Q. A mass of 6 kg is suspended by a rope of length 2 m from a ceiling. A force of 50 N is applied in the horizontal direction at the mid-point of the rope. The angel made by the rope, with the vertical, in equilibrium position will be (take $g = 10\ ms^{- 2},$ neglect the mass of the rope)
A) 90${^\circ}$
B) 60${^\circ}$
C) 50${^\circ}$
D) 40${^\circ}$
Explanation
(d) The three forces acting on the mass at location $A$ have been shown in figure. Since the mass is in equilibrium, therefore, the three forces acting on the mass must be represented by the three sides of a triangle taken in one order. Hence
[image:5_image285.emf]
$\frac{50}{SA} = \frac{6 \times 10}{SB}$ or $\frac{SA}{SB} = \frac{50}{60} = \frac{5}{6}$
or $\tan{\theta = \frac{SA}{SB} = \frac{5}{6} = 0.8333}$
$= \tan{40{^\circ}}$
$\therefore\theta = 40{^\circ}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.