See Fig . A mass of 6 kg is suspended by a rope of length 2 m from the celling. A force of 50 N in the horizontal direction is applied at the mid point P of the rope , as shoe . What is the angle the rope makes with the vertical in equilibrium ? (Take `g = 10 ms^(-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 ig . (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) co theta = T_(2) = 60 N `
`T_(2) sin theta = T_(2) = 50 N`
which gives that tan `theta = 5/6 ` or
` theta = tan^(-1) (5/6) = 40^(@)`
Note the answer does not depend on the length of the rope (assumed massless ) nor on the point at which the horizontal force is applied .