Concurrent tensile forces at a joint – three forces equally inclined A pin-joint in a frame is acted upon by three concurrent tensile forces P, Q, and R that are equally inclined to one another (120° apart). If P = 10 tonnes, what are Q and R for the joint to be in equilibrium?

Introduction / Context:At a pin-joint, concurrent forces must sum vectorially to zero for equilibrium. When three forces are equally separated in direction by 120°, symmetry allows a quick conclusion about their magnitudes.

Given Data / Assumptions:

• Three tensile forces (pulling away from the joint) P, Q, R.
• Azimuthal separation between each pair is 120°.
• P is given as 10 tonnes (metric tonnes).

Concept / Approach:For three coplanar forces 120° apart to be in equilibrium, their magnitudes must be equal. This follows from polygon of forces: a closed equilateral triangle is formed if and only if all three sides (magnitudes) are equal.

Step-by-Step Solution:Write vector equilibrium: P⃗ + Q⃗ + R⃗ = 0⃗.Geometric condition: directions at 0°, 120°, 240° imply symmetry.Therefore |P| = |Q| = |R|.Given P = 10 tonnes ⇒ Q = 10 tonnes, R = 10 tonnes.

Verification / Alternative check:Resolve along any two perpendicular axes; equal magnitudes 120° apart yield zero net components.

Why Other Options Are Wrong:

• (a), (b) give zero for one force, breaking symmetry and equilibrium.
• (c) and (d) provide scalar relations that do not guarantee vector balance.

Common Pitfalls:Adding magnitudes algebraically instead of as vectors; always use vector geometry for non-collinear forces.

Final Answer:Q = 10 tonnes and R = 10 tonnes