Question

Given the set up shown in the figure. Blocks $\text{A, B}$ and $\text{C}$ have masses $m_A=M$ & $m_B=m_C=m$. The strings are assumed massless and unstretchable and the pulley is frictionless. There is no friction between block $\text{B}$ and the support table, but there is friction between blocks $\text{B}$ and $\text{C}$ donated by a given coefficient $\mu$. Find the accelaration of block $\text{A}$.

• A. $(\frac{M+\mu m}{M-2m})g$
• B. $(\frac{M+2\mu m}{M-2m})g$
• C. $(\frac{M+2\mu m}{M+2m})g$
• D. $(\frac{M-2\mu m}{M+2m})g$

Answer 1

The correct option is D $(\frac{M-2\mu m}{M+2m})g$

Apply constraint equation on strings, i.e., length of strings is constant. Differentiate twice to get relation between acceleration. Let acceleration of blocks $\text{A, B}$ and $\text{C}$ be $a,b$ and $c$ respectively.


$l_1+l_2=$ constant

$\Rightarrow {\ddot{l}}_1+{\ddot{l}}_2=0$

Also, $l_3+l_4=constant$

$\Rightarrow {\ddot{l}}_3+{\ddot{l}}_4=0$

From which we get, $a=b=c$

Applying Newton's law on block $\text{A}$,

$Mg-T=Ma$ -------(1)

on block $\text{B}$,

$T-T_1-\mu mg=ma$ ------ (2)

on block $\text{C}$,

$T_1-\mu mg=ma$ ......... (3)

Adding eq(1) and (2) we get,

$T=2ma+2\mu mg=2m(a+\mu g)$

Substituting the value of $T$ in eq(1) we get,

$a=(\frac{M-2\mu m}{M+2m})g$

Hence , option (d) is the correct answer.