A mass of 3kg is suspended by a rope of length 2m from the ceiling. A force of 40 N in the horizonal 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 ceiling? (Neglect the mass of the rope, g=10m//s^(2))

A mass of 3kg is suspended by a rope of length 2m from the ceiling. A

1.	A mass of 3kg is suspended by a rope of length 2m from the ceiling. A force of 40 N in the horizonal 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 ceiling? (Neglect the mass of the rope, g=10m//s^(2))
Answer» Solution :Resolving the tension `T_(1)` into two mutually perpendicular components, we have `T_(1)cos theta=W=30N""T_(1)sin theta=40N` `THEREFORE tan theta=(4)/(3)(or)theta=tan^(-1)((4)/(3))=53^(@)` The tension in part of STRING attached to the ceiling `T_(1)=sqrt(W^(2)+F^(2))=sqrt(30^(2)+40^(2))=50N`