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

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(B) Let the two forces be $F_1$ and $F_2$,where $F_1$ is the larger force and $F_2$ is the smaller force.Given: $F_1 + F_2 = 18 \,N$ $(i)$Let the resu

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$13 \,N, 5 \,N$

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(B) Let the two forces be $F_1$ and $F_2$,where $F_1$ is the larger force and $F_2$ is the smaller force.Given: $F_1 + F_2 = 18 \,N$ $(i)$Let the resultant $R = 12 \,N$ be perpendicular to the smaller force $F_2$.The angle $\alpha$ between the resultant $R$ and the force $F_2$ is $90^\circ$.The formula for the direction of the resultant is $	an \alpha = \frac{F_1 \sin 	heta}{F_2 + F_1 \cos 	heta} = 	an 90^\circ$.This implies the denominator $F_2 + F_1 \cos 	heta = 0$,so $\cos 	heta = -\frac{F_2}{F_1}$.The magnitude of the resultant is $R^2 = F_1^2 + F_2^2 + 2F_1 F_2 \cos 	heta$.Substituting $\cos 	heta = -\frac{F_2}{F_1}$,we get $R^2 = F_1^2 + F_2^2 + 2F_1 F_2 (-F_2/F_1) = F_1^2 - F_2^2$.Given $R = 12$,so $144 = F_1^2 - F_2^2 = (F_1 + F_2)(F_1 - F_2)$.Using $(i)$,$144 = 18(F_1 - F_2)$,which gives $F_1 - F_2 = 8 \,N$ (ii).Adding $(i)$ and (ii): $2F_1 = 26 \,N \implies F_1 = 13 \,N$.Subtracting (ii) from $(i)$: $2F_2 = 10 \,N \implies F_2 = 5 \,N$.Thus,the magnitudes are $13 \,N$ and $5 \,N$.

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

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