Question

A pulley-block system as shown in the figure is released from rest. What is the magnitude of acceleration of $\text{COM}$ of pulley-block system? (Take $g=10{\text{m/s}}^2$)

• A. $7{\text{m/s}}^2$
• B. $5{\text{m/s}}^2$
• C. $3.77{\text{m/s}}^2$
• D. $10.5{\text{m/s}}^2$

Answer 1

The correct option is C $3.77{\text{m/s}}^2$


Tension $T$ is same throughout string length and tension in string connecting block $4\text{kg}$ will be $2T$


Applying string constraint relation:
$\Rightarrow l_1+l_2+l_3=$ constant
Differentiating twice w.r t time gives:
$-a-a+a^{\prime }=0$

Rate of decrease of length$=-ve$
Rate of increase of length$=+ve$
$\therefore a^{\prime }=2a....\left(i\right)$

From the FBD of blocks, we get the equation of dynamics in direction of acceleration as:


$9g-T=9\times 2a....\left(ii\right)$

$2T-4g=4a....\left(iii\right)$
Solving the equations $\left(ii\right)$ and $\left(iii\right)$ we get:
$a=\frac{140}{40}=3.5{\text{m/s}}^2$.
Hence acceleration of $9\text{kg}$ is $2a=2\times 3.5=7{\text{m/s}}^2$

Applying formula for $\text{COM}$ of system:
$a_{\text{CM}}=\frac{m_1{a}_1+m_2{a}_2}{m_1+m_2}$
Considering the downward direction as $+ve$:
$a_2=+7{\text{m/s}}^2$
$a_1=-3.5{\text{m/s}}^2$
$\Rightarrow a_{\text{CM}}=\frac{-(4\times 3.5)+(9\times 7)}{4+9}$
$\therefore a_{\text{CM}}=\frac{49}{13}=+3.77{\text{m/s}}^2$
Hence $\text{COM}$ of system will move downwards with magnitude of acceleration as $3.77{\text{m/s}}^2$.