Question

A $50kg$ person stands on a $25$ kg platform. He pulls on the rope which is attached to the platform via the frictionless pulleys as shown in the figure. The whole system remains in equilibrium. The force applied by the person on the rope is $(g = 10m/{s^2})$

Answer 1

In physics, Tension is the pulling force, transmitted force, restoring force or the action-reaction force acting on the body when it is hanged by the means of a string, a cable, a chain, etc. It is denoted by “T” and it is measured in Newton. Use the formula of Tension, $T = mg + ma$. Where “m” is the mass, “g” is the gravitational force and “a” is the acceleration.

Complete step by step answer:
Let us consider Mass of the man be $M = 50kg$
And the mass of the platform be $m = 25kg$
Let the force with which the man pulls the rope is equal to the tension in the rope be $= T$
And therefore the tension in the string attached to the platform is $= 2T$
Normal reaction acts on the man in the upward direction whereas on the platform in the downward direction.
Forces acting on the platform will be –
$Mg + N = 2T{\text{ }}.....{\text{(a)}}$
Forces acting on the man will be –
$mg = N + T{\text{ }}.....{\text{(b)}}$
Add equations (a) and (b) and simplify
$Mg + N + mg = 2T{\text{ + N + T }}$
Same term with the same sign cancel each other, therefore remove “N” from both the sides of the equation.
$Mg + mg = 3T \\\ \therefore (M + m)g = 3T \\\$
Place the given values –
$3T = (50 + 25)g \\\ \implies 3T = 75 \times 10 \\\ \implies T = \dfrac{{750}}{3} \\\ \therefore T = 250N \\\$

Hence, the required answer - The force applied by the person on the rope is $250N$.

Note:
The same problem can also be solved by using another method. Consider the man and the platform as the single system. Therefore upward force will be equal to the downward force.
$T + 2T = (50 + 25)g \\\ \implies 3T = 75g \\\ \implies T = 25g \\\ \implies T = 25 \times 10 \\\ \therefore T = 250N \\\$