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 ($Takeg=10m{s}^{-2}$)

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

Answer 1

Answer: (D)

Let $\theta$ be the angle made by the rope with the vertical in equilibrium. The free body diagram of $5kg$ block is as shown in Fig. (b).

In equilibrium
$T_2$ $=5g=5\times 10=50N$
The free body diagram of the point P is as shown in Fig. (c).

In equilibrium
$T_1$ $sin\theta =50N$ $\dots \left(i\right)$
$T_1$ $cos\theta =T_2=50N$ $...\left(ii\right)$
Dividing (i) by (ii), we get

$tan\theta$ $=\frac{50}{50}=1$
$\theta$ $=ta{n}^{-1}$ $\left(1\right)=45^{\circ }$