0
$\begingroup$

For this question I would like to define weight as a scale reading. I have a conceptual question: if I leap off a scale, according to my definition of weight the scale will claim I am heavier than normal as I am pushing off. But this is just a reflection of the surface of the scale providing a force greater than the force of gravity acting on me... which is what allows me to become airborne. In particular, if I were to substitute the force of gravity working on me from the scale reading, I would recover the total force on me and hence have a handle on my exact acceleration (after scaling by my mass). Do you agree? Just a basic sanity check. I would appreciate any thoughts (teaching freshman physics and I'm a bit disturbed by a book problem which describes such an event by implying that the lab scale reading would be the total acceleration read from the scale: I claim that would be impossible, unless you had a computer in the scale that knew your mass and the force of gravity; because in reality the scale only knows the upward force it is exerting on you, not the total force on you). May I ask for agreement or correction?

$\endgroup$
0
$\begingroup$

...implying that the lab scale reading would be the total acceleration

I think the implication here is that the reading of the scale will correspond to the weight of $m(g+a)$.

$g+a$ here, not just $a$, is being referred to as the total acceleration. So, the scale, does not need to know the total (net) force on you. This matches your understanding that your weight will appear greater than normal.

$\endgroup$
1
  • $\begingroup$ I totally agree with you. The weight reading from the scale will be m(g+a). And as you said, the scale does not need to know the total force on me. If there was a motion detector on the person jumping, it would read the 'a' from m(g+a). My difficulty is that the book seemed to claim that the acceleration data given was the "motion detector a" (and that the data was read off the scale). But I think that is just a mistake in the book, to the extent that the book was making the claim that I perceive. $\endgroup$ – okcapp Oct 23 '18 at 16:43
0
$\begingroup$

When you are standing on the scale, you exert a downward force of $mg$ on the scale and it exerts an upward force of $mg$ on you. The net force is zero and you are not accelerating.

Now imagine you are on the scale in an elevator. The elevator undergoes free fall. You are weightless. A force of -mg is exerted on you by gravity, but the scale reads zero. So you are accelerating at $-g$.

Next you leap off the scale and the scale reads, say, twice your weight. So the scale pushes up on you with a force of $2mg$. The force of gravity on you is $-mg$. The net force is $+mg$. Since your mass is constant, the scale is telling you that you are accelerating $+g$,

So I believe the scale can be used to give you your acceleration as a fraction or multiple of $g$

Hope this helps.

$\endgroup$
2
  • $\begingroup$ I agree with all of this. But this means that I have to put in the effort to subtract mg from the scale reading (and divine what remains by my mass) before I actually have my acceleration. Therefore, to the extent that the book claimed that the scale reading directly was a measurement of my acceleration, the book was wrong. I have no intention at all to make an issue out of a "typo" in a book, and I would never mention which book it is: My only issue is checking my claim before I discuss the problem with the students. $\endgroup$ – okcapp Oct 23 '18 at 16:49
  • $\begingroup$ @okcapp. I see what you mean. Maybe the book should have stated that the scale reading can be used to determine your acceleration, not that the reading is a measurement of your acceleration. $\endgroup$ – Bob D Oct 23 '18 at 17:12

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.