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Question

Given the setup shown in Fig. Block A, B, and C have masses $$m_A=M$$ and $$m_B= m_C=m$$. The strings are assumed massless and unstretchable, and the pulleys frictionless. There is no friction between blocks B and the support table, but there is friction between blocks B and C, denoted by a given coefficient $$\mu$$.
a. In terms of the given, find (i) the acceleration of block A, and (ii) the tension in the string connecting A and B.
b. Suppose the system is related from rest with block C. near the right end of block B as shown in the above figure. If the length L of block B is given, what is the speed of block C as it reaches the lift end of block B? Treat the size of C as small.
c. If the mass of block A is less than some critical value, the blocks will not accelerate when released from rest. Write down an expression for that critical mass. 
983143_47dd3085dcb84729b8fb7d8840267126.png


Solution

Apply constraint equation on strings, the length of strings is constant. Differentiate twice to get relation between of acceleration of block A, B, and C be a, b, and c, respectively.
 $$l_1 +l_2$$ = constant
and $$l_3 + l_4 $$ constant
$$l_1+ l_2 = 0 \Rightarrow |b| = |c|$$
$$l_3 + l_4 = 0 \Rightarrow |a| = |b|$$
From which we get a=b=c.
From FBDs Of A, B, and C   [Fig. (a)],
Writing equations of motion for block A:
$$mg - T = Ma$$                        (i)
For block B, $$T - T_1 -\mu mg = ma$$              (ii)
For block C, $$T- \mu mg = ma $$                       (iii)
Solving equations (i), (ii) and (iii), we get
$$a= (\dfrac{m-2\mu m}{m+2m})g$$                       (iv)
Putting a in Eq. (i) , we get 
$$ mg -T = M (\frac{M-2\mu m}{M+2m}) g \Rightarrow T = \frac{2mMg(1+\mu)}{(M+2m)}$$
b. As there is relative motion between blocks, we apply 
$$V_{rel}^2 = V_{rel}^2 +2a_{rel} S_{rel}$$
If system is released from rest, $$u_{rel} =0$$
$$v_{rel}^2 = 2a_{rel} S_{rel} \Rightarrow v_{rel} =\sqrt{2a_{rel} S_{rel}}$$
$$a_{rel} = 2a $$ and $$S_{rel} L$$
$$\Rightarrow v=\sqrt{\dfrac{4gL(M-2\mu m)}{(M+2m)}}$$
c. If blocks will not accelerate, then put a=0 in Eq. (iv) to get $$M=2\mu m$$.

1029119_983143_ans_d9bc28c9e74546529c81efdc5a567bd3.png

Physics

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