Question

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

• A. ${\tan }^{-1}\left(\frac{4}{5}\right)$
• B. ${\tan }^{-1}\left(\frac{5}{4}\right)$
• C. ${\tan }^{-1}\left(\frac{5}{6}\right)$
• D. None of these

Answer 1

The correct option is C ${\tan }^{-1}\left(\frac{5}{6}\right)$

Conditions of total equilibrium quint the sum of all

the external forces acting on the body is zero, and

the sum of all the torques from external forces $c_0$

zero - These two Conditions must be simultaneously satisfied in equilibrium

for horizontal equilibrium

$T_3=T_1\sin \theta$

$forverticalequilibrium,$

$T_2=T_1\cos \theta$

$Takingratios$

$\frac{T_3}{T_2}=\tan \theta$

for the equilibrium body

$T_2=mg$

$T_2=6\times 10N$

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

${\theta }_2$

$\theta ={\tan }^{-1}\left(\frac{5}{6}\right)$

$optionc={\tan }^{-1}\left(\frac{5}{6}\right)$