Question

A mass of $10kg$ is suspended by a rope of length $4m$, from the ceiling. A force $F$ is applied horizontally at the mid-point of the rope such that the top half of the rope makes an angle of $45°$ with the vertical. Then $F$ equals (Take $g=10m{s}^{-2}$ and rope to be mass less).

• A. $100N$
• B. $90N$
• C. $75N$
• D. $70N$

Answer 1

The correct option is A

$100N$

Explanation for correct option:

Step 1: Given data

Mass, $m=10kg$

Rope of length, $l=4m$

Angle, $\theta =45°$

Force, $F$

Step 2: Calculation

The horizontal component of force at point $P$,

As system is in equilibrium,

$T_1\sin 45°-F=0$

$$\begin{gathered} T_1\sin 45°=F \\ \Rightarrow \frac{T_1}{\sqrt{2}}=F \end{gathered}$$ $\_\_\_\_\left(1\right)$ [$\sin 45°$is horizontal component]

The vertical component of forces at point $P$,

$F-T_1\cos 45°=0$

$$\begin{gathered} T_1\cos 45°=100N \\ \Rightarrow \frac{T_1}{\sqrt{2}}=100N \end{gathered}$$ $\_\_\_\_\left(2\right)$ [$\cos 45°$ is vertical component]

Equating equation $\left(1\right)$ and $\left(2\right)$, we get

$F=100N$

Hence, option (A) is the correct answer.