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Vector Component

The sum of ma...

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 :

6 N

and

10 N

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8 N

and

8 N

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4 N

and

12 N

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2 N

and

14 N

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Solution

The correct option is A $6 N$ and $10 N$
$A + B = 16 . . . (1)$
$t a n α = \frac{B s i n θ}{A + B c o s θ} = t a n 90$
Thus, $A + B c o s θ = 0 \Rightarrow c o s θ = - \frac{A}{B} . . . (2)$
We also have: $8 = \sqrt{A^{2} + B^{2} + 2 A B c o s θ} . . . (3)$
Solving these we get, $A = 6 N$ and $B = 10 N$

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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 :

The correct option is A 6N and 10NA+B=16...(1)tanα=BsinθA+Bcosθ=tan90Thus, A+Bcosθ=0⇒cosθ=−AB...(2)We also have: 8=√A2+B2+2ABcosθ...(3)Solving these we get, A=6N and B=10N

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 :