A mass of 6kg is suspended by a rope of length 2m from a ceiling. A force of 60N 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 be (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 by the three sides of a triangle taken in one order. Hence,
`(60)/(SA)= (6 xx 10)/(SB) rArr (SA)/(SB)=1 because tan theta =1 rArr theta = 45^(@)`