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Q.

The resultant of two equal forces is 141.4N when they are mutually perpendicular. When they are inclined at an angle $120^{0},$ then the resultant force will be _____N.

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Detailed Solution

$R = \sqrt{P^{2} + Q^{2} + 2 P Q \cos θ} P = Q, θ = 90^{0}, R \sqrt{2} P, R = 141.4 N$
$141.4 = 1.414 P \Rightarrow P = Q = 100, θ = 120^{0}$
$R^{1} = \sqrt{P^{2} + Q^{2} + 2 P^{2} \cos θ}$
$R^{1} = \sqrt{{(100)}^{2} + {(100)}^{2} + 2 {(100)}^{2} (\frac{- 1}{2})}$
$\cos 120^{0} = \frac{- 1}{2} \Rightarrow R^{1} = 100 N$

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