Question

a body of mass 10 kg is suspended by a rope and is pulled to aside by means of a horizontal force so that the rope makes angle 60 degrees with the vertical rope. Then the tension force in the rope is​

Answer 1

T = 196N

Given:

mass(m) = 10kg

g = 9.8m/s^2

angle = 45°

horizontal force F acts on the mass

(Refer the diagram)

Resolving Tension into two components ie, Tcosθ and Tsinθ.

At equilibrium,

Tcosθ= mg .......equ(1) [ Cos60 = 1/2]

$\bold{\mathff{T\bigg(\dfrac{1}{2}\bigg) =mg}}$

T = 2mg

T= 2x10x9.8

T = 196N

Tsinθ = F .....equ(2) [sin 60=√3/2]

 $\bold{\mathff{T\bigg(\dfrac{\sqrt{3} }{2}\bigg) = F}}$

from equ(1) ,

2mg.√3/2 = F

√3mg = F

√3 x 10x9.8

or, F=98√3N