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Air Pressure in Elementary Dynamics Problems

  1. Mar 20, 2014 #1
    Hello all,
    I was thinking about situations in introductory courses or textbooks where pressure or buoyant forces are suddenly introduced in situations where they become relevant which lead me to think about the consequences of neglected said forces in more elementary problems.

    I'm curious as to how significant the effects of net pressure forces are on objects even in fairly straightforward situations such as statics problems. I'm thinking of three thought experiments involving, a simple object like, say a book that is at rest. I'm hoping someone can poke holes in my logic or clarify anything I'm missing.

    1. The book is held up, say by a string so that all of its surfaces are immersed in the air. The book will experience a net pressure force ("the buoyant force") but this should be pretty small given that the density of air itself is low and consequently, there isn't much variation in pressure over the height of the book.

    2. The book is pressed down against a table and the book is made of a material that allows a good seal to be made between the book and the table. Here, there would be a substantial noticeable downwards pressure force ("the suction cup effect") due to the fact that there is very little air underneath the book and a substantial downwards force from the air above.

    Where I'm confusing myself is in case 3.

    3. The book is allowed to sit on the table without making a good seal. This is the standard case that is usually treated in dynamics in which case the normal force is taken to be equal to the weight of the book as per Newton's first law. This can only be a valid approximation if there is no substantial net pressure force acting on the book.

    My question is: in this case, what's going on at the bottom boundary of the book? Are the book and the table completely separated by a very thin layer of air, such that, in effect we're in case 1? Or, does air get trapped in small pockets between the book surface and the table... if so, is that air still at atmospheric pressure, or would the pressure be higher?

    I keep thinking that case 3 must somehow be intermediate between the first two cases, but can't wrap my head around how we could easily neglect net pressure forces in this very simple scenario, which is essential to even determining the mass of objects using balances.

    Thanks for reading and your opinion!

    Alexander
     
  2. jcsd
  3. Mar 21, 2014 #2

    AlephZero

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    It doesn't matter much which option of 2 or 3 is correct, because there is air underneath the table and its upwards pressure is transmitted through the table to the book. Even if the book was on the ground and you assume the earth is impervious to air, the air pressure on the opposite side of the earth is transmitted through the earth to the bottom of the book.

    The "suction effect" of creating a partial vacuum under the book is a separate issue, but it is no different in principle from the forces acting on a vacuum tank.

    The resultant buoyancy force on the book is always the same, i.e the weight of the air displaced by the book.This is usually small enough to ignore, but in accurate weighing a correction is made for it, based on the measured air pressure at the time and place where the weighing was done.

    The air pressure also creates a compressive stress in the book that is equal to the air pressure (about 0.1MPa at sea level), but for almost all materials this is so small in comparison with the yield stress (for metals it is of the order of 1000 times smaller) that it can be ignored in practice.
     
  4. Mar 21, 2014 #3
    I think you might have case 3 and 2 mixed up slightly.

    My guess would be that the act of pressing down the book in case 2 squeezes out the air from between the crevices of the book-table interface but traps some air in pockets. When the pressing force is removed the volume of the pockets of air increases but because of the seal no air rushes back in to fill the extra volume. This means that the air pressure in the pocket drops with the increased volume.

    Since the pressure outside is greater then the pressure between the interface then as you correctly pointed out there is a net force acting down in case 2. So it creates a suction cup effect.

    Case 3, however, the air that might be trapped in the crevices hasn't expanded in volume and so is the same pressure as the ambient air pressure. So no suction effect occurs.

    On another note, the effects of the buoyant force and the suction effect are small enough in most cases to be able to safely ignore.
     
  5. Mar 21, 2014 #4
    Thank you both for your thoughtful replies.
    AlephZero, if you don't mind, I have a question regarding your response. You state:"It doesn't matter much which option of 2 or 3 is correct, because there is air underneath the table and its upwards pressure is transmitted through the table to the book." I'm a little confused about this statement.

    Surely, there is a fundamental physical difference to how the book (or another object) will respond when there is a strong suction seal versus when there is no such seal... a strong upwards force must be exerted to lift the object from the surface and break this seal which implies there must be a physical difference between case 2 and 3. Arguing that because a pressure force on the other side of the table that is transmitted through to the object in the case of a seal does not explain why we can't easily lift the object from the table.

    I recognize that, in most real cases pressure forces and buoyant forces are pretty insignificant. But, suppose I came up with a exceptionally sensitive digital scale (whose readings are essentially indicative of the normal force applied to the object). I would think that the scale would read a larger mass (i.e. larger normal force) in the case of the suction effect (case 2) versus the case where there is no such effect (case 3)... even if the difference is relatively small and insignificant, there should still be a difference. Or am I missing somethign fundamental here?

    Thanks again. It's appreciated.

    Alexander

    "
     
  6. Mar 21, 2014 #5
    If you had a sufficiently sensitive scale then you would also have to correct for the following possible additional forces:

    -elevation within the earth's gravity field
    -centrifugal forces caused by the earth's rotation
    -hydrodynamic forces from small air currents
    -temperature affecting the scale readings
    -tidal forces caused by the moon and sun and planets
    -small vibration forces caused by people, equipment, the building, passing vehicles, and plate tectonics
    -coriolis forces caused by these vibrations and by the motion around the sun and other stars in the galaxy
    -additional weight caused by traces of moisture or DNA on the book
    -static electricity
    -the earth's magnetic field
    -Van Der Waals forces between the book and the scale
    -additional gravitational forces caused by other large masses such as buildings, mountains, or black holes
    -relativistic effects for masses not at rest
    -the momentum impulse caused by light hitting the book
    -the uncertainty principle
     
  7. Mar 22, 2014 #6

    olivermsun

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    I think the responders are correct in that cases 2 & 3 are essentially the same when the book on the table is undisturbed. The downwards pressure force on the top of the book is equal by the upward pressure force on the bottom of the table (although most likely the table legs and top would support the additional weight even if this weren't true). The two objects will be "pushed together" more strongly and so there will be more compression stress within each object.

    When you start trying to lift a book which is "suctioned" to the table, that's when you will have to overcome the additional pressure force which tries to keep book and the table together.
     
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