2
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ Force = mass * acceleration. The scale can de-accelerate the mass quickly $\endgroup$ – Martin Beckett Mar 7 '18 at 4:52
  • 1
    $\begingroup$ Thin of the dp/dt for of force, an impulse. It adds or subtracts during dt. $\endgroup$ – anna v Mar 7 '18 at 4:55
  • $\begingroup$ Have you read about variable mass systems?? $\endgroup$ – Jnan Mar 7 '18 at 6:22
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.