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.

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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.