Question:easy

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

Show 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!
Updated On: Jul 4, 2026
  • \( 2\text{ P}\sin\left(\frac{\alpha}{2}\right) \)
  • \( 2\text{ P}\cot\left(\frac{\alpha}{2}\right) \)
  • \( 2\text{ P}\cos\left(\frac{\alpha}{2}\right) \)
  • \( 2\text{ P}\tan\left(\frac{\alpha}{2}\right) \)
Show Solution

The Correct Option is C

Solution and Explanation

Step 1: Place the two forces symmetrically about a common axis.
Instead of the cosine rule, place both forces of magnitude \( P \) symmetrically, one at angle \( +\frac{\alpha}{2} \) and the other at angle \( -\frac{\alpha}{2} \) from a central bisecting line, so the angle between them is still \( \alpha \).

Step 2: Add the components along the bisector.
Along the bisector direction (call it the x-axis), both forces contribute: \[ R_x = P\cos\left(\frac{\alpha}{2}\right) + P\cos\left(\frac{\alpha}{2}\right) = 2P\cos\left(\frac{\alpha}{2}\right) \]
Step 3: Add the components perpendicular to the bisector.
Perpendicular to the bisector, the two forces point in opposite directions and cancel exactly: \[ R_y = P\sin\left(\frac{\alpha}{2}\right) - P\sin\left(\frac{\alpha}{2}\right) = 0 \]
Since the resultant has no perpendicular component, its full magnitude is just the bisector component: \[ R = \sqrt{R_x^2 + R_y^2} = R_x = \boxed{2P\cos\left(\frac{\alpha}{2}\right)} \]
which matches option (C). This also shows directly why the resultant always points exactly along the bisector line of the two equal forces.
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