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Question: A weight \[\omega \;\] is suspended from the midpoint of a rope, whose ends are at the same level. I...

A weight ω  \omega \; is suspended from the midpoint of a rope, whose ends are at the same level. In order to make the rope perfectly horizontal, the force applied to each of its ends must be
(A) Less than ω  \omega \;
(B) Equal to ω  \omega \;
(C) Equal to 2ω  2\omega \;
(D) Infinitely large

Explanation

Solution

We draw the free body diagram of the following system to solve this question. We find the sum of forces acting upwards and downwards and equate them since there is horizontal force. Both the horizontal components of the two sides of the string get cancelled. So, the weight acting downwards is equal to the sum of upward forces.
The angle between the string and the x-axis should be zero for the body to be horizontal. Substituting this condition in the final equation we get the answer.

Complete step by step solution: The free body diagram of the system is

The weight is at the centre of the string so the tension is distributed equally on both sides of the string. The tension force on both the sides of the string is resolved into
T=Tcosθ+TsinθT = Tcos\theta + T\sin \theta
As shown in the diagram
The horizontal components are equal and opposite in direction so they get cancelled.
TcosθTcosθ=0Tcos\theta - Tcos\theta = 0
Since the body is at equilibrium the sum of upwards forces is equal to downwards force (weight of the body)
Forces acting upwards are the vertical components of the tension force
Tsinθ+Tsinθ=2TsinθT\sin \theta + T\sin \theta = 2T\sin \theta
Weight of the body is ω \omega \
Upward force is equal to downwards force
ω =2Tsinθ\omega \ = 2T\sin \theta
For the rope to be perfectly horizontal θ\theta should be 00^\circ
Substituting θ=0\theta = 0^\circ
ω =2Tsinθ\omega \ = 2T\sin \theta
ω =2Tsin0\omega \ = 2T\sin 0^\circ
T=ω 2sin0=ω 0T = \dfrac{{\omega \ }}{{2\sin 0^\circ }} = \dfrac{{\omega \ }}{0}
Anything by zero is infinite
Hence the force applied on each end should be infinitely large.

Option (D) infinitely large is the correct answer.

Note: The angle θ\theta is also equal to the angle between the tension force and the Horizontal which is the reason we can resolve the tension force with angle θ\theta
Students might go wrong by taking tensions in the two sides of the string with different values. The tension remains the same on both the sides because the weight is being suspended from the center.