Question

A mass of $6kg$ is suspended by a rope of length $2m$ from the ceiling . A force of $50N$ in the horizontal direction is applied at the mid-point $P$ of the rope. as shown. What is the angle the rope makes with the vertical in equilibrium?

Answer 1

Using sine law, we get,

$\frac{50}{\sin (180-\theta )}=\frac{60}{\sin (90+\theta )}$

$\frac{5}{\sin \theta }=\frac{6}{\cos \theta }$

$\tan \theta =\frac{5}{6}$

$\Rightarrow \theta ={\tan }^{-1}\frac{5}{6}$

$\therefore \theta ={39.8}^{\circ }$