A mass of 6kg is suspended by a rope of length 2m from a celling a force of 50N in the horizontal direction is applied at the midpoint of the rope as shown in the figure what is the angle the rope makes with the vertical is equalibrium take `g=10 ms^(-2)` neglect 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^@`