Member Force Under a Hanging Load
Mediumtrussesmethod of joints
A 10 kN vertical load hangs from a joint supported by two identical members, each inclined 45° from the horizontal and symmetric about the vertical. What is the axial force in each member?
- A7.07 kN
- B5 kN
- C10 kN
- D14.1 kN
Show solutionHide solution
Answer: A. 7.07 kN
Method of joints at the loaded joint: ΣFy = 0 → 2 · F · sin 45° = 10 → F = 10 / (2 · 0.707) = 7.07 kN. Horizontal components cancel by symmetry. 5 kN forgets the sine; 14.1 kN uses 10/sin45° without splitting between two members.
kN
Practice this deckUntimed, one problem at a time, with your progress saved.
Related problems
- Member Force in a Symmetric Pin Joint · Trusses
- Member Force in a Triangular Truss · Trusses
- Zero-Force Members in a Truss · Trusses
Keep studying
- All Statics & Equilibrium practice problems
- FE Civil Statics: formulas and examples
- Check your work with the Truss Calculator