I have attached the question and the correct option given in book is (c) I.e 2F/3mg. When solving firstly I considered the system as a whole and found the acceleration. Then, I made FBD of m and got the required answer. But when I tried to make FBD of 2m, I applied one horizontal force F(due to rod attached to pulley) and umg(due to friction), I got a different value of u. But when I applied only umg on 2m FBD, I got same answer as that I got in FBD of m. My question is which forces are acting on 2m FBD?
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Here was ur mistake. The force F is trying to pull the block of mass m so the friction opposes relative motion between them and stops it. When on drawing the FBD of 2m the friction force is the only force that causes acceleration in the bigger block. If u consider the 2 blocks as a system,then u must know frictional forces becomes internal and they exist in action reaction pairs
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