0
$\begingroup$

From what I understand, weighing scales measure the normal force acting on the person on the scale. They do this by measuring the displacement of a spring with a known spring constant. Shouldn't the weight of the person pressing down on the scales be responsible for this displacement, not the normal force?

$\endgroup$
  • $\begingroup$ The normal force is the force exerted by the person's feet on the scale. From a free body diagram on the person, this is also equal to his weight. His weight is the force that the earth exerts on him, not the scale. $\endgroup$ – Chet Miller Dec 22 '18 at 17:21
1
$\begingroup$

From what I understand, weighing scales measure the normal force acting on the person on the scale.

You understand correctly.

There are many types of scales and a lot of them actually measure the value of a downward force.
(Scales are often calibrated in kilograms by placing known masses onto the scales or by using a value of the local gravitational field strength and the known characteristics of the spring the force can be converted into a reading of mass.)

A downward force equal to your weight, compresses a spring inside the scale and that spring then exerts an upward force on you creating a force (the normal reaction) which is equal in magnitude to your weight so the scale is in static equilibrium.

Whilst on the scales if you decided to jump up the pan would start to move down, the spring would be compressed more and exert a greater upward force than your weight because you are in the process of accelerating upwards.
So in this case the force on you due to the scales is greater than your weight and the reading on the scale (the value of the normal reaction) will be the value of the force that the scale exerts on you.

If you were standing on the scales in an elevator which was in free fall the reading on the scales (the value of the normal reaction) would be zero.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ How about scales with two pans: one with the sample and the other with reference weights. Apart from the possibility of a tiny variance in gravity between the two pans, aren't these measuring mass? E.g. if I take this type to the Moon, it will say that I am 70kg. $\endgroup$ – badjohn Dec 22 '18 at 20:40
  • $\begingroup$ @badjohn Yes that type is comparing masses via the comparison of their weights but the OP specifically mentioned a spring type balance and that is why I wrote about type. $\endgroup$ – Farcher Dec 22 '18 at 21:04
  • $\begingroup$ So he did - sorry. $\endgroup$ – badjohn Dec 22 '18 at 21:12
  • $\begingroup$ @badjohn You have nothing to be sorry about having asked a perfectly reasonable question. $\endgroup$ – Farcher Dec 22 '18 at 21:48

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.