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Question

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


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

The correct option is A $$6N$$ and $$10N$$
$$\displaystyle A + B = 16\,         ...(1)$$
$$\displaystyle tan \alpha = \frac {B\,sin\,\theta} {A + B\,cos\,\theta} = tan\,90$$
Thus, $$\displaystyle A + B\,cos\,\theta = 0 \Rightarrow cos\theta= - \frac {A} {B}         ...(2)$$
We also have: $$\displaystyle 8 = \sqrt {A^2 + B^2 + 2AB\,cos\,\theta}         ...(3)$$
Solving these we get, $$A=6N$$ and $$B=10N$$

Physics

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