Question

A mass of 6 kg is suspended by a rope of length 2m from a celling. A force of 50N is applied in horizontal direction at the mid point of the rope. What is the angle of the rope with the verticle in equilibrium:-

A. `tan^(-1)((4)/(5))`
B. `tan^(-1)((5)/(4))`
C. `tan^(-1)((5)/(6))`
D. None of these

Answer 1

Correct Answer - C
`T cos theta=60`
`T sin theta=50`
`therefore tan theta=(5)/(6)`