A question of physics

[ríomhaire]

I was just thinking: So if you had a very large, very tall object that one half was very smooth in order to give it a minimal surface area, and the other half had a huge surface area due to many folds, grooves, etc. would the weight of the object change as you change it's orientation, because of atmospheric pressure? Ie, would it weigh less when the half with the higher surface area was underneath than when the area with the lower surface area was underneath?

Note: Anyone who doesn't know the difference between mass and weight don't post in this thread.

 

[Tarkus]

I can't see how that could be, because pressure is felt in all directions, so the position of the object wouldn't matter. What you described would only happen if pressure was only felt in a negative vertical direction I think.

I'm problably wrong though.

 

[soulslicer]

Mass is constant. Pressure changes. Weight changes (Not weight in direct physics terms, but like if you place it on a scale. If there is a presence of fluidity or air or some upthrust force, stability will cause the area with sharper edges to be lighter, because the upthrust = vpg, where v represents the volume of air or liquid displaced. Thus, greater grooves, more area displaced, greater upthrust, Newton's third law, smaller reading of weight)

That is all.

 

[Tarkus]

Mass is constant. Pressure changes. Weight changes.

That is all.

Well sure, but if you change the placement of a certain object on earth, at a constant pressure, it's weight won't change

 

[soulslicer]

Well sure, but if you change the placement of a certain object on earth, at a constant pressure, it's weight won't change

I edited my post. Weight does not change in physics terms. Oh and if the object is a metal object, I was thinking..and it cuts the magnetic field of whatever planet it is, and a current is produced, and if it is the area where an e-field is present (naturally occurs in some places, times, conditions yes..), that would mean if the sharper edge area was to face the orientation sidewards, there would be a greater flow of current due to greater flux linkage, and a greater opposing electrical force, causing it to seem lighter. Yeah..

 

[Solaris]

No, weight = mass x 9.81 on earth

 

[xcellerate]

One end of the object is more massive than the other, meaning the CoG is not in the middle. And since the the force of gravity is one of those fancy /r^2 equations, the object will technically weigh less if you rotate it so the heavier end is up. Just imagine it at an extreme level, pretend you're holding a massless stick that's 30 miles long, you're end on earth has a tennis ball on the end, and the end poking out into space has a bowling ball on it. The stick will weigh tennis ball + % of bowling ball. Where as if you flipped it, the stick would weigh bowling ball + % of tennis ball. But with your example the difference will be about as extreme as how you become lighter when you walk into your house because the roof is pulling up on you.

or did I misunderstand what you were asking?

 

[Godron]

It might do because of Xcellerate's reason, but not because of air pressure I don't think. I think every neuton of air pressure pushing one way would be cancelled out by air pressure force in the opposite direction. I think it's the same as how if you had a hollow object full of compressed gas, the gas would exert a force outward in all directions, but the net force would be nil, and would not move the object at all.

 

[Cole]

Ah yes! but if the tree was 5ft tall and could blast off the ground and go 99.999999..........(infinite 9's)% of the speed of light
how tall would the tree look to you? why?
how much energy would the tree have?
Relative to the tree, how fast does it go through time?
If the tree could measure the speed of light relative to itself, how fast would it measure the speed of light?

Real questions!!!!

Orientation wouldn't affect it. Weight = Mass & Gravity.

 

[xcellerate]

Weight = Mass & Gravity.

but gravity changes the farther you are away from the earth. Honestly I'm just splitting hairs here, unless this was the most ridiculously extreme experiment with nearly astronomical numbers the change in weight would be virtually immeasurable.

 

[No Limit]

How the hell do you people know this shit?

 

[jverne]

my take would be yes if would weight less due to the bigger displacement of air on the rough side (read about flotation in liquids), but that would be so theoretically insignificant that it wouldn't rally matter.

were going too much into detail, yes gravity is not the same with height or horizontal position, even pressure isn't.

 

[ríomhaire]

I'm taking weight to be perceived weight rather than
W=mg
in the same way that trainee astronauts on the Vomit Comet are called weightless even though gravity is still acting on them but they appear to have no weight due to other forces cancelling out the force of gravity.

Note on attached pic:
When I saw larger/smaller atmospheric pressure I of course mean that there would be a larger force caused by the atmospheric pressure due to the larger surface area rather than a larger pressure. I only just realised that I worded it incorrectly but I've already attached the pic so I can't be arsed to change it now.

 

[Dan]

The extra surface area makes no change on the net air pressure it feels. That only depends on the volume and the altitude. If this object was getting lift from wind or something as you weighted it, then the rough surface would matter.

The reason that surface area doesn't matter is that you are basically integrating a closed surface, so whatever extra air pressure you have pushing up on the surface, will always be counteracted by extra pressure pushing down from the point above it, assuming a constant pressure field.

If you want to think of it this way, try drawing a vertical line through any part of the object. Now imagine that you travel along this line. Going from the bottom to the top, whenever you transition from air to object, the object has a pressure force pushing up. Whenever you transition from the object back to air, the object has a pressure force pushing down. The number of transitions from air to object must equal the number of transitions from object to air.

So assuming that we are in air, and barring any winds, large altitude changes, magnetic effects, or God pulling up on it, the objects weight will (mass*g - volume*density_of_air)

 

[Ennui]

I'll get my roommate in this thread and he'll take care of everything

 

[ríomhaire]

The extra surface area makes no change on the net air pressure it feels. That only depends on the volume and the altitude if this thing is really big. If this object was getting lift from wind or something as you weighted it, then the rough surface would matter.

The reason that surface area doesn't matter is that you are basically integrating a closed surface, so whatever extra air pressure you have pushing up on the surface, will always be counteracted by extra pressure pushing down from the point above it, assuming a constant pressure field.

If you want to think of it this way, try drawing a vertical line through any part of the object. Now imagine that you travel along this line. Going from the bottom to the top, whenever you transition from air to object, the object has a pressure force pushing up. Whenever you transition from the object back to air, the object has a pressure force pushing down. The number of transitions from air to object must equal the number of transitions from object to air.

So assuming that we are in air, and barring any winds, large altitude changes, magnetic effects, or God pulling up on it, the objects weight will (mass*g - volume*density_of_air)

I hadn't thought of that. Good man.

 

[Dan]

Yeah, I just thought of an easier way to explain it. The upwards pressure, will be the air pressure multiplied by the bottom surface area as seen if you took an orthographic projection of the object from below. The downwards air pressure will be the air pressure multiplied by the surface area of the orthographic projection from above. Obviously, since these two projections are on the same axis, they must be exactly the same. Well there will be a slight difference because air pressure drops as you go up, but you can see that the bumpiness of the surface doesn't matter.

 

[L3N!N]

IT WORKS.

 

[spunge]

haven't read much of the thread or payed much attention but, wouldn't more surface area end up with the same amount of pressure in all directions, up down etc. since all the "in's" need "out's"

 

[ríomhaire]

Dan already pointed that out, yes.

 

[Tarkus]

And I've said that like in the second post