Question

Two equal forces each of magnitude P acting at a point with an angle of \( (\alpha) \), then the resultant force is

• A. \( 2\text{ P}\sin\left(\frac{\alpha}{2}\right) \)
• B. \( 2\text{ P}\cot\left(\frac{\alpha}{2}\right) \)
• C. \( 2\text{ P}\cos\left(\frac{\alpha}{2}\right) \)
• D. \( 2\text{ P}\tan\left(\frac{\alpha}{2}\right) \)

Hint:

The resultant of two equal vectors always bisects the angle between them. Geometrically, this splits the vector addition parallelogram into two congruent isosceles triangles. Projecting the vectors along this central bisection line gives \( P\cos(\alpha/2) + P\cos(\alpha/2) = 2P\cos(\alpha/2) \) directly!

Answer 1

The Correct Option is C

Concept: According to the Parallelogram Law of Vector Addition, when two concurrent forces \( F_1 \) and \( F_2 \) act at a single point with an included angle \( \alpha \) between their vectors, the magnitude of their combined net resultant force vector \( R \) is determined using the geometric cosine formula: \[ R = \sqrt{F_1^2 + F_2^2 + 2F_1F_2\cos\alpha} \] We will substitute the specific conditions given in this problem and simplify using standard trigonometric half-angle identity mappings.

Step 1: Substituting the equal force values into the general formula.
The problem states that both component forces are equal in magnitude. Let us define: \[ F_1 = P \quad \text{and} \quad F_2 = P \] Substituting these values into our general equation for the resultant force magnitude: \[ R = \sqrt{P^2 + P^2 + 2(P)(P)\cos\alpha} \]

Step 2: Combining like algebraic terms.
Sum the individual squared component terms under the radical sign: \[ P^2 + P^2 = 2P^2 \] Multiply out the final product term under the radical sign: \[ 2(P)(P)\cos\alpha = 2P^2\cos\alpha \] Now replace these simplified parts back into the main radical equation: \[ R = \sqrt{2P^2 + 2P^2\cos\alpha} \]

Step 3: Factoring out common expressions.
We can factor out the common multiplier expression \( 2P^2 \) from both terms inside the radical: \[ R = \sqrt{2P^2(1 + \cos\alpha)} \quad \cdots (1) \]

Step 4: Applying trigonometric identities.
Recall the standard double-angle trigonometric identity for cosines: \[ \cos(2\theta) = 2\cos^2\theta - 1 \quad \Rightarrow \quad 1 + \cos(2\theta) = 2\cos^2\theta \] By substituting \( 2\theta = \alpha \), which means \( \theta = \frac{\alpha}{2} \), we get the half-angle formula: \[ 1 + \cos\alpha = 2\cos^2\left(\frac{\alpha}{2}\right) \] Now substitute this trigonometric relation directly back into equation (1): \[ R = \sqrt{2P^2 \cdot \left[2\cos^2\left(\frac{\alpha}{2}\right)\right]} \]

Step 5: Simplifying the radical to find the final result.
Multiply out the scalar quantities under the square root: \[ R = \sqrt{4P^2\cos^2\left(\frac{\alpha}{2}\right)} \] Taking the clear square root of each factor independently: \[ R = 2P\cos\left(\frac{\alpha}{2}\right) \] This derived expression matches Option (C).