Topic: Engineering Mechanics
The bar AC, 10m long supports a load of 6000 N as shown in figure. The cable BC is horizontal and 5m long. Forces in the cable and the bar are
- A. 6928 N and 3464 N
- B. 3464 N and 6928 N
- C. 3464 N and 3464 N
- D. 6928 N and 6928 N
Answer: B. 3464 N and 6928 N
Explanation: From the geometry of right-angled triangle ABC, $\sin\theta = \frac{BC}{AC} = \frac{5}{10} = 0.5 \Rightarrow \theta = 30^{\circ}$ (angle with vertical). Balancing vertical forces at C: $F_{AC}\cos 30^{\circ} = 6000 \Rightarrow F_{AC} = \frac{6000}{\sqrt{3}/2} = 6928.2~N$ (Compression). Balancing horizontal forces: $F_{BC} = F_{AC}\sin 30^{\circ} = 6928.2 \times 0.5 = 3464.1~N$ (Tension). Hence, forces in the cable and bar are 3464 N and 6928 N respectively.