Question

a force is inclined at 30 degrees to the horizontal.if its rectangular component in the horizontal direction be 50N,find the magnitude of the force and its vertical component

Answer 1

Explanation:

It is given that,

Angle between the horizontal and the force, θ = 30°

Force acting in horizontal direction, F = 50 N

We have to find the magnitude of force and its vertical components. We know that the horizontal component is F cosθ and vertical component is F sinθ.

So, $Fcos(30)=50$

F = 57.7 N

Vertical component, $Fsin\theta=57.7\times sin(30)$

$Fsin\theta=28.85\ N$

Hence, this is the required solution.

Answer 2

Answer & Explanation:

GIVEN: Fx = 50 N ; θ = 30° ; Fy = ?

Fx = F Cosθ wherein θ = 30°

50 = F×√3/2

F = 50×2/√3

= 100/√3

= 57.73 N

Fy = F Sinθ

= 57.73×1/2

= 28.86 N.

Therefore, the magnitude of force = 57.73 N and the vertical component = 28.86 N.