Why don't weighing machines show the weight of the atmosphere? The school experiment of hanging two balloons, one filled with air, other empty, on a ruler, show that air has weight.
A person has weight. It's shown on a weighing machine. A pencil has weight. It's shown on a sensitive weighing machine.
None of these machines show the weight of air even under an open sky. Why?
I researched online. Found the following:

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*Air column under an open sky when standing on the surface of the Earth is heavy enough to be shown even on the large weighing machine.

So, both of my weighing machines should show the weight of of the atmosphere.


*Air pressure inside the weighing machines balance the air pressure / weight outside.

That do not seems right. Air inside machine do not change by taking it out under open sky. Its pressure obviously cannot change thus. How do it balance both the air weight in room and under open sky?
Also, if it do where the extra energy came from?
Also, air molecules in the weighing machines have to vibrate more to have more pressure, right? How else would the increase in pressure manifest / happen? More vibration means more temperature because thats what temperature is. We observe no such thing.
I am aware of action reaction phenomena. If I push wall, wall push me back with equal and opposite force. However, this cannot solve this because then the weighing machines would not show weight of me and pencil.


*An answer to this question here says pressure is equalized on all sides.

I do not see how this can help even if true my weighing problem. Even if I am squeezing my weighing machines at all sides the upside (where scale is) would show the weight. The body of the weighing machine made of steel and thick enough can handle that pressure at sides.
The pressure is not equal at all sides. (i) There is no pressure from ground direction (upwards). (ii) Sides of my yard where I stand under open sky are much shorter than air length above me (100 km vs 20 ft).
I am careful in changing only one variable. The location of the machines. My friend did the opposite experiment with me at the same time and about 2 feet from me. He got himself weighted under open sky and in the room. His weight comes equal at both places. Why?
I can see that it is something with confining. Air in a balloon is exerting less pressure upwards to counter its weight. May be due to equal sideways and upwards dimensions. The air-filled balloon is somewhat like a ball in my experiment, more air pressure is going sideways than upwards. I filled an oval-shaped balloon with air, hang it with my hand long side up. It quickly fall to side. So, less long dimension get more pressure. Length of atmosphere upwards is just 100 km, breadth is 40,000 km at least at one side wherever you are on earth and on both sides if you are at equator. So, this do seems like the reason why air show no weight on my weighing machine. The pressure of air upwards is more than its weight downwards. Is that it?
This does not explain why sun has fusion though. Where do the pressure comes from if gravity does not work that way? You see if air on earth given column of just 100 km can cancel its own weight plasma in sun which is much more vibrant and have much larger dimension to move into would easily cancel its own weight. Then there would not be enough pressure due to gravity (weight) to reach temperature of millions of degrees of centigrade to start fusion.
Why do observation show air is weightless?
 A: For a gas in a container, if volume is increased - ofcourse cubicly because its volume - then surface area increase squarely. However, this do not mean now cubicly less number of molecules hit the walls in same time.
Consider your country expanding in all 3 dimensions such that all distances are now doubled. If you travel between two cities your distance is now only linearly larger, not squarely or cubicly.
If your room is made 27 times bigger in volume while keeping the ratios between length, breadth and height, then whereever you are in the room it will take you 3 times than before to reach the nearest wall. Randomly moving objects in the room will hit the walls 3 times less often.
Molecules now take linearly more time to hit the walls.
Its because gas is a fluid and molecules are too small and too few compared to volume, they do not block each others' path. Infact an ideal gas is taken as having zero-sized molecules.
We ofcourse assume that each molecule is doing the hitting with equal force than any other molecule. So its just number of molecules that hit that we have to keep track of. We always have sufficient number of molecules, due to very small size of them, in these kinds of experiments, for this assumption (really averaging) to work.
By increasing the volume we decrease the totality of force the molecules are hitting the walls, only linearly. Surface area that take the hits is increased squarely. Since pressure is defined as force per unit area the pressure decrease cubicly therefore proportional to change in volume.
P = F/A
If F decrease, P decrease. If A increase, P decrease.
Two containers of different volumes having number of molecules of same gas proportional to the volumes would have same pressure on their walls.
PV = nkT
Suppose 1 unit of volume have 40 molecules.
For volume = 10 units
P = (40 x 10 x k x T) / (10)
  = 40kT

