A mass of 6 kg is suspeded by a rope of length 2 m from a ceiling.A force of 60 N is applied in the horizontal direction at the mid-point of the rope. The angle made by the rope, with the vertical, in equilibrium position will he (Take g = 10 `ms^(-2)` , neglect the mass of the rope)

Options

A. `90^(@)`
B. `60^(@)`
C. `30^(@)`
D. `45^(@)`

Answer

D

Explanation

Since the mass is in equilibrium, therefore, the three forces acting on the mass A must be represented the three sides of a triangle taken in one order. Hence,
`(60)/(SA)=(6xx10)/(SB)rArr(SA)/(SB)=1` `therefore tantheta=1 rArrtheta=45^@`