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 same points , the rope deviated at an angle of ${45}^0$ 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. 70 N
• D. 140 N

Answer 1

The correct option is B 100 N

Given$;-$

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 $(g=10m/s^2)$

We have the rope being deviated to ${45}^0$ so$,$

$F/sin(180-\theta )$ and $Mg/sin(90+\theta )$

$F/sin(180-45)=Mg/sin(90+45)$

$F/sin45=Mg/cos45$

$F/1/\surd 2=Mg/1/\surd 2$

$F=M\times g$

$F=10\times 10$

$F=100N$

Hence,

option $\left(B\right)$ is correct answer.