Question

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

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

Answer 1

Answer: (D)

Using constraint theory,
$l_1+l_2+l_3=$ constant
$\Rightarrow v_1+2v_2+v_3=0$
Take downward as positive and upward as $-ve.$
So $12+2(-4)+v_3=0$
$v_3=$ velocity of pulley $P=-4m{s}^{-1}$
$=4m{s}^{-1}$ in upward direction
${\vec{v}}_{AP}=-{\vec{v}}_{BP}$
$\Rightarrow {\vec{v}}_{AP}=-\left({\vec{v}}_B-{\vec{v}}_P\right)$
${\vec{v}}_B={\vec{v}}_P-{\vec{v}}_{AP}=-4-(3)=-7m{s}^{-1}$
i.e., block $B$ will moves up with speed $7m{s}^{-1}$.