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 (in newton) 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 (in newton) applied is (g = 10 ms-2) (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/24784db4deccacb2dad069fa5a572d86-.png" /> At equilibrium, tan 45° = (mg/F) = (100/F) ∴ F = 100N

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 (in newton) applied is (g=10ms−2)

At equilibrium, tan 45∘=Fmg​=F100​ ∴F=100N

At equilibrium,
tan $4 5^{\circ} = \frac{m g}{F} = \frac{100}{F}$
$∴ F = 100 N$