Question

A $50\text{kg}$ person stands on a $25\text{kg}$ platform. He pulls the rope which is attached to the platform through the frictionless pulleys as shown in the figure. If the platform moves upwards at a steady velocity, find the force with which man pulls the rope. Take $g=10{\text{m/s}}^2$.

• A. $325\text{N}$
• B. $175\text{N}$
• C. $250\text{N}$
• D. $750\text{N}$

Answer 1

The correct option is C $250\text{N}$

Let the tension in the rope be $T$ everywhere throughout the length of the string. So, the tension in the string connecting the platform to the adjacent pulley will be $2T$ as pulley and string are massless.

Steady velocity means, for the system of (platform+man):
$\overrightarrow{a}=\frac{\overrightarrow{dv}}{dt}=0...\left(i\right)$

From the FBD of system of (platform+man), applying the equilibrium condition in vertical direction as dictated by Eq. $\left(i\right)$:
$2T+T=(25+50)g$
$\therefore T=\frac{750}{3}=250\text{N}$

At the point where the man is holding the string, $T$ is acting upwards on him, so by Newton's $3^{rd}$ law, he must have pulled the rope downwards by a force, $T=250\text{N}$.