Question

A 10 kg body is suspended by a rope is pulled

by means of a horizontal force to make 60°
by rope to vertical. The horizontal force is
1) 10 kgwt
2) 30 kgwt
3) 10/3 kgwt 4) 30/3 kgwt​

Answer 1

The correct option is;

3) 10·√3 kgwt

Step-by-step explanation:

The parameters given are;

Mass of body, m = 10 kg

Direction of applied force = Horizontal

Angle of inclination of the rope with the vertical, θ = 60°

The component of the tension in the rope equivalent to the force is given by the relation;

T·sin(θ) = F

The component of the rope tension still under gravitational pull is given by the relation;

T·cos(θ) = m·g

Therefore, we have;

`(T \cdot sin(\theta))/(T \cdot cos(\theta)) = (F)/(m \cdot g)`

Which gives;

`(sin(\theta))/(cos(\theta)) = tan(\theta) = (F)/(m \cdot g)`

Therefore;

F = tan(θ) × m×g

Where:

g = Acceleration due to gravity

tan(θ) = √3

m = 10 kg

∴ F = 10·√3 × g = 10·√3 kg-wt.