Question

An arrangement of pulley, string and blocks is shown in figure. Given $m_1=4\text{kg},m_2=2\text{kg}$ and $m_3=4\text{kg}$. All surfaces are frictionless, string and pulleys are massless. Find the acceleration of block $m_1$. Assume $g=10{\text{m/s}}^2$.

• A. $\frac{36}{11}{\text{m/s}}^2$
• B. $\frac{40}{11}{\text{m/s}}^2$
• C. $\frac{36}{7}{\text{m/s}}^2$
• D. $\frac{40}{7}{\text{m/s}}^2$

Answer 1

The correct option is D $\frac{40}{7}{\text{m/s}}^2$

Assuming the acceleration of pulley and blocks, as shown in the figure.
.
Applying string contraint :
$l_1+l_2=$constant
On double differentiating with time gives:
$-a_0+a_p=0\Rightarrow a_p=a_0...\left(i\right)$
($-ve$ and $+ve$ sign are used for decrease and increase of length of string respectively.)

Similarly,
$l_3+l_4=$ constant
Double differentiate w.r.t time
$\Rightarrow +(a_1-a_p)+(a_2-a_p)=0$
Or, $a_1+a_2-2a_p=0$
$\therefore 2a_p=2a_0=a_1+a_2....\left(ii\right)$

From the FBD of blocks and massless pulley:


Equation of dynamics in direction of acceleration;
$2T=m_1{a}_0$
$\Rightarrow 2T=4a_0...\left(iii\right)$


$m_{2}g-T=m_2{a}_1$
$\Rightarrow 20-T=2a_1...\left(iv\right)$
and $m_{3}g-T=m_3{a}_2$
$\Rightarrow 40-T=4a_2...\left(v\right)$

($2\times$Eq.$iv$) $+$ Eq.$v$ gives:
$80-3T=4(a_1+a_2)$
Putting $a_0$ from Eq. $\left(ii\right)$ $80-3T=4\left(2a_0\right)=8a_0...\left(vi\right)$

Adding $2\times$Eq. $\left(vi\right)$ and $3\times$ Eq. $\left(iii\right)$ gives:
$160=28a_0$
$\therefore a_0=\frac{40}{7}{\text{m/s}}^2$