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

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Question

(A) Let the two forces be $A$ and $B$,where $A < B$. Given $A + B = 16$ (equation $i$).Since the resultant $R = 8\, N$ is perpendicular to the smaller

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let the two forces be $A$ and $B$,where $A Since the resultant $R = 8\, N$ is perpendicular to the smaller force $A$,the angle $\alpha$ between $R$ and $A$ is $90^\circ$.The formula for the direction of the resultant is $	an \alpha = \frac{B \sin 	heta}{A + B \cos 	heta} = 	an 90^\circ$.This implies $A + B \cos 	heta = 0$,so $\cos 	heta = -A/B$ (equation $ii$).The magnitude of the resultant is $R^2 = A^2 + B^2 + 2AB \cos 	heta$.Substituting $\cos 	heta = -A/B$ into the resultant formula: $8^2 = A^2 + B^2 + 2AB(-A/B) = A^2 + B^2 - 2A^2 = B^2 - A^2$.Thus,$B^2 - A^2 = 64$,which factors to $(B - A)(B + A) = 64$.Since $B + A = 16$,we have $(B - A)(16) = 64$,so $B - A = 4$.Solving $A + B = 16$ and $B - A = 4$ gives $2B = 20$,so $B = 10\, N$ and $A = 6\, N$.

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 the minimum force,then the forces are:

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