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Today, 15:30

A 36.9-kg crate rests on a horizontal floor, and a 64.9-kg person is standing on the crate. Determine the magnitude of the normal force that (a) the floor exerts on the crate and (b) the crate exerts on the person.

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  1. Today, 16:08
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    FN = 997.64 N : Force that the floor exerts on the crate

    Fcp = 636.02N, vertical and upaward

    Explanation:

    Wp: person Weigth (N)

    Wc: crate Weigth (N)

    FN: Normal force

    Calculated of the weight

    Wc = (36.9kg) (9.8 m/s²) = 361.62 N

    Wp = (64.9kg) (9.8 m/s²) = 636.02N

    (a) Force that the floor exerts on the crate

    For free body diagram of the crate+person

    ∑Fy=0

    FN-Wc-Wp = 0

    FN-361.62 N-636.02N = 0

    FN = 997.64 N : Force that the floor exerts on the crate

    (b) For free body diagram of the crate

    Fpc: Force of the person on the box

    ∑Fy=0

    FN-Wc-Fpc = 0

    997.64 N-361.62 N = Fpc

    Fpc = 636.02N : Force of the person on the crate (vertical and downaward)

    Newton's third law or principle of action and reaction:

    Force of the person on the crate = - Force of the crate on the person

    Fpc = - Fcp

    Fcp:Force of the crate on the person

    Fcp = 636.02N, vertical and upaward
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