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The scale weighs everything on top, including the air. So if you turn on the suction cups, and they suck vacuum (i.e., air is taken out), the scale will give a fraction less weight (I didn't calculate it, but it will probably be a few ng to a few µg. That scale sees not at all, but this was about the idea.

But isn't the air under your feet gone, so that the upward force also disappears immediately and you (or at least what the scale indicates) actually becomes heavier? The buoyant force is equal to the weight of the displaced substance. But does this equal the weight that has been lost because there is also less air under your feet or is the effect of an upward force larger? (the effects would be very small but that doens't matter).

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    $\begingroup$ This is an experiment you can do for the price of a suction cup! Why not let reality answer you, then try to fit theory to results? -With help from here if you're unsatisfied with your attempt. $\endgroup$
    – g s
    Commented Jul 17, 2022 at 17:13
  • $\begingroup$ The scale does NOT weigh the air on top of it. All of the air above that scale exerts a force of 2100 lb/ft^2 at sea level! $\endgroup$
    – David White
    Commented Jul 18, 2022 at 1:56

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A person in air is subject to a buoyant force equal to the weight of the air they displace. A person who is neutrally buoyant when swimming underwater has the same density as water. Their volume is therefore $V=m/\rho_\text{person}≈m/\rho_\text{water}$, and the buoyant force on them in air is $F_B = mg\frac{\rho_\text{air}}{\rho_\text{person}}$. Buoyancy reduces your weight on a scale by about 0.12%.

In water you have probably noticed that having a lungful of air increases the buoyant force on you and makes you tend to float better. This is because a lungful of air reduces your average density. Likewise, an evacuated suction cup reduces your average density compared to an air-filled suction cup.

However, the material which makes up the suction cup is heavier than the air which the evacuated cup displaces, so adding suction cups will increase the force you exert on the scale, which the scale reports as your weight. (A person with evacuated suction cups would exert less force on a scale than a person with air-filled suction cups, by the weight of the air.) To decrease your weight using buoyancy, you need to attach an accessory which is actually positive-buoyant in air, like a helium balloon. The non-existence of a positive-buoyant “vacuum balloon” is a famous engineering problem.

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  • $\begingroup$ But doesn't the vacuum between feet and scale decrease the upward force so you become heavier? (In my example you can forget the extra weight of the cups itself; before and after you made them vacuum the mass is the same (exept the smal amount of air; but you can leave that to)) $\endgroup$
    – Marijn
    Commented Jul 18, 2022 at 17:40
  • $\begingroup$ Vacuum doesn’t exert a force. The scale is calibrated so that it reads “zero” when the air is pushing down on it. If you put some object on the scale that gets in the way of the air, the air pushes down on that object instead — except slightly less, because the air pressure at the top of the object is a little less. The difference is the buoyant force. Proving that the buoyant force is independent of shape is a homework exercise. $\endgroup$
    – rob ♦
    Commented Jul 18, 2022 at 18:20
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I think any real suction cup of appropriate size to fit under your feet will be heavier than its volume of air at sea-level pressure, but let's say you'd have suction cups made from a massless rigid material - in that case yes, the scales would indeed show you a slightly lower weight and the reason for that is buoyancy. With the massless suction cups on your feet and the air evacuated from them, you would displace a larger volume of air at otherwise the same mass, resulting in a larger upwards buoyancy force while the gravitational force would stay the same - thus the scales would read a slightly smaller weight.

In simpler terms: the evacuated massless suction cups would behave the same as a helium balloon and thus slightly pull you upwards.

Given a perfect vacuum in the suction cups, weight difference that the scales would measure would correspond exactly to the weight of additionally displaced air, so kg/l of air at current pressure multiplied by the volume of the suction cups. Eureka!

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It would make no difference.

Note that if a person stands on a scale part of him will be in contact with the scale whether the suction cups are on or off. Let us suppose he has suction cups on both feet that are always on. But sometimes he stands on one foot.

First a caveat. Air has mass and therefore weight. It gets confusing when you try to make sense of that weight because you don't put a piece of air on a scale to weigh it. Air is all around. The mass of a cubic meter of air is about 1 kg, so its weight is 9.8 N. This is small enough that we will ignore it. We could do this problem again taking this into account. It would give the same result.

Start with a scale that has nothing on it. It shows a weight of $0$. Such a scale has a plate where you put objects Air pressure exerts tremendous forces on the plate. Air on top presses down. Air on the bottom presses up equally hard. The net force from air is $0$. The plate itself weighs something, but the scale is built to show that weight as $0$.

A person with suction cups stands on scale with a built in trap door. The trap door opens and the person begins to fall. Air pressure presses on all sides. The net force of air pressure is $0$. The total force is the force of gravity, or his weight, $W$.

He stands on the trap door or the scale with no air under the suction cups. The suction cups cover regions $R_L$ and $R_R$ of the door. Normally, air would be exerting a force $F_L$ on $R_L$ and $F_R$ on $R_R$. Note that if the door was open, air would exert the same force upward on the bottom of the suction cups.

To hold the person still, the door must exert an upward force of $F_L+R_R+W$. The person exerts an equal and opposite downward force $F_L+R_R+W$ on the door.

Suppose he picks up his right foot. Now the downward force he exerts on the scale is $F_L+W$. But air now presses on $R_R$ with a force $F_R$. So the downward force on $R_L$ and $R_R$ is the same.

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  • $\begingroup$ Well actually if the material of the suction cups was massless, the scales would indeed show a slightly smaller weight. Reason: Buoyancy. Since the person with the (massless) suction cups would have a larger volume and thus displace more air than the one without at equivalent mass, the first one would have greater buoyancy, which is an upwards force and thus the scales would measure a slightly smaller weight. $\endgroup$
    – matthias_buehlmann
    Commented Jul 18, 2022 at 0:51
  • $\begingroup$ I was ignoring issues like the weight of air. $\endgroup$
    – mmesser314
    Commented Jul 18, 2022 at 2:57
  • $\begingroup$ but that's exactly what it is about when dealing with suction cups $\endgroup$
    – matthias_buehlmann
    Commented Jul 18, 2022 at 10:10
  • $\begingroup$ I misread the problem. I thought you meant a flexible rubber cup that could be flattened onto the scale. Trying to pull it away creates a vacuum. That takes a lot of force, so they stick you to the scale. Not useful for sticking to a scale, but helpful for climbing a glass wall. You had in mind a non-flexible cup that you could pump down like a bell jar in a vacuum chamber. Rob's answer is right. $\endgroup$
    – mmesser314
    Commented Jul 18, 2022 at 14:18

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