# Does Atmospheric Pressure Affect Weight Of Objects

The atmospheric pressure is around 100KPa which is 100,000 N/m2. If the gravitational acceleration is g = 10 m/s2, then it would be around 10 tons of weight of air per square metre!

1. Let's say that we have a thin sheet of metal weighing 10 kg. It is placed outdoors on top of a weight scale facing the sky.

2. The length and width are both 1 m and the thickness is negligible. Then we take the reading from the weight scale which is X.

3. Then place the sheet of metal vertical to the weight scale (i.e. perpendicular to the ground). Take the reading from the weight scale with is Y.

The question that I want to ask is does X-Y = 10 tons? I know this sounds crazy, but I couldn't understand why we apply atmospheric pressure when calculating fluids but not solids.

No. Air can get under the sheets of metal you have described, so the atmospheric pressure, which pushes in all directions equally, pushes up as hard as it pushes down. There is no net force.

If you had a perfectly sealed $$1m^2$$ suction cup and tried to pull it away from the surface, thereby creating a $$1m^2$$ cross section of vacuum under the suction cup, the atmosphere would indeed resist you with 10 tonnes of force. You would need a 10 tonne rated power crane to pull it away... or a pin to slide under the edge of the suction cup and break the seal before lifting it by yourself without effort. Note that this would be the case no matter which direction you were pulling, as long as there was air on one side and vacuum on the other. A suction cup on the ceiling takes as much force to remove (less the weight of the cup itself) as a suction cup on the floor.

From the Bernoulli equation for a constant-density fluid,

$$P + \frac12 \rho v^2 + \rho g h =\text{constant}$$

you can see that the “vertical pressure” $$\rho g h$$ at some point is the same as the weight of a vertical column of the fluid divided by the area of that column. A cubic meter of air at sea level has a mass of about a kilogram. If the atmosphere had constant density with altitude, you could get $$10^5\rm\,Pa$$ at sea level from a pile of stationary air with height $$10^4\rm\,m$$. A better model of the atmosphere has the density falling off exponentially as you rise; however the “scale height” of this exponential is in fact approximately ten kilometers.

That is to say, the sea-level pressure of $$10^5\rm\,N/m^2$$ occurs because a meter-squared column of air, from Earth’s surface to space, has a mass of about $$10^4\rm\,kg$$.

In your question about a mass on a scale, you don’t ordinarily observe an effective-mass difference between horizontal and vertical objects. If you were somehow able to remove the air between your horizontal plate and the scale, most scale designs wouldn’t change the effective mass either. However, if you remove the air between the scale and the plate, then try to pull the plate and the scale apart, you would find that it takes ten tons of effort to pry them apart. Normal people refer to this effect as “suction,” but it is actually the weight of the atmosphere trying to push its way into gaps. Note that suction forces can point in directions other than “down.”

The primary way that the atmosphere affects objects’ weight is via buoyancy. Consider a helium balloon, which has positive mass: if you tie it to a scale it will float up instead of pushing down, so its effective weight is negative. For objects with densities much greater than air (such as water, $$\rho_\text{water} = 10^3\,\rho_\text{air}$$), the buoyancy effect is frequently negligible.

The answer to your main question is yes. There is a reduction in the weight of all objects due to their buoyancy in the air, not just helium balloons, as there would be if they were submerged in water, except that the air weighs only approximately $$1.2kg/m^3$$ compared with water which weighs $$1000kg/m^2$$ so the buoyancy is correspondingly less.

But the answer to your second question is no. X-Y is not 10 tonnes. You have considered only the air pressure acting down on the metal sheet and forgotten the pressure of the air acting underneath it. If X-Y was 10 tonnes, the metal would have a very large negative buoyancy, but in reality there is an upward buoyancy equal to the difference in the pressure underneath the plate and the downward pressure above it. If the thickness really was negligible then the buoyancy would be negligible, but in general the buoyancy on an object, which reduces its net weight, is equal to the weight of the fluid (in this case air) that is displaced by the object (The Archimedes Principle).

Pressure in a fluid (liquid or gas) acts equally in all directions, so the air pressure doesn't add to the weight of your metal. However, there is a small reduction in the weight of the plate due to buoyancy.

To get some solid numbers, let's assume that your plate is made of iron. The density of iron is ~$$7800\,\rm kg/m^3$$, that of dry air at $$20°$$C is ~$$1.204\,\rm kg/m^3$$. Hence the $$10\rm kg$$ plate is ~$$1.282\,\rm mm$$ thick. The same volume of air has a mass of ~$$1.544\,\rm g$$. So the effect of buoyancy on your plate is quite small, but easily measurable with sensitive scales.

Not the pressure but the density of air impacts the weight of objects due to buoyancy.

• This is wrong. The buoyant force is caused by the pressure difference between the top and the bottom of an object. The famous result that the magnitude of the buoyant force works out to the weight of the displaced fluid comes from the Bernoulli equation for the pressure; there is a buoyant force on an object of any density immersed in a fluid of any density.
– rob
Jan 20 at 0:20
• @rob density impacts weight via the pressure gradient induced by the gravitational field Jan 20 at 7:34