Question

A massless string passing over a friction less pulley and carries two masses $m_1\&m_2$ $\left(m_1>m_2\right)$ at its ends. If $g$ is the acceleration due to gravity then the thrust on the pulley is $\left(m_1>m_2\right)$

• A. $\left(\frac{4m_1{m}_2}{m_1+m_2}\right)g$
• B. Zero
• C. $\left(\frac{2m_1{m}_2}{m_1+m_2}\right)g$
• D. $\left(\frac{2m_1{m}_2}{m_1{m}_2}\right)g$

Answer 1

Answer: (A)

Equations of motion are
$m_{1}g-T=m_{1}a\dots (i)$
$T-m_{2}g=m_{2}a\dots (ii)$
From eq (i) $T=-m_{1}a+m_{1}g$
$-m_{1}a+m_{1}g-m_{2}g=m_{2}a$
$m_{1}g-m_{1}a=m_{2}a+m_{2}g$
$g\left(m_1-m_2\right)=a\left(m_1+m_2\right)$
Putting the value of $a$ in eq (i) we get
$m_{1}g-T=\frac{m_1\left(m_1-m_2\right)g}{\left(m_1+m_2\right)}$
$m_{1}g-T=\frac{\left(m_1^2-m_1{m}_2\right)g}{\left(m_1+m_2\right)}$
$\Rightarrow \left(m_1+m_2\right)\left(m_{1}g-T\right)$
$=\left(m_1^2-m_1{m}_2\right)g$
$\therefore m_1^{2}g-m_{1}T+m_1{m}_{2}g-m_{2}T=m_1^{2}g-m_1{m}_{2}g$
$m_1^{2}g-m_{1}T+m_2{m}_{1}g-m_{2}T=m_1^{2}g+m_1{m}_{2}g=0$
$-T\left(m_1+m_2\right)+2m_1{m}_{2}g=0$
$\frac{2m_1{m}_2}{m_1+m_2}g=T$
Thrust on the pulley is $2T$
$\therefore$ Thrust is equal to
$\frac{4m_1{m}_{2}g}{m_1+m_2}$