Question

A mass of 6kg is suspended by a rope of length 2m from ceiling. A force of 50N in the horizontal is applied at mid point of the rope. The angle rope makes with vertical in equilibrium is

Answer 1

hope the answer helps

Answer 2

$\bold{\theta=\tan ^{-1} \frac{5}{6}}$  is the angle rope which makes with vertical in equilibrium.

Given:

Mass  = 6 kg

Force = 50N

Length of the rope = 2 m

To find:

Angle = ?

Solution:

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

$T_2 = 6 \times 10 = 60 N$

There are three forces acting on the mass tension 1, $T_1$, tension 2, $T_2$  and the force acting horizontally that is 50 N

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

Hence $T_{1} \cos \theta=T_{2}=60 \mathrm{N}$

And $T_{1} \sin \theta=50 N$

$\begin{array}{l}{\frac{50}{\sin \theta}=\frac{60}{\cos \theta}} \\ {\tan \theta=\frac{5}{6}} \\ {\theta=\tan ^{-1} \frac{5}{6}}\end{array}$

Hence angle θ will be $\bold{\theta=\tan ^{-1} \frac{5}{6}.}$