A mass of 10 kg is suspended vertically by a rope from the roof. When a horizontal force is applied on the rope at some point, the rope deviated at an angle of 45° at the roof point. If the suspended mass is at equilibrium, the magnitude of the force applied is (g = 10 ms -2)

Question

A mass of 10 kg is suspended vertically by a rope from the roof. When a horizontal force is applied on the rope at some point, the rope deviated at an angle of 45° at the roof point. If the suspended mass is at equilibrium, the magnitude of the force applied is (g = 10 ms -2) (A) 200 N (B) 100 N (C) 140 N (D) 70 N

Tardigrade Admin · Accepted Answer

at equation tan 45° = (100/F) F = 100 N

Q. A mass of $10 k g$ is suspended vertically by a rope from the roof. When a horizontal force is applied on the rope at some point, the rope deviated at an angle of $4 5^{\circ}$ at the roof point. If the suspended mass is at equilibrium, the magnitude of the force applied is $(g = 10 m s^{- 2})$

200 N

100 N

140 N

70 N

at equation
$tan 4 5^{\circ} = \frac{100}{F}$
F = 100 N

Q. A mass of 10kg is suspended vertically by a rope from the roof. When a horizontal force is applied on the rope at some point, the rope deviated at an angle of 45∘ at the roof point. If the suspended mass is at equilibrium, the magnitude of the force applied is (g=10ms−2)