Block B, of mass $$mB=0.5kg$$, rests on block A, with mass $$mA=1.5kg$$, which in turn is on a horizontal tabletop (as$$ shown in $$figure). The coefficient of kinetic friction between block A and the tabletop is μ$$k=0.4$$ and the coefficient of static friction between block A and block B is μ$$S=0.6$$. 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 (in$$ $$kg) that block C can have so that blocks A and B still slide together when the system is released from rest?

Question

Block B, of mass $$mB=0.5kg$$, rests on block A, with mass $$mA=1.5kg$$, which in turn is on a horizontal tabletop (as$$ shown in $$figure). The coefficient of kinetic friction between block A and the tabletop is μ$$k=0.4$$ and the coefficient of static friction between block A and block B is μ$$S=0.6$$. 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 (in$$ $$kg) that block C can have so that blocks A and B still slide together when the system is released from rest?

Infinity Learn · Accepted Answer

For block A to remain at rest with respect to block B, a≤μsg. Let us assume $$a=msg$$ for mass of C to be largest. The tension in the cord is $$thenT=mA+mBa+$$μ$$kgmA+mB=mA+mBa+$$μkgThis tension is related to the mass mC (largest) $$byT=mC(g$$−$$a)Solving$$ for mC $$yieldsmC=mA+mB$$μ$$s+$$μ$$k1$$−μ$$s=(1.5+0.5)(0.6+0.4)1$$−$$0.6=5kg$$

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