Question

A 10 kg body is suspended by a rope is pulled

Answer 1

Answer:

The correct option is;

3) 10·√3 kgwt

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;

$\dfrac{T \cdot sin(\theta)}{T \cdot cos(\theta)} = \dfrac{F}{m \cdot g}$

Which gives;

$\dfrac{sin(\theta)}{cos(\theta)} = tan(\theta) = \dfrac{F}{m \cdot g}$

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.