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Last updated date: 04th Dec 2023
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MVSAT Dec 2023

A rope of mass 5 Kg is moving vertically with an upward force of 100 N acting at the upper end and a downward force of 70 N acting at the lower end. The tension at the mid-point of rope is.

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Hint: In the question it is given that the rope with mass is moving upwards by action of force. We can analyse that there will be gravitational force or weight will be acting along with given external force. As the rope is moving upwards that force will be greater in action.
We can use formula $ {F_1} = {F_2} + mg + ma$

Complete answer:
In this question we are given with
Mass of the rope is 5 Kg
The force acting in upward direction is 100N
The force in downward direction is 70N
It is known from the law of motion that all the forces must balance each other in order to attain equilibrium.
Using the formula in accordance with the law of motion we get,
{F_1} = {F_2} + mg + ma \\
   \Rightarrow 100 = 70 + (5 \times 10) + 5 \times a \\
   \Rightarrow 30 - 50 = 5 \times a \\
From the above equations we will get the solution as
$ a = 4m/{s^2}$
It is in the downward direction
But, we are asked to find the tension at the midpoint.
Let, the mass of the half rope be 2.5 Kg.
Using the laws of motion we have
  {F_1} = {F_2} + mg + ma \\
   \Rightarrow T = 70 + 2.5g - 2.5 \times 4 \\
   \Rightarrow T = 85N \\
So, we get the tension at the midpoint of rope as 85N.

Note: A string or rope is often idealized as a single dimension, having length but being massless with no cross section or also known as zero cross section. If there are no bends in the string, as occur with pulleys. The tension is a constant along the string, equal to the magnitude of the forces applied by the ends of the string.
Every force has an equal and opposite reaction.