When a mass is dropped onto something like a bathroom scale, the reading on the scale temporarily exceeds the actual weight of the mass. How do I explain this using forces and a force body diagram? Also, let's say instead of a mass and a scale, its just a person, a ball, and a scale. The person is standing on the scale with the ball in hand and throws it up in the air. When the person catches the ball, should the scale also read a value greater than the weight of the human and the ball combined? Is the reasoning for this the same as the mass and scale example?

Edit: Could the explanation be that at the instantaneous moment when the mass comes in contact with the scale, there is an instantaneous force caused by the impulse?

  • 1
    $\begingroup$ Force = mass * acceleration. The scale can de-accelerate the mass quickly $\endgroup$ Mar 7, 2018 at 4:52
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    $\begingroup$ Thin of the dp/dt for of force, an impulse. It adds or subtracts during dt. $\endgroup$
    – anna v
    Mar 7, 2018 at 4:55
  • $\begingroup$ Have you read about variable mass systems?? $\endgroup$
    – Jnan
    Mar 7, 2018 at 6:22

1 Answer 1


Forces must always balance. A force is required to support a stationary mass on a bathroom scale. An additional force is required to effect the deceleration of a mass if it has vertical downward motion as it makes contact with the bathroom scale surface.

Dropping the mass onto the bathroom scale: $$F = mg + ma \tag1$$ where m is the kg mass of the mass dropped on the bathroom scale, g is gravitational acceleration, and $$a = \Delta v/\Delta t \tag2$$ where $v$ and $t$ are velocity and time. The maximum $a$ determines the maximum force indicated on the scale. The heavier the ball, the harder the scale surface and the stiffer the scale's spring, the higher $F$ will be (figure below).

When the person is catching the ball, the person is as the stationary mass above, and the ball has a stationary and decelerating component. See figure below.

mass dropped onto scale


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