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

Answer: (D)

$10 \, \text{N}$

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).