For volume = 500 units
P = (500 x 10 x k x T) / (500)
  = 10kT

So, as long as temperature is same in both containers, different volumes of same gas exert same pressure.
(Note that this is different exercise than above, with same results. In the first exercise, we take one container and increase its volume. We see how its pressure changed. Changing the volume change temperature there. In second exercise we dont change any volume. So there is no change in temperature).
This is how air under the weighing machine is able to compensate against air above the weighing machine as far as pressure is concerned.
When your machine is in room the air under it dont put same pressure upwards as when the machine is out in open. The air under the machine is confined at one side by the machine and at other by ground. However, it still has four other walls. In the room its the walls of the room. In the open its the walls of your yard. The volume dont matter though as explained in above paragraphs. This is to answer your confusion about apparently same thing being equal to two different things. Both sides of equation change when you take the machine out in open.
This also explains why you dont suddenly feel increase in air pressure whenever you go out in open from a room.
If you totally seal air under your machine then you have decreased its volume. This will increase the pressure its exerting in all directions including the upward direction. The pressure in upward direction would always match with pressure of air above in downward direction. Sealing the air under the machine therefore do not change pressure upwards.
You can make a box of as light material as you want. It will never break as long as it has same number of molecules of air in it as in equal volume of air outside the box.
Zepellins had passengers area thats not air tight. Windows kept open for days'  long flight. Inner pressure equal to outer pressure ofcourse. But the point is, inner pressure would still be same as outer pressure if windows were closed. Its same as if you close windows and doors of your room pressure dont increase. Zepellins were slow moving.
Modern airline planes have more air pressure inside than outside. One reason is they fly at higher altitude where air per unit volume is less than at ground where plane's windows and doors are shut. If a hole happen in such a plane then air will rush out but not all the air will leave. As soon as air inside has same number of molecules than air outside air rush outside will stop. Then it will be normal air flowing as air flow from your room window plus some because of movement of the plane.
None of this explain why the weight of air is not shown on the machine. Air pressure dont compensate for air weight. Air pressure at one side precisely balance the air pressure at other side.
A: All your examples (humans, pencils, etc) are solid bodies. Air is a fluid. It acts differently on surfaces. Basically, unlike solids, it does not preserve the direction of the applied force. A solid, when pushed down on, transmits this force to lower objects as a force pointed downwards. Fluids act differently. For instance, if you take a U-shaped tube, fill it with water and push on the water down in one arm (e.g. through a movable piston), the water will produce/transmit an upward force in the other arm.
Similarly, as @gsomani said, when you try to weigh an air column, you have to make sure that that column has no access to the other side of the weighing plate, as in that case, in addition to the downward force it exerts on the plate from above, it will also push on the plate from beneath (cf. with the U-tube example). Provided the plate is not too thick these two forces will cancel each other almost perfectly.
If you really want to weight an air column, you have to make sure it has access only to one side of the probing surface. One possible way is to use for this purpose the surface of the water in the U-tube discussed above. Make sure the air doesn't push on the other side of the U-tube by sealing it, and voila, you a weighing scales for air. If you do this, you will find out that the entire atmospheric column from the surface of the Earth up to stratosphere and above weighs roughly as much as 10 meter-high column of water with the same footprint area. 10 meters-high U-tube is not an easily thing to maintain, so you can make it shorter by using more dense fluid, like mercury. This is exactly what Toricelli (a student of Galileo's) did to literally "weigh atmosphere". He found that the atmospheric column weighs roughly as mercury column of $\approx 780$mm high (with the same footprint area).
A: I have a partial answer.
For ideal gas, pressure is PV = nkT
where V is volume, n is number of molecules of the gas, k is a constant and T is temperature in kelvin.
Approximating behavior of air under scale (weighing machine) to the equation we get a value for upward pressure of air.
The value do not change when you take the scale outside. The value depends only on volume under the scale, number of molecules in that volume and temperature. None of this change by taking the scale outside.
If one wants to be pedantic he can argue that temperature inside and outside room vary. It dont vary significantly though -  especially when measured in kelvin. Also, the same experiment done in winter can be repeated in summer, with same result, when where temperature is larger switch between inside room and outside.
Since the upward pressure of air under scale is not enough to compensate downward pressure of air above the scale, neither in room nor outside, this cannot be the reason of not showing of weight of air by the scale. There is simply not enough upward pressure. Downward pressure of air is always much, much larger.
We can therefore rule out all explanations  based on pressure gradiant. Note that pressure gradiant is only between upper surface and lower surface of the scale. Each parts of a surface get same pressure as any other part of the same surface.
A: It is because air is not a single column just above the weighing machine. There is air all around it including below the weighing machine which leads to net upward buoyancy force.
The buoyancy force is very small compared to its weight due to air being much less dense(about a factor of 1000) than weighing machine. It would be equal to 10g for a 10kg weighing machine.
So, atmospheric pressure, in fact, causes net upward force (and not downward).
A: They ideally do measure the weight of the atmosphere.
But this measurement is set to 0 so that it is convenient for us to use.
If you took a scale to a vacuum, where there is no air to measure its weight, and set that measurement to 0 - Then when you get it back to normal conditions, under the air, it would measure the weight of air.
But this is useless in our life, so we set the initial reading to zero.
However for mechanical scales with plates, as someone already explained, air can get under the plate, and it exerts the same force pushing the plate upwards as the air above it exerts pushing it downwards, so atmospheric pressure would not matter all that much.
As to how this works, lets take a look at this -
Heres a block of air, stationary. There is force exerted on it from all directions. The force from up is the weight of the air above it. But this does not push the block of air down? Therefore there must be some some force acting upwards to balance that weight. And this is a properties of fluids and pressure, where in the deeper you get in a fluid , the more pressure is exerted to balance the fluids weight.

Now imagine that we replace this block of air with a block of steel. The force the air exerted on it is the same as what was exerted on the block of air, so everything pretty much balances out in this case. (For a more dense fluid, bouyant force would be considerable, but in the case of air it is negligible when replaced with steel)
And then if we consider this block of steel to be the plate, the plate does not move even if you change the atmospheric pressure. (apart from the effect gravity would have on it regardless of air)
