Question

A block fof mass 5kg is suspended by a massless rope of length 2 `m` from the ceilling. A force of 50 `N` is applied in the horizontal direction at the midpoint `P` of the rope, as shown in the figure. The angle made by the rope with the vertical in equilibrium is (Take `g=10ms^(-2)m` .

A. `30^(@)`
B. `40^(@)`
C. `60^(@)`
D. `45^(@)`

Answer 1

Correct Answer - D
Let `theta` be the angle made by the rope with the vertical in equilibrium. The free body diagram of 5 kg block si as shown in Fig. (b).
In equilibrium
`T_(2)=5g=5xx10=50N`
The free body diagram of the point P is as shown in Fig. (c.)
In equilibrium `T_(1)sin theta=50N ....(i)`
`T_(1)cos theta=T_(2)=50N` ...(ii)
Dividing (i) by (ii), we get `tan theta=(50)/(50)=1`
`theta=tan ^(-1)=45^(@)`