Two-force resultants at 120°: Two forces act at 120°. The larger force is 50 kgf. If the resultant is perpendicular to the smaller force, determine the magnitude of the smaller force.

Introduction / Context:

Vector addition of two forces can be simplified using components. Here, the resultant's orthogonality to one of the forces (the smaller) provides a clean dot-product condition that determines the unknown magnitude without trigonometric law-of-cosines algebra.

Given Data / Assumptions:

• Forces F_small = P (unknown) and F_large = 50 kgf.
• Angle between them = 120°.
• Resultant R is perpendicular to the smaller force.
• Planar (2D) force system.

Concept / Approach:

Let the smaller force lie along the +x axis: S = (P, 0). The 50 kgf force at 120° has components L = 50 (cos 120°, sin 120°) = (−25, 25√3). Then R = S + L = (P − 25, 25√3). The perpendicularity condition is S · R = 0.

Step-by-Step Solution:

Verification / Alternative check:

Law of cosines approach: impose R ⟂ S ⇒ R^2 = L^2 + S^2 (since R is perpendicular to S, triangle is right-angled at S). With L = 50 and S = 25, R^2 = 50^2 + 25^2 = 3125, consistent with the component result.

Why Other Options Are Wrong:

• 20, 30, 35, 40 kgf do not satisfy the perpendicularity (dot-product) condition for a 120° spacing with a 50 kgf companion force.

Common Pitfalls:

• Using 60° instead of 120° for components, which flips the sign of the x-component.
• Forgetting that cos 120° = −1/2 and sin 120° = √3/2.

Final Answer:

25 kg