The sum of the magnitudes of two forces acting at a point is 18 N and the magnitude of their resultant is 12 N. If the resultant is at $90{^\circ}$ with the smaller force, the magnitude of the forces in N are

Question

Dr. Abubacker · Accepted Answer

(c) Let smaller force be $F_{1}$. Resultant R of the forces is at $90{^\circ}$ to AB,
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$	herefore\ R^{2} + F_{1}^{2} = F_{2}^{2}$ in $\mathrm{\Delta}ABC$
$or\ \ \ (12)^{2} = F_{2}^{2} - F_{1}^{2}\ \ \ \ \ \ \ \ \ldots\ldots(i)$
$or\ \ 144 = \left( F_{2} - F_{1} ight)\left( F_{2} + F_{1} ight)$
$but\ F_{1} + F_{2} =$18 N (given) ….(ii)
$	herefore\ F_{2} - F_{1} = \frac{144}{18} = 8\ \ \ldots..(iii)$
From Eqs. (ii) and (iii), $F_{1} = 5,\ \ \ F_{2} = 13$
Hence, forces are 5 N and 13 N.

Q. The sum of the magnitudes of two forces acting at a point is 18 N and the magnitude of their resultant is 12 N. If the resultant is at $90{^\circ}$ with the smaller force, the magnitude of the forces in N are

Explanation

Reviewed by Dr. Abubacker

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