If the resultant of two forces of magnitudes | vedclass

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(C) The formula for the resultant $R$ of two forces $P$ and $Q$ acting at an angle $	heta$ is given by $R^2 = P^2 + Q^2 + 2PQ \cos 	heta$. Given $R

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$2$

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(C) The formula for the resultant $R$ of two forces $P$ and $Q$ acting at an angle $	heta$ is given by $R^2 = P^2 + Q^2 + 2PQ \cos 	heta$. Given $R = \sqrt{7}Q$ and $	heta = 60^\circ$. Substituting these values: $(\sqrt{7}Q)^2 = P^2 + Q^2 + 2PQ \cos 60^\circ$. $7Q^2 = P^2 + Q^2 + 2PQ(1/2)$. $7Q^2 = P^2 + Q^2 + PQ$. Rearranging the terms: $P^2 + PQ - 6Q^2 = 0$. Factoring the quadratic equation: $P^2 + 3PQ - 2PQ - 6Q^2 = 0$. $P(P + 3Q) - 2Q(P + 3Q) = 0$. $(P - 2Q)(P + 3Q) = 0$. Since $P$ and $Q$ are magnitudes of forces,they must be positive,so $P + 3Q 
eq 0$. Therefore,$P - 2Q = 0$,which implies $P = 2Q$. Thus,$P/Q = 2$.

If the resultant of two forces of magnitudes $P$ and $Q$ acting at a point at an angle of $60^\circ$ is $\sqrt{7}Q$,then $P/Q$ is

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