Question 47
Equilibrium of ForcesIn the diagram above, a rod 50 cm long of uniform cross-section is suspended horizontally on a fulcrum, F, by the action of two forces. What is the weight of the rod?
- A.120N
- B.200 N
- C.40 N
- D.80 N.
Show the answer and explanation
Answer: D (80 N.)
Explanation
The rod's weight W acts at its middle (25 cm), 10 cm right of F. Moments about F: 200 × 10 = 40 × 30 + W × 10, so 2000 = 1200 + 10W and W = 80 N.