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If we remove the atmosphere, will our weight be the same ? So why isn't our weight = gravity + pressing force from atmosphere. The air is pressing us! Bonus question: do we need to include atm when calculating buoyancy?

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Pressure acts in all direction on us. Pressure pushes us from the top, it pushes up on us from the bottom. It pushes us from the left and on the right! So unless the object is significantly high and you require the measurement to a high degree of accuracy, you can ignore pressure from atmosphere. I cannot generalise and say that we can ignore atm, but in most cases since buoyancy is the weight of water displaced, we can ignore atm. This is because buoyancy is the difference in pressure that causes this upthrust, and both the "top" and "bottom" pressure terms have the same atm term, which cancels out.
weight=m*g atm pressure has nothing to do with it nor with bouyant force

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@mr.singh is correct. Weight is the force due to gravity. Thus, weight does not change with changing air pressure. Apparent weight, however, will change insignificantly. Look at my picture, and assume the gravitational field is just as strong for both people. The dude on the mountain carries the weight of less air, and so a very accurate scale will register his weight and the air's. The dude on the island holds up a lot more air, so a scale would measure his weight and even more air. If the guy on the mountain climbed down, he might gain apparent weight on the way! I don't know what air densities and gravitational fields exist on mountains versus islands, so I assumed the gravitational field was the same.
|dw:1374704756634:dw|That one's too tricky for me to say for certain.
The atmospheric pressure will be negligible, I'd think, but I can conjure up the math.
can't* conjure up the math.
Archimedes Principle applies to any body displacing a fluid (gas) There is a buoyant force in the atmosphere that is why lighter than air objects float (hot air , methane) A human who weights about 170 lbs displaces about 2.5 cubic ft .(67 L) of air. That amount of air is about 40 gms. equivalent to a buoyant force of about 0.1 lbs. a 0.06% correction if I did my math correct
The correction is not based on the pressure of the atmosphere on the object. just on the weight of the mass of substance displaced.
Gravitational force of attraction is, \[F=\frac{ GMm }{ R ^{2}}\]. The mass of Earth is a constant (M=6*10^24Kg) and so is Gravitational constant (G=6.67*10^-11). the mass of earth is calulated only upto it's land surface level. It doesn't include the atmosphere. The mass of a substance purely depends on the content (m=our body). The distance between the bodies (R) is not influenced by the presence or absence of atmosphere. So the atmosphere has not impact on our weight ! @Dmitt
The presents of the atmosphere has no impact of the force of gravity true but the apparent mass is reduce because of the buoyant force of the atmosphere. How do you explain the floating of balloon in the atmosphere? They have mass and weight when deflated but it disappears when inflated and becomes negative. I like to make a correction to my previous post. Air is composed of diatomic molecules and thus has twice the mass I gave. Thus our 170 lb human displacing 67 L of atmosphere will be buoyed by a force equal to the weight of about 80 gm of air or about 0.2 lbs as read on a scale.
Hi! I'm not sure I know a lot of this, but would the apparent weight of a person be \(\sf{increased}\) if he or she is in a less dense substance, @gleem ? This is in response to "but the apparent mass is reduce because of the buoyant force of the atmosphere." I'm not sure! Thanks!
It may have been ill advised to use the term effective mass. If the fluid were less dense the effect would become smaller. The fact is that the scale weight of an object is greatest in a vacuum compared to in the atmosphere. Gravity is counteracted by this buoyant force and of course it is dramatic in water.
|dw:1374783468455:dw|So, the person will have the greatest apparent weight in a vacuum, because there is no buoyant force. I think I found out I used the term "apparent weight" incorrectly. Thanks, @gleem!
Another picture, that doesn't involve... Well, that person under miles of water in that last pic is no longer with us. It's okay, that person was fictional. No offense meant to fictional characters.|dw:1374784009861:dw|So apparent weight is defined by the normal force. Or, maybe, the force in the direction of the decreasing gravitational field that has the normal force as a component, like in the case of an incline. My point was that the person was carrying the weight of the atmosphere. If that was so, then the normal force on the person would be greater than that of the person alone. Even if that was so, the apparent weight would probably be attributed to the person and the air... But maybe my problem in reality is not understanding fluids. If a scale is placed at the bottom of the ocean - just a solid spring scale, what would it register, assuming it was calibrated out of the water, at sea level, on my island?
Granite has a density of 169 lbs/cu ft So a 1 cu ft block in water totally submerged is buoyed by a force equal to the weight 1 cu ft of water. So for salt water that is 64.4 lbs. Thus totally submerged the 1 cu ft of granite will weight 104.6 lbs.

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