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 thye angle made bby the rope with vertical in equilibrium.
The free body diagram of `5kg` block is 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)sintheta=50N`
`T_(1)costheta=T_(2)=50N`
Dividing (i) by (ii), we get
`tantheta=(50)/(50)=1`
`theta=tan^(-1)(1)=45^(@)` and and