Question

Two crates of mass 75 kg and 110 kg, are in contact and rest on a horizontal surface. A 730 N force is exerted on 75Kg crate . If the coefficient of kinetic friction is 0.15 , calculate the acceleration of the system, and he force that each crate exerts on the other.

Answer 1

Dear Student ,
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/s²

The force that the 75 kg crate exerts on the 110 kg crate causes the 110 kg crate to accelerate at 2.638 m/s².
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.
Regards