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 ms-2)m

Let θ 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 equilibriumT2 = 5 g = 5 x l0 = 50 NThe free body diagram of the point P is as shown in Fig. (c).In equilibriumT1 sin θ = 50 N--------(i)T1 cos θ = T2 = 50 N------(ii)Dividing (i) by (ii), we gettan θ = 5050 = 1θ = tan-1(1) = 450