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3-1.Vectors
hard
The sum of two forces acting at a point is $16\, N.$ If the resultant force is $8\, N$ and its direction is perpendicular to minimum force then the forces are
A$6\, N$ and $10\, N$
B$8\, N$ and $ 8\, N$
C$4\, N$ and $12\, N$
D$2\, N$ and $14 \,N$
Solution
(a) $A + B = 16$ (given) …$(i)$
$\tan \alpha = \frac{{B\sin \theta }}{{A + B\cos \theta }} = \tan 90^\circ $
$A + B\cos \theta = 0 ⇒ \cos \theta = \frac{{ – A}}{B}$…$(ii)$
$8 = \sqrt {{A^2} + {B^2} + 2AB\cos \theta } $…$(iii)$
By solving eq. $(i),$ $(ii)$ and $(iii)$ we get $A = 6\,N,$ $B = 10\,N$
$\tan \alpha = \frac{{B\sin \theta }}{{A + B\cos \theta }} = \tan 90^\circ $
$A + B\cos \theta = 0 ⇒ \cos \theta = \frac{{ – A}}{B}$…$(ii)$
$8 = \sqrt {{A^2} + {B^2} + 2AB\cos \theta } $…$(iii)$
By solving eq. $(i),$ $(ii)$ and $(iii)$ we get $A = 6\,N,$ $B = 10\,N$
Standard 11
Physics
