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Block $M_a$ lies on top of block $M_b$. Initially at rest, the two blocks are pulled by a mass-less string passing over a pulley.

The pulley is accelerated at rate $A$. Block $M_b$ slides friction-less on the table surface while there is a constant friction force $f$ between the two blocks.

Express the tension on the string connecting the two blocks.

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My friends told me that the answer is $T = \frac{(M_a+M_b)A}{2}$, but I don't get it. My friends just ignores the friction between the two objects and treats them as a whole object. Also, I'm wondering if the tension for $M_a, M_b$ is the same when friction between two objects exists. (Not with the pulley of course)

What is the correct solution? I have no idea.

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closed as off-topic by John Rennie, sammy gerbil, Yashas, Kyle Kanos, Rory Alsop Mar 30 '17 at 14:03

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I think your friend (and Utkarsh also) is assuming that f is large enough that the blocks do not move relative to each other. In which case you can treat the blocks as being nailed together to form one single block with mass $M_a + M_b$ which is accelerated by a force 2T, and the solution follows.

However the question does not actually state that the blocks do not move relative to each other. So a better (i.e. more general) solution would be to assume that the upper block can slide across the lower block, and then derive an expression for T as a function of A, $M_a$, $M_b$ and f. Note that in this case the accelerations of the two blocks are not equal.

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  • $\begingroup$ How would I set the equation if the blocks can slide across each other? $\endgroup$ – zxcvber Mar 29 '17 at 23:33
  • $\begingroup$ Write down the forces acting on each block. (Note that the tension in the string, T, has the same value for each block and acts in the same direction, whereas the frictional force f has the same value but acts in opposite directions on the two blocks.) Then use F=ma to write down the equation of motion for each block. You don't know the accelerations of the blocks at this point, but you do know that the sum of the two accelerations is 2A . So you can eliminate the unknown accelerations and find one equation in T, Ma, Mb and f. Re-arrange this equation to find an expression for T. $\endgroup$ – gandalf61 Mar 30 '17 at 7:54
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Of course tension is same due to constraint ,hence we get 2 equation from free body diagram MaA = T - f And MbA = T + f. (You can take f in any direction)

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