two crates, of mass 75kg and 110 kg are in contact and at rest on a horizontal surface. a 730 N force is exerted on a 75kg crate. if the coefficient of kinetic friction is 0.15, calculate (a) the acceleration of the system and (b) the force that each crate exerts on the other.
...?
Friction force = µ * mass * g Friction force of 75 kg = 0.15 * 75 * 9.8 = 110.25 Friction force = 0.15 * 110 * 9.8 = 161.7
(a) the acceleration of the system: Total force = total mass * acceleration Total force = 730 – (110.25 + 131.7) 730 – (110.25 + 131.7) = 185 * acceleration Acceleration = [730 – (110.25 + 131.7)] ÷ (75 + 110) = 2.638 m/s^2
The force that the 75 kg crate exerts on the 110 kg crate causes the 110 kg crate to accelerate at 2.638 m/s^2. Force = 75 * 2.638 = 197.85 N
Newton’s 3rd Law: Every action has an equal action in opposite direction!!
The force that the 110 kg crate exerts on the 75 kg crate = 197.85 N in the opposite direction. If the force that the 110 kg crate exerts on the 75 kg crate is +197.85 N, the force that the 75 kg crate exerts on the 110 kg crate is -197.85 N.
