Question

# The sum of magnitudes of two forces acting at a point is $$16\,N$$. If the resultant force is $$8\,N$$ and its direction is perpendicular to smaller force, then the forces are:

A
6N and 10N
B
8N and 8/N
C
4N and 12N
D
2N and 14N

Solution

## The correct option is A $$6\,N$$ and $$10\,N$$The correct answer is A.Given,The magnitude of two n forces is $$16N$$The resultant force is $$8N$$so,$$X+Y 16(X< Y)$$ and R is the resultant force perpendicular to X.According to the Pythagoras theorem,$$\Rightarrow R^2=Y^2-X^2=(Y-X)(Y+X)=16(Y-X)$$$$\Rightarrow (Y-X)=\dfrac{R^2}{(Y+X)}=\dfrac{8^2}{16}$$$$\Rightarrow\dfrac{64}{16}=4$$$$(Y+X)=16$$ and $$(Y-X)=4$$By solving a linear equation in two variable we get,$$Y=10N$$  and $$X=6N$$Physics

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