Question

In the arrangement shown in figure pulley and strings are ideal. End $A$ of string connected to pulley $P_1$ is moved upwards with acceleration $a_A=2m/s^2$ while end $B$ of another string shown in figure is moved up with acceleration $a_B=1m/s^2$. Block $C$ of mass $1kg$ is moving up with acceleration $1m{s}^2$. If block $D$ to which string are connected symmetrically moves such that its orientations remains same, then choose the correct statements. (assume $g=10m/s^2)$

• A. Acceleration of block $D$ is $\frac{1}{2}$ $m/s^2$ upwards
• B. End $A$ is pulled with force of $22N$
• C. Mass of block $B$ is $\frac{1}{2}kg$
• D. acceleration of block $B$ is $\frac{7}{3}m{s}^{-2}$ upwards

Answer 1

Answer: (B), (D)

The correct options are
B End $A$ is pulled with force of $22N$
D acceleration of block $B$ is $\frac{7}{3}m{s}^{-2}$ upwards



Using constraints of lengths in the diagram
${\stackrel{..}{l}}_1+{\stackrel{..}{l}}_2+{\stackrel{..}{l}}_4=0$
here
${\stackrel{..}{l}}_1=a_b-a_{p_3}=1-1=0(\because a_{p_3}=a_c=1m{s}^{-2})$
${\stackrel{..}{l}}_2=(a_{p_2}-a_{p_3})=(a_{p_2}-1)$
${\stackrel{..}{l}}_3=(a_{p_2}-a_D)$

$\Rightarrow 0+(a_{p_2}-1)+(a_{p_2}-a_D)=0\Rightarrow a_D=2a_{p2}-1............\left(1\right){\stackrel{..}{l}}_3+{\stackrel{..}{l}}_5=0$
here
${\stackrel{..}{l}}_3=a_A-a_{p2}=2-a_{p2}$
${\stackrel{..}{l}}_5=a_A-a_D=2-a_D$
$\Rightarrow 2-a_{p2}+(2-a_D)=04-a_D-a_{p2}=0\text{Substituting from}\left(1\right)4-2a_{p2}+1-a_{p2}=0a_{p2}=\frac{5}{3}{a}_D=\frac{10}{3}-1=\frac{7}{3}m{s}^{-2}$

Acceleration of block $B=a_D=\frac{7}{3}m{s}^{-2}$
For C,
$2T-m_{c}g=m_c{a}_c$
$2T-1\left(10\right)=1\left(1\right)$
$T=\frac{11}{2}N$

Tension at end A is
$4T=22N.\text{Also,}T=T_1,$
Since orientation of D should not alter.

FBD of B


$2T-T-m_{B}g=m_B{a}_{B}2T-T-m_{B}g=m_B\left(\frac{7}{3}\right)\frac{37m_B}{3}=\frac{11}{2}{m}_B=\frac{33}{74}kg$