Question

Two forces F₁ and F₂ are acting at point A as shown in. The angle between the two forces is 50°. It is found that the resultant R is 500 N and makes angles 20° with the force F₁ as shown in the figure. Determine the forces F₁ and F₂.

Answer 1

Let ∆ ABC be the triangle of forces drawn to some scale. In this 

∠BAC = α = 20° 

∠ABC = 180 – 50 = 130° 

∴ ∠ACB = 180 – (20 + 130) = 30° 

Applying sine rule to ∆ ABC, we get

\(\cfrac{AB}{sin\,30^\circ}\) = \(\cfrac{BC}{sin\,20^\circ}\) = \(\cfrac{500}{sin\,130^\circ}\)

AB = 326.35 N 

and BC = 223.24 N. 

Thus F₁ = AB = 326.35 N 

and F₂ = BC = 223.24 N