Question:medium

The resultant of two forces, magnitude of each is equal to ‘P’ and the acting angle between them is $60^{\circ}$, is

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For two equal forces $P$ at angle $\theta$, the resultant can be simplified to $R = 2P \cos(\theta/2)$. For $\theta = 60^{\circ}$: $R = 2P \cos 30^{\circ} = 2P \left(\frac{\sqrt{3}}{2}\right) = \sqrt{3}P$. This shortcut saves time during exams!
Updated On: Jul 1, 2026
  • $\sqrt{2} \text{ P}$
  • $\sqrt{3} \text{ P}$
  • $2 \text{ P}$
  • $\sqrt{5} \text{ P}$
Show Solution

The Correct Option is B

Solution and Explanation

1. The Law of Cosines for Resultant: For two forces $F_1$ and $F_2$ acting at an angle $\theta$, the magnitude of the resultant $R$ is given by: $$R = \sqrt{F_1^2 + F_2^2 + 2F_1F_2 \cos \theta}$$

2. Applying Given Values: We are given that:

• $F_1 = P$

• $F_2 = P$

• $\theta = 60^{\circ}$
Substituting these into the formula: $$R = \sqrt{P^2 + P^2 + 2(P)(P) \cos 60^{\circ}}$$

3. Mathematical Calculation: Since $\cos 60^{\circ} = \frac{1}{2}$: $$R = \sqrt{2P^2 + 2P^2 \left(\frac{1}{2}\right)}$$ $$R = \sqrt{2P^2 + P^2}$$ $$R = \sqrt{3P^2}$$ $$R = \sqrt{3} \text{ P}$$ The magnitude of the resultant force is exactly $\sqrt{3}$ times the magnitude of the individual forces.
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