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7 September, 06:03

You take a ride in a fast elevator to the top of a tall building and ride back down while standing on a bathroom scale. During which parts of the ride will your apparent and real weights be the same? During which parts will your apparent weight be less than your real weight? More than your real weight? Sketch freebody diagrams to support your answers.

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  1. 7 September, 07:37
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    Your apparent weight will be the same as your actual weight when the elevator is going up or down at a constant speed. Its apparent weight will be less than its actual weight when the elevator accelerates while moving down. Its apparent weight will be greater than its actual weight when the elevator decelerates while moving down.

    Explanation:

    When we enter an elevator, according to its movement we can feel different sensations. Recalling that according to Newton's 1st Law, the body, by inertia, tends to maintain its state, be it at rest or from MRU. And according to the fundamental principle of Dynamics, the resulting force (FR) can be calculated by FR = m. a, where m is the mass of the body and a is the acceleration developed by it.

    When the lifter is stopped or going up and down at constant speeds, the normal force applied to our feet is equal to our weight force, because the only acceleration we are feeling is gravity. The resulting force between normal and weight is zero. FR = 0 - > FN = P. In this case, our apparent weight is equal to our real weight.

    When the elevator is descending and decelerating, apply an upward force to stop. FN> P - > FR = FN - P as a result, your apparent weight becomes greater than your actual weight.

    When the elevator is descending in an accelerated way, its resultant is facing downwards. P> FN - > FR = P - FN as a result, your apparent weight will be less than your actual weight.

    P = real weight

    FN = normal force

    FR = resultant force
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