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
KEAMKEAM 2007
Solution:
Let smaller force be $ {{F}_{1}} $ .
Resultant R of the forces is at $ 90{}^\circ $ to AB,
$ \therefore \,\,{{R}^{2}}\,+F_{1}^{2}\,=F_{2}^{2} $ in $ \Delta \,ABC $
or $ {{(12)}^{2}}=F_{2}^{2}-F_{1}^{2} $ ...(i)
or $ 144=({{F}_{2}}-{{F}_{1}})({{F}_{2}}-{{F}_{1}}) $
but $ {{F}_{1}}+{{F}_{2}}=18\,N $ (given)... (ii)
$ \therefore $ $ {{F}_{2}}-{{F}_{1}}=\frac{144}{18}=8 $ ....(iii)
From Eqs. (ii) and (iii),
$ {{F}_{1}}=5,\text{ }{{F}_{2}}=13 $
Hence, forces are 5 N and 13 N.
