Calculate the minimum force F required to lift up a 0.5 m uniform rod of weight 30 N at the 50 cm mark.
A
30 N
B
18 N
C
15 N
D
20 N
Verified Correct Answer: C.
Step-by-Step Mathematical Solution
Given: Length of rod = 0.5 m = 50 cm; weight W = 30 N acts at the centre of gravity (25 cm from either end)
Force F is applied at the 50 cm mark (one end)
Taking moments about the other end: F × 50 = 30 × 25
F = \(\frac{30 \times 25}{50}\) = 15 N
Source: WAEC / JAMB / NECO Classroom Archive (Record #78297)