Forces of magnitude $3P$ and $2P$ act at a point and their resultant is $R$. If the force $3P$ is doubled to $6P$ while $2P$ remains unchanged, the new resultant becomes $2R$. What is the angle between the original two forces?

Question

Accepted Answer

120 degrees. Let the angle between the forces be $\theta$. Then $R^2 = (3P)^2 + (2P)^2 + 2(3P)(2P)\cos\theta$ and $(2R)^2 = (6P)^2 + (2P)^2 + 2(6P)(2P)\cos\theta$. Solving these equations gives $\cos\theta = -1/2$, so $\theta = 120^\circ$.

Forces of magnitude \(3P\) and \(2P\) act at a point and their resultant is \(R\). If the force \(3P\) is doubled to \(6P\) while \(2P\) remains unchanged, the new resultant becomes \(2R\). What is the angle between the original two forces?

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