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A book on top of a table

  1. May 31, 2005 #1
    A book on top of a table....

    A book on earth that is not moving is sitting on top of a table. There is a force pulling the book towards the ground (gravity) and a normal force in the opposite direction. But what is the NET force excerted by the atomosphere? I also have a follow up question after a few responses.
     
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  3. May 31, 2005 #2

    Meir Achuz

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    That depends on the surfaces of the book and the table.
    For the usual book and table, the surfaces are rough enough that air comes between the book and the table. Then, the atmospheric pressure cancels.
    However, if that were a rubber suction cup with no air between the cup and the table,
    the atmospheric force would be large. In American units it would be 15 pounds times the area of the suction cup in sq. inches.
     
  4. May 31, 2005 #3

    SGT

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    By Archimedes law, the atmosphere would exert a force upwards equal to the weight of the displaced air. A book with a volume of 1 dm3 would receive a force upwards of approximately 0.01N.
     
    Last edited by a moderator: May 31, 2005
  5. Jun 2, 2005 #4
    Please go easy on the newbie...

    Since there is not a perfect seal, the force (pressure x area) would be almost identical on both the top and bottom of the book (differing only by the pressure gradient between the top and bottom). The forces around the sides would cancel out. Thus the net upward force exerted by the atmosphere would be the difference in pressure across the book (mean density of air across the interval * thickness of book) times the area of the cover.
     
  6. Jun 2, 2005 #5
    If there were a perfect seal, however, the net atmospheric force would be in the downward direction at a magnitude of force = atmospheric pressure at top of book * area of cover.
     
  7. Jun 2, 2005 #6
    What if the book was replaced with a trapazoidal object, with the smaller base of the trapazoid on the table on the larger base facing upwards. What is the net force from the atomosphere.
     
  8. Jun 2, 2005 #7

    Doc Al

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    The buoyant force depends only the weight of the displaced air; thus it depends on the volume of the object but not on the shape. (Assuming no perfect seal, of course. :wink: )
     
  9. Jun 2, 2005 #8
    This is true if you assume constant air density versus elevation. Even if a density gradient is allowed, however, the difference would be very negligible. The net force would be very slightly upwards.
     
    Last edited: Jun 2, 2005
  10. Jun 4, 2005 #9
    Shouldn't any object in air be treated the same way as an object submerged in water?
    The upward force is equal to the displaced amount of fluid.
     
  11. Jun 5, 2005 #10

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    The upward force is equal to the weight of the displaced amount of fluid.
    That is exactly what everybody has been saying in different manners, since my previous post.
     
  12. Jun 5, 2005 #11
    That's true as long as the fluid exerts a force on the entire surface area (i.e. not attached to or sealed against another object). The summation of pressure*area across the entire surface area is a more robust solution, but looking at the weight of the fluid displaced is a good enough answer in most cases.
     
  13. Jun 5, 2005 #12

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    In the case of a book on a table we must assume that there is air contacting the whole surface area. It is not entirely true, but a good enough approximation.
    Since the air is compressible we must assume that the density varies along the height of the book. The weight of the fluid is the sum off the various elementary areas along the height. This is equivalent to the summation of the pressure*area and is also equal to the weight of the fluid.
     
  14. Jun 5, 2005 #13
    How does the net upward force change if the book had some regions that were sealed, let's say half of the book does not let air in and the other half of the surface does. This is a very heavy book with a special cover. Would the net force still be upwards?
     
  15. Jun 6, 2005 #14

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    Probably not. The force upwards is roughly the pressure at the lower surface times the area of this surface that is in contact with the air minus the pressure at the upper surface times the whole area of the surface.
    The difference in pressure is tiny, but the difference in area may be appreciable, so we may approximate the downward force by
    F = P*ΔA
    where P is the mean pressure and ΔA the area difference.
     
  16. Jun 6, 2005 #15
    I have nothing to add except to say that this is a great thread for where I am at right now with some projects/demo's I am working on for my high school physics class. It's been 20 years since my fluids class in college. Great stuff !
     
  17. Jun 6, 2005 #16
    Building on the topic...

    Have you ever had a cup of lemonade on a glass table with a pool of condensation around the base of the cup? (in other words, a seal)

    There is a "sticking force" that you have to overcome to release the cup. Is that due to the pressure * area integral that has been discussed (similar to a perfect seal), or is it due to an adhesive force from the water?
     
  18. Jun 7, 2005 #17

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    It's due to adhesion between the molecules of water and those of the cup and of the table.
    Water, being a fluid, transmits entirely the athmospheric pressure, so there is no seal between the cup and the table.
     
  19. Jun 7, 2005 #18
    That makes sense. How do you calculate adhesive forces? I've never dealt with them before.
     
  20. Jun 7, 2005 #19

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    You must obtain a table of the adhesion force coefficient between water and the material of the cup and of the table. I suppose the unit will be N/m2. Multiplying the contact area by the coefficient you get the force.
    Notice that the adhesion force between water and glass is stronger than the cohesion force between water molecules. So, when you pull the cup, the water volume will break. Part will adhere to the cup and part to the table. The force you make is in order to break the cohesion between water molecules.
    If instead of a glass table you had a wooden one, there would not be breaking effort. The water would simply adhere to the cup (less some water who would still wet the table).
     
  21. Jun 10, 2005 #20
    Can't find a table of values for cohesive force of water. Any ideas ? I have made an attempt and will continue to search.

    I am testing the pull out strength of a pair of (0.75 inch thick) lexan plates. I place a drop of water between the plates and then press and turn them until most of the air bubbles are out. I have been able to lift 85% of the max load that atmospheric pressure would allow us to. The plates are advertised as Magdeburg plates. Similar to Magdeburg spheres where you draw a vacuum inside a split sphere with machined and lubricated surfaces.

    My observation is that the water does create a seal so that when the plates are loaded there is a large pressure differential between the center and the outside region (the air/vacuum accumulates at the center due to a slight bending of plates). I am loading 150 pounds on 4 inch diameter plates. I would hypothesize that there is a pressure gradient along the water seal. starting at one atmoshere and dramatically dropping off. Once I get the cohesion value I can (perhaps) confirm my hypothesis? I also plan to make up some plates with air ports at different radius' so that I can get vacuum readings under load.

    I teach Physics at a high school in Maine. The industrial arts teacher recently got some 5/4 inch lexan that I plan to use. Yee Hah !
     
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