In the given figure the blocks A, B and C weigh 4 kg, 6 kg and 8 kg respectively. The coefficient of sliding friction between any two surfaces is 0.5. The force F required to slide the block C with constant speed is N. (Use g = 10 m/s2) [figure]
Answer
210
Solution
Normal reactions: A on B: N1 = 4 10 = 40 N; B on C: N2 = (4 + 6) 10 = 100 N; ground on C: N3 = (4 + 6 + 8) 10 = 180 N
Sliding friction ( = 0.5) acts at every surface:
f1 = 0.5 40 = 20 N (A–B), f2 = 0.5 100 = 50 N (B–C), f3 = 0.5 180 = 90 N (C–ground)
B and C are tied by the string over the fixed pulley, so when C moves to the left, B moves to the right with the same constant speed.
Block B (no acceleration): T = f1 + f2 = 20 + 50 = 70 N
Block C (no acceleration): F = T + f2 + f3
F = 70 + 50 + 90 = 210 N
Answer: 210