Conceptual Issue Dealing With Atmospheric Pressure

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
4 replies · 2K views
ScareCrow271828
Messages
22
Reaction score
0

Homework Statement


A 1m^3 steel cube is to be floated on mercury (specific gravity=13.6). On each side of the cube there is a 5mm clearance. Assume that the container is open to atmospheric pressure at 100kPA. Find the mass of the mercury.

Homework Equations


P=F/A

The Attempt at a Solution



I sum-mated the forces in the vertical direction and combined that with an equation for buoyancy. Then I solved for the height of the mercury around the cube, found the volume of the mercury and multiplied it by its density.

My question is: is it right to assume that atmospheric pressure has cancelling effects because it pushes down the steel cube but also pushes down the mercury which would push the steel cube back up? This makes sense conceptually to me, but I am having trouble understanding that since P∝1/A. It seems like by that equation atmospheric pressure would have different effects on the block and on the mercury.
 
Physics news on Phys.org
ScareCrow271828 said:
atmospheric pressure has cancelling effects because it pushes down the steel cube but also pushes down the mercury
Yes, it cancels. Can you go into more detail on why you think it would not cancel?
 
  • Like
Likes   Reactions: Chestermiller and ScareCrow271828
haruspex said:
Yes, it cancels. Can you go into more detail on why you think it would not cancel?

Thank you! That was my intuition I am trouble understanding how atmospheric pressure acts as a pressure -or at least has units of pressure (Pa)- but in this case seems to act independent of area. By P=F/A shouldn't pressure be dependent on area. Or more specifically Inversely proportional to area?

Ive always thought of pressure as kind of a deform-able blanket that covers and weighs everything evenly so it makes sense to me conceptually. I am just having trouble understanding it mathematically. Wouldn't there be a greater force due to the atmosphere if the pressure was over a greater area since F∝PA?

Another question if you could:
Mass flow rate is equal to the product of density, velocity,and area or ⋅m=ρA∨. Therefore if area is decreased velocity has to increase. The increase in velocity makes sense to me but it seems like that would also increase the pressure since pressure is dependent on the force that molecules exert on surfaces. How would shooting molecules at a higher speed not increase that force? But by Bernoullis equation ∨∝1/P? How does increasing the velocity and therefore decreasing the pressure make any sense conceptually?

Sorry for the simple questions
 
ScareCrow271828 said:
Wouldn't there be a greater force due to the atmosphere if the pressure was over a greater area since F∝PA?
Yes, there is a greater force over the upper surface of the cube than there is on the surrounding margin of mercury. But the force on the mercury is transmitted straight down to the bottom end of that margin, i.e. to the level of the base of the cube. At that point, the pressure infinitesimally above the level must equal that immediately below, and that pressure must also be the same all across the underside of the cube. So the force on the underside of the cube due to the atmosphere is the same as that on the top surface.
ScareCrow271828 said:
How would shooting molecules at a higher speed not increase that force?
The movement of the molecules does not create the pressure difference. Rather, the pressure difference accelerates the molecules. If there is a reduction in area in a pipe, the molecules must accelerate as they transit to the narrower section, so the pressure must be greater before the narrowing than after. If it widens again later, the pressure difference goes the other way to slow them down again.
 
  • Like
Likes   Reactions: ScareCrow271828
haruspex said:
Yes, there is a greater force over the upper surface of the cube than there is on the surrounding margin of mercury. But the force on the mercury is transmitted straight down to the bottom end of that margin, i.e. to the level of the base of the cube. At that point, the pressure infinitesimally above the level must equal that immediately below, and that pressure must also be the same all across the underside of the cube. So the force on the underside of the cube due to the atmosphere is the same as that on the top surface.

The movement of the molecules does not create the pressure difference. Rather, the pressure difference accelerates the molecules. If there is a reduction in area in a pipe, the molecules must accelerate as they transit to the narrower section, so the pressure must be greater before the narrowing than after. If it widens again later, the pressure difference goes the other way to slow them down again.
Great! Thank you very much!