Question

A block of mass $5kg$ is suspended by a massless rope of length $2m$ from the ceiling. A force of $50N$ is applied in the horizontal direction at the midpoint $P$ of the rope, as shown in the figure.
The angle made by the rope with the vertical in equilibrium is (Take $g=10m{s}^{-2}$)

• A. ${30}^0$
• B. ${40}^0$
• C. ${60}^0$
• D. ${45}^0$

Answer 1

The correct option is D ${45}^0$

Let $\theta$ be the angle made by the rope with the vertical in equilibrium. Let $T_1$ and $T_2$ be the tension in the rope as shown in figure.

The free body diagram of $5kg$ block is as shown in fig (b).
In equilibrium, $\Rightarrow T_2=5g=5\times 10=50N$

The free body diagram of the point $P$ is as shown in fig (c).
In equilibrium, $\Rightarrow T_{1}sin\left(\theta \right)=50N$ ..............1
$\Rightarrow T_{1}cos\left(\theta \right)=T_2=50N$ ..............2

Dividing equation 1 by 2, we get
$tan\left(\theta \right)=\frac{50}{50}=1$
$\Rightarrow \theta =ta{n}^{-1}={45}^0$