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.

Answer 1

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

$a=\left(\frac{m-2\mu m}{m+2m}\right)g$ (iv)

Putting a in Eq. (i) , we get

$mg-T=M\left(\frac{M-2\mu m}{M+2m}\right)g\Rightarrow T=\frac{2mMg(1+\mu )}{(M+2m)}$

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 $S_{rel}L$

$\Rightarrow v=\sqrt{\frac{4gL(M-2\mu m)}{(M+2m)}}$

c. If blocks will not accelerate, then put a=0 in Eq. (iv) to get $M=2\mu m$.