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Weight of nothing in an elevator (Posted on 2018-09-06) Difficulty: 3 of 5
Take an electronic kitchen scale, a scale with a spring dial, and a balance scale into an elevator.

Put nothing on them. Indeed, press the tare button, or adjust them so that they all read zero.

When the elevator accelerates upwards, what happens?

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Spring dial scale: The plate above the spring has mass m. The spring is normally compressed such that its restoring force is balanced against mg, where "g" is the acceleration of gravity. The acceleration "a" of the elevator adds to g to put more downward force on the plate, making this force m(g+a). The spring will compress sufficiently to balance this added force. The displacement will swing the dial to positive weight. For example, if the elevator were to accelerate at "g" (like a jet), the spring would contract twice as much, and the scale would display the weight of the plate above it.

Balance scale: There will be a uniform increase of force on both trays m(a+g) which will cancel. (No registered weight.) If it is a triple beam balance (common in school labs), again the same downward force (ma) is added to the mass distributions on each side of the fulcrum, thus maintaining equilibrium. 

Electronic scale: The tray is held up by pegs at its four corners to distribute the load uniformly. In normal use, placing an object in the center of the tray will deform the tray with an added center force, and the warped tray will impinge on the load cell beneath the tray. The strain gauge therein (consisting of metal tracks bonded to a printed circuit board) will likewise flex, deforming the tracks and changing their resistance, which is measured electronically with a Wheatstone Bridge circuit. That is in normal use. 

As for the case of an electronic scale in the accelerating elevator: acceleration will apply added force uniformly to the tray and will not thus deform the center and so no added strain will occur. (Zero added weight). 

One caveat: four pegs around a tray do not actually completely distribute the load of the tray uniformly. (Metal is not infinitely stiff). Just as a dense enough tray will deform and sag in the middle, enough acceleration on the tray will make it begin to sag in the middle and register strain on the load cell and strain gauge. This deformation is not due to a non-uniform mass distribution on the tray, but rather the non-uniform support under the tray, and so an added weight measured will be far less than "ag" and will depend on the stiffness of the tray.

Also, see the additional comments added later in this thread in the case there is a tray located above the flat surface of the unit. 


Edited on September 7, 2018, 12:46 am
  Posted by Steven Lord on 2018-09-06 11:02:47

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