Question

In the arrangement shown in $Fig.6.337$ 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)

$7M{s}^{-1}$ in upward direction

Given,

$v_{A\to P}=3m/s(\downarrow )$

Here, we are taking downward as positive.

Equating,

$l_1+l_2\times 2+l_3=$ constant

Differentiating w.r.t time,

$\frac{d{l}_1}{dt}+\frac{2d{L}_2}{dt}+\frac{d{l}_3}{dt}=0$

$\Rightarrow v_1+2v_2+v_3=0$

$\Rightarrow 12+2(-4)+v_3=0$

$\therefore v_3=-4{ms}^{-1}$

Also,

$V_{B\to P}=-V_{A\to P}$

$V_{B\to P}=-3$

$\Rightarrow V_B=-3+V_P$

$\Rightarrow V_B=-3-4$

$\therefore V_B=-7m{s}^{-1}$

$\therefore V_B=7m{s}^{-1}$ in upward direction