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30 May, 21:42

An elevator (mass 4500 kg) is to be designed so that the maximum acceleration is 0.0680 g. what is the maximum force the motor should exert on the supporting cable?

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  1. 30 May, 21:50
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    For this case what you should do is to start from the definition that force is equal to mass by acceleration.

    We have then:

    f = m * a

    where,

    m = 4500Kg

    a = 0.0680g

    g = 9.81 m / s ^ 2

    Substituting the values:

    f = (4500) * ((0.0680) * (9.81))

    f = 3001.86 N

    answer:

    the maximum force the motor should exert on the supporting cable is

    f = 3001.86 N
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