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 ${\mathrm{ms}}^{-2}$)m
 

Let $\mathrm{\theta }$ be the angle made by the rope with the vertical in equilibrium.

The free body diagram of 5 kg block is as shown in fie.(b)

In equilibrium
${\text{T}}_2 = 5 \mathrm{g} =$5 x l0 = 50 N
The free body diagram of the point P is as shown in Fig. (c).

In equilibrium

${\mathrm{T}}_1 \sin \mathrm{\theta } = 50 \mathrm{N}$--------(i)
${\mathrm{T}}_1 \cos \mathrm{\theta } = {\mathrm{T}}_2 = 50 \mathrm{N}$------(ii)
Dividing (i) by (ii), we get
$\tan \mathrm{\theta } = \frac{50}{50} = 1$

$\mathrm{\theta } = {\tan }^{-1}\left(1\right) = {45}^0$