Two forces are such that the sum of their mag | vedclass

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(C) Let $P$ be the smaller force and $Q$ be the greater force. According to the problem:$P + Q = 18$......$(i)$Given that the resultant $R = 12 \; N$

Vedclass · Accepted Answer

$5 \; N, 13 \; N$

Vedclass · Answer

(C) Let $P$ be the smaller force and $Q$ be the greater force. According to the problem:$P + Q = 18$......$(i)$Given that the resultant $R = 12 \; N$ is perpendicular to the smaller force $P$,the angle between $R$ and $P$ is $90^{\circ}$.Using the formula for the direction of the resultant: $	an \alpha = \frac{Q \sin 	heta}{P + Q \cos 	heta}$.Since $\alpha = 90^{\circ}$,$	an 90^{\circ} = \infty$,which implies $P + Q \cos 	heta = 0$,so $Q \cos 	heta = -P$......$(ii)$The magnitude of the resultant is given by $R^2 = P^2 + Q^2 + 2PQ \cos 	heta$.Substituting $R = 12$ and $Q \cos 	heta = -P$ into the equation:$12^2 = P^2 + Q^2 + 2P(-P)$$144 = P^2 + Q^2 - 2P^2$$144 = Q^2 - P^2$$144 = (Q - P)(Q + P)$Since $Q + P = 18$,we have $144 = (Q - P)(18)$,which gives $Q - P = 8$......$(iii)$Adding $(i)$ and $(iii)$: $2Q = 26 \implies Q = 13 \; N$.Subtracting $(iii)$ from $(i)$: $2P = 10 \implies P = 5 \; N$.Thus,the magnitudes of the forces are $5 \; N$ and $13 \; N$.

Two forces are such that the sum of their magnitudes is $18 \; N$ and their resultant is $12 \; N$,which is perpendicular to the smaller force. Then the magnitudes of the forces are:

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