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    brijmohan123456's Avatar
    brijmohan123456 Posts: 41, Reputation: 1
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    #1

    Feb 12, 2013, 02:38 PM
    Reaction R in lift problem?
    In one of the physics book I read the following article on variation of weight in lift.I request uyou also please see it & answer my query.
    It was written that
    "In a lift the actual weight of the person=mg.This acts on the weighing machine which offers a reaction R given by the weighing machine & this is the apparent weight of the person".
    After this the variation of R was given in different case.Here I want to ask that as R is the reaction force given by machine in response to actual weight of the person=mg,so it must be same but how it can be different.I was not able to understand mathematically so can anyone please explain it to me logically.
    thank you
    ebaines's Avatar
    ebaines Posts: 12,131, Reputation: 1307
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    #2

    Feb 12, 2013, 03:04 PM
    Since you mention a lift I assume that the question has to do with how R changes under vertical acceleration or deceleration. Keep in mind the famous equation: . The forces acting on the person's feet are his weight pressing down (weight = mg) and the reaction force of the scale pressing up (R).

    Case 1: If R = mg then the sum of forces is 0, and from F=ma that means acceleration is zero, and so either the lift is stationary or traveling at a constant rate of speed.

    Case 2: If R is greater than mg then the sum of forces is positive in the upward direction and the man must therefore have positve acceleration upward. Convesely - if the man has positive acceleration then R > mg. Note that positive acceleration could mean either his upward velocity is increasing or his downward velocity is decreasing.

    Case 3: If R < mg then the reaction force is not sufficient to hold the weight of the man and he accelerates downward. Conversely - when the car accelerates downward R < mg.

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