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 is deviated by an angle of 45∘ at the roof point. If the suspended mass is at equilibrium, the magnitude of the force is? (Assume g=10ms−2)

Answer 1

We are given a hanging mass that is pushed by a force at a certain angle. To compute this force when the mass is in equilibrium, we first build the free body diagram to understand what forces are operating on the mass and in which direction.

External force on the mass is F

Tension due to the rope is T

Mass of the body is m=10 kg

The angle by which the rope is moved is θ = 45∘

 

We balance the forces on the mass by taking their horizontal and vertical components. From the FBD, we can see that:

F = Tsinθ —  1

And mg=Tcosθ — 2

 

Putting the values in 2, we get:

10×10 =T cos45 =T√2

Also, from Eq. 1:

F = Tsin45 = T√2

Comparing the two equations, we get:

F = 10 × 10 = 100 N

 

 

The magnitude of force is 100 N