Friction Part 4

A block A of mass M is placed on horizontal table. Another block B of mass m is places on block A as shown in figure. Now F force is applied on A. The conditions are explained below.

1. If there is no friction between A and B and between A and table, then A will move with acceleration aA = . There is no frictional force to move B on A, so B remain rest.

In this case B moves with acceleration aA with respect to A in the opposite direction of F.

If L is the length of A, then B takes t time to fall from A, then L =  =    

t = .

2. Friction presents in between A and B only then both will move with acceleration a = . The force on block B is F/ = ma = .  The coefficient of static friction between A and B is . The limiting friction between A and B is f = mg. B will not slide on A till F/ < f i.e. F/ < mg  or,  < mg or, F < g(M+m). Therefore, if F < g(M+m), then both the bodies move with same acceleration aA = aB = .

If F > g(M+m), then both the bodies move with different acceleration.

The net force on A is F – f = F – mg. The acceleration of A is aA = .

The net force on B is f = mg. The acceleration of B is aB =  = g.

The acceleration of B with respect to A is aBA = aB – aA = g – = – []

Negative sign indicates that B accelerates backward with respect to A. If L is the length of A, then B takes t time to fall from A.

L =

 t =  = .

3. The coefficient of static friction between A and B is . Friction present in between block A and table. The coefficient of static friction between A and table is . The limiting friction between A and B is f = mg. The limiting friction between A and table is f/ = (M+m)g. Block A will move if F > f/. The acceleration of A is  =  = .

All the cases of (1) and (2) are valid with resultant force F – (M+m)g.

If F < f/, then the system remain rest. The friction between A and B is zero and the friction between A and table is F.

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