Question

A 1 kg mass is suspended from the ceiling by a rope of length 4m. A horizontal force 'F' is applied at the mid point of the rope so that the rope makes an angle of 45° with respect to the vertical axis as shown in figure. The magnitude of F is

• A. \( \frac{10}{\sqrt{2}} \, \text{N} \)
• B. \( 1 \, \text{N} \)
• C. \( \frac{1}{10 \times \sqrt{2}} \, \text{N} \)
• D. \( 10 \, \text{N} \)

Answer 1

The Correct Option is D

To resolve this issue, we must examine the forces acting on the system. A horizontal force \( F \) is applied at the rope's midpoint, causing it to form a \( 45^\circ \) angle with the vertical, as illustrated in the accompanying image.

1. The rope's tension can be separated into vertical and horizontal components. Let \( T_1 \) represent the tension in the rope.
2. The vertical component of \( T_1 \) must counteract the mass's weight:

\(T_1 \cos(45^\circ) = mg\)

Given \( m = 1 \, \text{kg} \) and \( g = 10 \, \text{m/s}^2 \), we have:

\(T_1 \times \frac{1}{\sqrt{2}} = 1 \times 10\)

\(T_1 = 10 \sqrt{2} \, \text{N}\)

1. The horizontal component of tension equals the force \( F \):

\(T_1 \sin(45^\circ) = F\)

Substituting the value of \( T_1 \):

\(10 \sqrt{2} \times \frac{1}{\sqrt{2}} = F\)

\(F = 10 \, \text{N}\)

1. Therefore, the magnitude of force \( F \) is \( 10 \, \text{N} \).

Consequently, the correct answer is \( 10 \, \text{N} \).