When we stand on a weighing scale the reading we get is in $\mathrm{kg}$. Does it refers to mass or weight?

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    $\begingroup$ Rule of thumb, if you think the measurement device would say something different on the Moon as compared to Earth, it's probably weight and not mass. $\endgroup$
    – Jim
    Commented Jun 6, 2013 at 14:30
  • $\begingroup$ Essentially a duplicate of physics.stackexchange.com/q/43195/2451 $\endgroup$
    – Qmechanic
    Commented Jun 6, 2013 at 14:39
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    $\begingroup$ Depends on how your "scale" works. Spring deflection and strain-transducer based systems read force. Balance arm devices and harmonic oscillators read mass. @Jim's rule is very good. $\endgroup$ Commented Jun 6, 2013 at 16:11
  • $\begingroup$ Oddly this is at once too vague (what exactly do you mean by scale) and too specific to be very localized (once you specify a mechanism the answer is trivial). I'm going to close it for now, but am open to having my mind changed if someone thinks I'm way off base here. $\endgroup$ Commented Jun 6, 2013 at 20:41
  • $\begingroup$ It's actually the mass you observe on a weighing scale. A weighing scale feel your normal force. This force in Newtons is calibrated to kg. So if we put g=10 m/s^2, the scale is calibrated to 1 kg for 10 N of force. $\endgroup$
    – UKH
    Commented Apr 14, 2016 at 3:20

2 Answers 2


I disagree with Will. In all cases I could conceive, the scale directly measures the Normal force acting on you. For example, if you are in an accelerating elevator, the scale would read whatever your calculated normal force is.

Since I'm currently studying Chemistry, I would like to add that chemists make no distinction between mass and weight. In fact, there is a conversion factor from $\mathrm{kg}$ to $\mathrm{lb}$: $$1\: \mathrm{kg} = 2.2046\: \mathrm{lb}$$

Of course, they never have to deal with cases where the normal force is NOT equal to the gravitational force acting on the object; still, it's incorrect to think of mass and weight (normal force) as interchangeable. It's almost as bad as thinking momentum and velocity are the same thing!

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    $\begingroup$ I'm not sure if this is officially standard, but it's fairly common to consider the pound (lb) a unit of mass, and the pound-force (lbf) as the corresponding unit of force. $\endgroup$
    – David Z
    Commented Jun 6, 2013 at 6:11
  • $\begingroup$ The official imperial unit for mass is the slug. Newton's second equation takes units of slugs and ft/s/s, thus $1_{lbf}=1_{slug}*1\frac{ft}{sec^2}$ $\endgroup$
    – Greg
    Commented Jun 6, 2013 at 6:17
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    $\begingroup$ "In all cases I could conceive" Then you didn't conceive of a beam balance. Nor of a vibratory or centrifugal measurement. It is really only things that depend on static compression for which the answers depend on the local value of $g$. $\endgroup$ Commented Jun 6, 2013 at 20:26
  • $\begingroup$ @dmckee I was referring to the kind of scale you stand on; all possible cases include when you are at rest, moving upward/downward at constant velocity, accelerating upward or downward. Perhaps I should have referred to the weight as the 'apparent weight,' that is, the weight that you feel. Indeed, when you are pulling g's in a plane, it is the normal force that affects how heavy you feel; the force of gravity remains constant. And you're right, I didn't think of a triple beam balance (I think it does in fact measure mass); furthermore, I've never heard of vibratory/centrifugal measurement. $\endgroup$
    – Greg
    Commented Jun 6, 2013 at 21:28
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    $\begingroup$ I guessed and was perhaps being intentionally difficult, but the question was bugging me. In orbit you can still measure mass by strapping the subject between two spring and observing the period of SHM, likewise you can measure the thrust on the bearing of a centrifuge to find the mass of a cargo. (You can probably think up several more schemes, but I know those have been tried..) $\endgroup$ Commented Jun 6, 2013 at 22:05

It measures the force your body exerts on the scale due to gravity. That is, it measures your weight force $F_w = mg$. A low-tech example of this is a spring scale which uses the scale displacement, $x$, due to your weight force and the known spring constant, $k$, to determine your mass via

$kx = mg \implies m = \frac{k}{g} x$

The scale then reads out this mass, $m$.

More technically, the scale measures the normal force acting on you from the scale pushing up on your feet. So, if you are accelerating with respect to the Earth's surface (example: elevator) or, you are under influence of a different acceleration due to gravity, $g\neq g_{E}$, (where $E$ stands for surface of Earth) or a combination of the two, the scale will not read out your mass, but taking the mass readout $m_{scale}$ and multiplying it by $g_E$ will tell you the normal force $F_N$.

So, starting again, in general take a situation where the acceleration due to gravity is $g$ and you are also experiencing an acceleration $a$ (where positive $a$ means acceleration away from the gravitating mass) we have a normal force

$$F_N = m~(g+a)$$

now the scales takes this weight and divides by $g_{E}$ to give its reading, so we have

$$m~(g+a) = m_{scale}~g_E~~\implies~~m_{scale} = \frac{g+a}{g_E} m = \frac{F_N}{g_E}$$

This is the value that the scale will display, which we see, isn't your mass unless $g+a = g_{grav}$.

We see that the answer initially given at the top is the limiting case of $g=g_E$ and $a = 0$, and have now given a more general answer.

I hope this clarifies any problems.

  • $\begingroup$ You should be explicit that the scale is calibrated (assumes) the acceleration of gravity at the surface of the Earth. That is, the scale measures force and then factors out what it assumes is the acceleration of gravity to compute the mass. $\endgroup$ Commented Jun 6, 2013 at 6:09
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    $\begingroup$ Will, I think that the non-constancy of $g$ was exactly the point of the original poster. When you move to a place with a different gravitational acceleration $g$, then - because the scale really measures $x=mg/k$, not $m$ directly, you will see $mg$, the weight, and not the mass (assuming that at least $k$, the spring constant, is constant). So a scale with a spring may pretend to measure the mass but it actually measures the weight and it will give wrong values for the mass on the Moon, for example, and because this dichotomy was the reason behind the question, your answer is just wrong. $\endgroup$ Commented Jun 6, 2013 at 6:56
  • $\begingroup$ I think Will's answer could be better written, but it is basically correct and the downvote is unfair. Will correctly says It measures the force your body exerts on the scale due to gravity. His final statement The scale then reads out this mass, m is correct for terrestial scales used on the Earth's surface. Maybe the limitations in his final statement could be explicitly stated. $\endgroup$ Commented Jun 6, 2013 at 7:18
  • $\begingroup$ Perhaps it could be but it certainly should have been, John, because this is what the question seems to be about! ;-) $\endgroup$ Commented Jun 6, 2013 at 9:59
  • $\begingroup$ Well, I think the point of the question is completely up to interpretation until the asker of the question tells us what he/she precisely means. I admit that I interpreted the question rather simply, being about measurements on Earth and gave my answer accordingly. I also agree that the measurement is of normal force, but I thought that with the simplicity of the question, I would give a simple answer. So yes, I could have been more general with my answer as it doesn't take into account possible measurements in an elevator or on the moon. $\endgroup$
    – Will
    Commented Jun 6, 2013 at 13:46

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