- #1

Buffu

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## Homework Statement

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Question :- Find the acceleration of block of mass ##M##. The coefficient of friction between blocks is ##\mu_1## and between block and ground is ##\mu_2##.

free body diagram at the end.

Variables :-

##f_1## - friction between blocks.

##f_2## - friction between block and ground.

##N_2## - Normal reaction between blocks.

##N_1## - Normal reaction between block and ground.

##T## - Tension in the string.

##a## - acceleration of the blocks since string is in-extendable.

##M < m##

## Homework Equations

Since small block will move down, and big one right.

$$ma = mg - (f_1 + T) \qquad (1)$$

$$T - (N_2 + f_2) = Ma \qquad (2)$$

$$N = Mg + f_1 \qquad (3)$$

If big block moves right then, pulleys will move right and small block will move right. Thus big block is non inertial frame. So we add the adequate pseudoforces to balance out unbalanced forces.

##N_2 = ma \qquad (4)##

## The Attempt at a Solution

##(1) + (2)##

$$a = mg - f_1 - N_2 - f_2 $$

$$a = mg - N_2 - ( f_2 + f_1 )$$

$$a = mg - N_2 - ( \mu_1 N_2 + N_1 \mu_2 )$$

substituting ##(3)## and ##(4)##

$$a(m + M) = mg - ma-(ma\mu_1 + \mu_2(Mg + f_1))$$

$$a(2m + M + \mu_1) = mg - \mu_2(Mg + ma\mu_1)$$

$$a(m(2 + \mu_1 + \mu_1 \mu_2) + M) = g(m - \mu_2 M)$$

$$a = {g(m - \mu_2 M)\over (m(2 + \mu_1 + \mu_1 \mu_2) + M)}$$

But the given answer is ##a = {[2m - \mu_2(M + m)]g\over M + m[5 + 2(\mu_1 + \mu_2)]}##.

I think the given solution is incorrect, because i can't find any error in my calculations but i am not sure.

Please check my free body diagrams in the picture below and my equations. I think any error in those would be sufficient for me to proceed ahead.

My kindest Thanks for your time and help.

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