ale mu_(s)g` Let as assume `a = m_(s)g` for mass C in the largest The tension in the cord is then
`T = (m_(A) + m_(B)) a + mu_(k)g(m_(A) + m_(B))`
`= (m_(A) + m_(B)) (a + mu_(k)g)`
The tension is relted to the mass `m_(c)` (largest by)
`T = m_(C) (g - d)`
Solving `m_(c)` yiclds
`m_(c) = ((m_(A) + m_(B))(mu_(s) + mu_(k)))/(1 - mu_(s))`
`= ((0.5 + 0.5)(0.6 +0.4))/(1 - 0.6)= 5kg`