Question

A block of mass 5 kg is suspended by a mass less 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 $g = 10ms^{- 2}$)

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

Answer 1

Answer: (D)

$45^{o}$

Answer 2

Let $\theta$be the angle made by rope with the vertical in equilibrium

The free body diagram of 5 kg block is as shown in fig (2)

In equilibrium $T_{2} = 5g = 5 \times 10 = 50N$

The free body diagram of the point p is as shown in fig (3)

In equilibrium $T_{1}\sin\theta = 50N....(i)$$$T_{1}\cos\theta = T_{2} = 50N....(ii)$$

Dividing (i) by (ii) we get

$\tan\theta = \frac{50}{50} = 1$

$\theta = \tan^{- 1}{}(1) = 45{^\circ}$