Question

Two equal, non parallel forces acting at a point on a body. The square of the resultant is found to be three times the product of the forces, what is the angle between them.

Answer 1

Given, P = Q & R² = 3PQ = 3P²

we have R² = P² + Q² + 2PQ cosθ

i.e. 3P²= P² + P² + 2P² cosθ

i.e. 3P² = 2P² + 2P² cosθ

1P² = 2P² cos θ

\(\frac{1}{2}\)= cose

⇒ θ = cos^-1 \((\frac{1}{2})\) = 60°