The resultant of forces P and Q is R.

Question

The resultant of forces $\vec P$ and $\vec Q$ is $\vec R$. If $\vec Q$ is doubled then $\vec R$ is doubled. If the direction of $\vec Q$ is reversed, then $\vec R$ is again doubled. Then $P^2:Q^2:R^2$ is:

Accepted Answer

$R^2=P^2+Q^2+2PQ\cos\theta$ \ldots (1); $(2R)^2=P^2+4Q^2+4PQ\cos\theta$ \ldots (2); $(2R)^2=P^2+Q^2-2PQ\cos\theta$ \ldots (3)

Adding (1) and (3): $5R^2=2P^2+2Q^2$. Combining with (2) and (3) to eliminate $\cos\theta$ leads to:

$$\frac{P^2}6=\frac{Q^2}9=\frac{R^2}6 \implies P^2:Q^2:R^2=2:3:2$$

Solutions · 1