ln the arrangement shown in figure, mA = mB = 2 kg. String is massless and pulley is frictionless. Block B is resting on a smooth horizontal surface, while friction coefficient between blocks A and B is μ = 0.5 The maximum horizontal force F that can be applied so that block.A does not slip over the block B is

Maximum value of friction on A, fmax = μmg = 0.5 × 2 × 10 = 10 NHence, the block B is subjected to this frictional force in the forward direction. Since the block A does not slip on B, both of them move with a common acceleration. The acceleration of B is the same as the acceleration of A. To determine acceleration of 'a', we find that mBamax = fmaxamax = 102 = 5 m/s2Fmax = (mA+mB)amax = 4 ×5 = 20 N