Question

The pulleys in the diagram are all smooth and light. The acceleration of block A is $a$ upwards and the acceleration of block C is $f$ downwards. The acceleration of block B is

• A. $\frac{1}{2}(f-a)\text{upwards}$
• B. $\frac{1}{2}(a+f)\text{downwards}$
• C. $\frac{1}{2}(a+f)\text{upwards}$
• D. $\frac{1}{2}(a-f)\text{upwards}$

Answer 1

The correct option is A $\frac{1}{2}(f-a)\text{upwards}$

Let $a_B$ be the acceleration of block B.
Since the string is inextensible,
$l_1+l_2+l_3+l_4=L$ (where L is the total length of the rope)
Differentiating twice
$\Rightarrow \frac{d^2{l}_1}{d{t}^2}+\frac{d^2{l}_2}{dt}+\frac{d^2{l}_2}{dt}+\frac{d^2{l}_4}{dt}=0$
Assuming upward acceleration to be positive and downward negative.
$\Rightarrow (-a-0)+(-a_B-0)+(-a_B-0)+(f-0)=0$
$\Rightarrow f-a=2a_B$
$\Rightarrow a_B=\frac{(f-a)}{2}$
Since it is positive therefore upwards.