Two equal, non parallel forces acting at a point on a body. The square

1.	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» 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°

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»

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°