Question

In the system shown in figure, masses of the blocks are such that when system is released, the acceleration of pulley $P_{1}$ is $a$ upward and acceleration of block $A$ is $a_{1}$ upward. It is found that the acceleration of block $C$ is same as that of $A$ both in magnitude and direction $\left(a_{1} > a > a_{1} / 2\right)$. Then

• A. Acceleration of $B$ is $\left(2 a-a_{1}\right)$ upward
• B. Acceleration of $D$ is $\left(2 a+a_{1}\right)$ downward
• C. Acceleration of $B$ with respect to $C$ is downward
• D. Acceleration of $B$ with respect to $D$ is upward

Answer 1

Answer: (A), (B), (C), (D)

The acceleration of block $A$, w.r.t pulley $P_{1}$;
$a_{A P_{1}}=\left(a_{1}-a\right) \uparrow$
$a_{B P_{1}}=\left(a_{B}-a\right) \downarrow$
The acceleration of block $B$, w.r.t pulley $P_{1}$;
But, $\left|\vec{a}_{AP_{1}}\right|=-\left|\vec{a}_{B P_1}\right|$
$a_{1}-a \equiv a_{B}+a$
Hence, $a_{B}=2 a-a_{1} \uparrow$
The acceleration of block $C$ w.r.t pulley $P_{2}$;
$a_{C P_{2}}=\left(a_{c}+a\right) \uparrow$
The acceleration of block $D$ w.r.t pulley $P_{2}$;
$a_{D P_{2}}=\left(a_{D}-a\right) \downarrow$
$a_{C}+a=a_{D}-a$
$a_{D}=\left(2 a+a_{1}\right) \downarrow$
Hence, $\vec{a}_{B C}=\vec{a}_{B}-\vec{a}_{c}=2\left(a_{1}-a\right) \downarrow$
and $\vec{a}_{\beta D}=\vec{a}_{\Delta}-\vec{a}_{D}=4 a \uparrow$