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Q.

Two forces 3 N and 2 N are at an angle $θ$ such that the resultant is R. The first force is now increased to 6 N and the resultant become 2R. The value of $θ$ is

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a

$30^{0}$

b

$90^{0}$

c

$60^{0}$

d

$120^{0}$

answer is D.

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Detailed Solution

A = 3 N, B = 2 N then R = $\sqrt{A^{2} + B^{2} + 2 ABcosθ}$

R = $\sqrt{9 + 4 + 12 cosθ}$

Now A = 6 N, B = 2 N then

2R = $\sqrt{36 + 4 + 24 cosθ}$

from (i) and (ii) we get $cosθ = - \frac{1}{2}$

$∴ θ {= 120}^{0}$

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