Example

Question:

See Fig. 4.8. 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, as shown. What is the angle the rope makes with the vertical in equilibrium? (Take \( g = 10~\text{m/s}^2 \)). Neglect the mass of the rope.

Solution:

Consider equilibrium of the weight \( W \):
\( T_2 = 6 \times 10 = 60~\text{N} \)

Now consider equilibrium of point P: three forces—tensions \( T_1 \) and \( T_2 \), and horizontal force 50 N.
Vertical: \( T_1 \cos\theta = T_2 = 60~\text{N} \)
Horizontal: \( T_1 \sin\theta = 50~\text{N} \)
So, \[ \tan\theta = \frac{50}{60} = \frac{5}{6} \implies \theta = \tan^{-1}\left(\frac{5}{6}\right) = 40^\circ \] The angle is \( 40^\circ \).