Question

A block of mass $5 \,kg$ is suspended by a massless rope of length $2\, m$ from the ceiling. A force of $50\, N$ 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 ($Take \,g=10 m\,s^{-2}$)

• 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 $5 \,kg$ block is as shown in Fig. (b).

In equilibrium
$T_{2}$ $=5 \,g=5\times10=50 \,N$
The free body diagram of the point P is as shown in Fig. (c).

In equilibrium
$T_{1}$ $sin\, \theta=50 N$ $\quad\ldots\left(i\right)$
$T_{1}$ $cos\, \theta=T_{2}=50\, N$ $...\left(ii\right)$
Dividing (i) by (ii), we get

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