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

Given m= 6kg
length = 2 m
force =50 N
g=10`ms^(-2)`
As shown in the diagram there are three forces acting on the midpoint p of the rope Suppose the rope makes an angle `theta` with the verticel in equilibrium resolving the forces horizontally and vertically we get
`T_(1) sin theta =T_(3) = 50 N`
`T_(1) cos theta T_(2) = 6kg wt = 60N`
Dividing (1) by (2) we get tan `theta `
`=(5)/(6) ` (or) `theta tan ^(-1) ((5)/(6))`
`= tan^(-1) (0.833)`
`=theta = 39 .8^(@)`