See Fig. A mass of 6 kg is suspended by a rope of length 2 m from the ceiling. A force of 50 N in the horizontal direction is applied at the mid- point P of the rope, as shown. What is the angle the rope makes with the vertical in equilibrium ? (Take `g = 10 m s^(-2)` ). Neglect the mass of the rope.

Explanation

Figures (b) and (c) are known as free-body diagrams. Figure (b) is the free-body diagram of W and Fig. (c) is the free-body diagram of point P. Consider the equilibrium of the weight W.
Clearly `T_2 = 6 xx 10 = 60 N`
Consider the equilibrium of the point P under the action of three forces - the tensions `T_1` and `T_2`, and the horizontal force 50 N. The horizontal and vertical components of the resultant force must vanish separately :
`T_1 cos theta = T_2 = 60 N`
`T_1 sin theta = 50 N`
which gives that
`tan theta = 5/6 or theta = tan^(-1) (5/6) = 40^@`