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 solve this problem, we need to analyze the forces acting on the system where a horizontal force \( F \) is applied at the midpoint of the rope, causing it to make a \( 45^\circ \) angle with the vertical. The setup is shown in the image above.

1. The tension in the rope can be resolved into two components: vertical and horizontal. Let \( T_1 \) be the tension in the rope.
2. The vertical component of tension \( T_1 \) must balance the weight of the mass:

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

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

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

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

1. The horizontal component of tension provides 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. Thus, the magnitude of the force \( F \) is \( 10 \, \text{N} \).

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

Answer 2

We are given that a mass of 1 kg is suspended by a rope, and a horizontal force F is applied at the midpoint of the rope, making an angle of 45° with the vertical.

Let \( T_1 \) be the tension in the rope at the point of application of the force and \( T_2 \) be the tension in the vertical section of the rope.

Step 1: Resolving Forces

1. The horizontal force \( F \) is balanced by the horizontal component of the tension \( T_1 \):
2. The vertical component of the tension \( T_1 \) balances the weight of the mass:

Step 2: Solving for \( F \)

From \( T_1 \cos 45^\circ = T_2 \), we have:

\[ T_1 \cos 45^\circ = 10 \, \text{N}. \]

Since \( \cos 45^\circ = \frac{1}{\sqrt{2}} \), we find:

\[ T_1 \times \frac{1}{\sqrt{2}} = 10 \quad \implies \quad T_1 = 10\sqrt{2}. \]

Now, substituting into the equation \( T_1 \sin 45^\circ = F \):

\[ 10\sqrt{2} \times \frac{1}{\sqrt{2}} = F \quad \implies \quad F = 10 \, \text{N}. \]

Thus, the magnitude of the force is 10 N, and the correct answer is Option (4).