Question

A mass of $6$ kg is suspended by a rope of length $2$m from 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 be :

(Take $g=10m{s}^{-2}$, neglect the mass of the rope)

• A. ${53}^{\circ }$
• B. ${60}^{\circ }$
• C. ${30}^{\circ }$
• D. ${45}^{\circ }$

Answer 1

The correct option is D ${45}^{\circ }$

Given,

Mass $=6kg$

Force $=60N$

Length of the rope $=2m$

Equilibrium of the given weight with gravitational force acting on it gives,

$T_2=6\times 10=60N$

There are three forces acting on the mass tension 1 $\left(T_1\right)$ and tension 2 $\left(T_2\right)$ and the force acting horizontally, that is, $60N$

The resultant force will be vertical and horizontal components acting separately vanish to become one force

Hence,

$T_{1}cos\theta =T_2=60N$

and

$T_{1}sin\theta =60N$

$\frac{60}{sin\theta }=\frac{60}{cos\theta }$

$tan\theta =1$

$\theta =ta{n}^{-1}1={45}^0$