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21 January, 05:41

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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  1. 21 January, 08:20
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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.
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