Force in an inclined light member supporting a weight: A light member is hinged to a vertical wall and supports a weight W at its free end. If the member makes an angle θ with the (upper) wall, find the axial force induced in the member.

Introduction / Context:
Light two-force members (like links or rods) carry only axial force. When such a member supports a vertical load at its free end, the component of the axial force along the vertical direction must balance the applied weight. Determining the axial force is a frequent step in truss and frame analysis.

Given Data / Assumptions:

• Member is light (self-weight neglected) and hinged at the wall: a classic two-force member.
• Weight W acts vertically downward at the free end.
• Member makes an angle θ with the horizontal (upper wall), hence its vertical component is T sin θ.
• Static equilibrium; no other external forces.

Concept / Approach:
For a two-force member, internal force is purely axial with magnitude T. Vertical equilibrium at the free end requires the vertical component of T to equal W. Horizontal components are reacted at the wall hinge.

Step-by-Step Solution:

Verification / Alternative check:
As θ approaches 90°, sin θ → 1 and T → W (near vertical member). As θ gets small, sin θ decreases and T grows large, matching the mechanical intuition of a near-horizontal tie carrying high axial force.

Why Other Options Are Wrong:

• W sec θ, W tan θ: incorrect trigonometric relations for the required vertical balance.
• W cos θ, W sin θ: these are components, not the required axial magnitude.

Common Pitfalls:
Using horizontal angle vs. vertical angle inconsistently; forgetting the member is a two-force member so the internal force is axial only.

Final Answer:
W cosec θ