Question

In the arrangement shown in the figure at a particular instant, the roller is coming down with a speed of $12\, m s ^{-1}$ and $C$ is moving up with $4\, ms ^{-1}$. At the same instant, it is also known that w.r.t. pulley $P$, block $A$ is moving down with speed $3\, m s ^{-1}$. Determine the motion of block $B$ (velocity) w.r.t. ground.

• A. $4\, m s ^{-1}$ in downward direction
• B. $3\, ms ^{-1}$ in upward direction
• C. $7\, ms ^{-1}$ in downward direction
• D. $7\, ms ^{-1}$ in upward direction

Answer 1

Answer: (D)

Using constraint theory,
$l_{1}+l_{2}+l_{3}=$ constant
$\Rightarrow v_{1}+2 v_{2}+v_{3}=0$
Take downward as positive and upward as $-ve.$
So $12+2(-4)+v_{3}=0$
$v_{3}=$ velocity of pulley $P=-4\, m s ^{-1}$
$=4\, m s ^{-1}$ in upward direction
$\vec{v}_{A P}=-\vec{v}_{B P}$
$\Rightarrow \vec{v}_{A P}=-\left(\vec{v}_{B}-\vec{v}_{P}\right)$
$\vec{v}_{B}=\vec{v}_{P}-\vec{v}_{A P}=-4-(3)=-7\, ms ^{-1}$
i.e., block $B$ will moves up with speed $7\, ms ^{-1}$.