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

Question

A mass of 6 kg is suspended by a rope of length 2 m from a ceiling. A force of 50 N in the horizontal direction is applied at the mid-point of the rope. The angle made by the rope with the vertical, in equilibrium is (A)  50°  (B)  60°  (C)  30°  (D)  40°

Tardigrade Admin · Accepted Answer

tan θ =(50/6g)=0.83 or tan text θ =tan text 40° or θ =40°

Q. A mass of $6 k g$ is suspended by a rope of length $2 m$ from a ceiling. A force of $50 N$ in the horizontal direction is applied at the mid-point of the rope. The angle made by the rope with the vertical, in equilibrium is

$50^{\circ}$

$60^{\circ}$

$30^{\circ}$

$40^{\circ}$

$5.5^{\circ}$

$tan θ = \frac{50}{6 g} = 0.83$
or $t an θ = t an 40^{\circ}$ or $θ = 40^{\circ}$

Q. A mass of 6kg is suspended by a rope of length 2m from a ceiling. A force of 50N in the horizontal direction is applied at the mid-point of the rope. The angle made by the rope with the vertical, in equilibrium is

50∘

60∘

30∘

40∘

5.5∘

tanθ=6g50​=0.83 or tanθ=tan40∘ or θ=40∘