Question

Two masses $5$ kg and $3$ kg are suspended from the ends of an unstretchable light string passing over a frictionless pulley. When the masses are released, the force on the pulley due to string connecting $5$ kg and $3$ kg body is:

• A. $30N$
• B. $75N$
• C. $15N$
• D. $60N$

Answer 1

The correct option is B $75N$

Lets do an Free body diagram of $m_1=5kg$

$T$ is acting upwards whereas m₁g is acting downwards$.$

$m_{1}g>T$ $(m_1>m_2)$

Free body diagram of $m_2=3kg$

$T$ is acting upwards whereas $m_{2}g$ acts downwards$.$

$m_{2}g<T$

m₁g-T=m₁a----------------$\left(i\right)$

T₂-m₂g=m₂a--------------$\left(ii\right)$

add $\left(i\right)$ and$(ii$)

$m_{1}g-m_{2}g=m_{1}a+m_{2}a$

$g(m_1-m_2)=a(m_1+m_2)$

$10\left(2\right)=a\left(8\right)$

$20=8a$

$20/8=a$

$a=2.5m/s^2$

For finding $T$

$T-m_{2}g=m_{2}a$

$T-30=7.5$

$T=37.5N$

Therefore Force on pulley by rope $=2\times Tension$

Force on pulley$=37.5N\times 2$

force$=75N$

Hence,

option $\left(B\right)$ is correct answer.