The sum of two forces acting at a point is 16 | vedclass

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Question

(a) $A + B = 16$ (given) &hellip;$(i)$

$	an \alpha = \frac{{B\sin 	heta }}{{A + B\cos 	heta }} = 	an 90^\circ $

$A + B\cos 	heta = 0 &

Vedclass · Accepted Answer

$6\, N$ and $10\, N$

Vedclass · Answer

(a) $A + B = 16$ (given) &hellip;$(i)$

$	an \alpha = \frac{{B\sin 	heta }}{{A + B\cos 	heta }} = 	an 90^\circ $

$A + B\cos 	heta = 0 &rArr; \cos 	heta = \frac{{ - A}}{B}$&hellip;$(ii)$

$8 = \sqrt {{A^2} + {B^2} + 2AB\cos 	heta } $&hellip;$(iii)$

By solving eq. $(i),$ $(ii)$ and $(iii)$ we get $A = 6\,N,$ $B = 10\,N$

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

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