Question

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

• A. $\sin ^{-1}\left(\frac{3}{5}\right)$
• B. $\sin ^{-1}\left(\frac{5}{6}\right)$
• C. $\tan ^{-1}\left(\frac{4}{5}\right)$
• D. $\tan ^{-1}\left(\frac{5}{6}\right)$

Answer 1

Answer: (D)

At equilibrium, $T_{1} \sin \theta=50 N$
$T_{1} \cos \theta=T_{2} \quad $
$\Rightarrow \tan \theta=\frac{50 N }{T_{2}}=\frac{50 N }{60 N }$
or $\theta=\tan ^{-1}\left(\frac{5}{6}\right)$