In the diagram shown the table top is rough and coefficient of friction between any of the blocks and table top is μ. The blocks A and B are on the verge of slipping. If position of the blocks A and B are interchanged and the system is released from rest, then acceleration of the blocks is

Question

In the diagram shown the table top is rough and coefficient of friction between any of the blocks and table top is μ. The blocks A and B are on the verge of slipping. If position of the blocks A and B are interchanged and the system is released from rest, then acceleration of the blocks is

Infinity Learn · Accepted Answer

In the first case, $$mg=$$μ×Mg⇒$$m=$$μMIn the second case, resultant driving force on the blocks $$=Mg$$−μ$$mg=Mg(1$$−μ$$2)$$∴ Acceleration, $$a=Mg(1$$−μ$$2)(M+m)=Mg(1$$−μ$$2)M(1+$$μ$$)=g(1$$−μ$$)$$

In the diagram shown the table top is rough and coefficient of friction between any of the blocks and table top is $μ$ . The blocks A and B are on the verge of slipping. If position of the blocks A and B are interchanged and the system is released from rest, then acceleration of the blocks is

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