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;
[tex]\dfrac{T \cdot sin(\theta)}{T \cdot cos(\theta)} = \dfrac{F}{m \cdot g}[/tex]
Which gives;
[tex]\dfrac{sin(\theta)}{cos(\theta)} = tan(\theta) = \dfrac{F}{m \cdot g}[/tex]
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.
