The sum of two forces is 18 N and the result | vedclass

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Question

(A) Let the two forces be $P$ and $Q$,where $Q > P$. Given that the sum of the forces is $P + Q = 18 \ N$. Let the resultant $R = 12 \ N$ be perpendic

Vedclass · Accepted Answer

$13 \ N, 5 \ N$

Vedclass · Answer

(A) Let the two forces be $P$ and $Q$,where $Q > P$. Given that the sum of the forces is $P + Q = 18 \ N$. Let the resultant $R = 12 \ N$ be perpendicular to the smaller force $P$. The angle between the resultant $R$ and the force $P$ is $90^o$. The formula for the direction of the resultant is given by $	an \alpha = \frac{Q \sin 	heta}{P + Q \cos 	heta}$,where $	heta$ is the angle between the forces. Since the resultant is perpendicular to $P$,$	an 90^o = \infty$,which implies $P + Q \cos 	heta = 0$,or $\cos 	heta = -\frac{P}{Q}$. Using the magnitude formula for the resultant: $R^2 = P^2 + Q^2 + 2PQ \cos 	heta$. Substituting $\cos 	heta = -\frac{P}{Q}$: $12^2 = P^2 + Q^2 + 2PQ(-\frac{P}{Q}) = P^2 + Q^2 - 2P^2 = Q^2 - P^2$. $144 = (Q - P)(Q + P)$. Since $Q + P = 18$,we have $144 = (Q - P) 	imes 18$,so $Q - P = 8$. Solving the system $Q + P = 18$ and $Q - P = 8$: Adding the equations: $2Q = 26 \Rightarrow Q = 13 \ N$. Subtracting the equations: $2P = 10 \Rightarrow P = 5 \ N$. Thus,the magnitudes of the two forces are $13 \ N$ and $5 \ N$.

The sum of two forces is $18 \ N$ and the resultant,whose direction is at right angles to the smaller force,is $12 \ N$. The magnitudes of the two forces are:

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