A mass of 3kg is suspended by a rope of length 2m from the ceiling.A force of 40 N in the horizontal direction is applied at midpoint P of the rope as shown. What is the angle the rope makes with the vertical in equilibrium and the tension in part of string attached to the celling ? (Neglect the mass of the rope, `g = 10 m//s^(2)`).

Explanation

In Equilibrium, `T_(2)=W=30 N`
Resolving the tension `T_(1)` into two mutually perpendicular components, we have
`T_(1)cos theta = T_(2)=30 N`
`T_(1)sin theta = 40 N`
`tan theta = (4)/(3) rArr theta = 53^(@)`
The tension in part of string attached to the ceiling
`T_(1)=sqrt(W^(2)+F^(2))`
`= sqrt(30^(@)+40^(@))N=50N`