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Pressure and transmission of pressure in liquids

  1. Apr 13, 2012 #1
    I understand that pressure is F/A. But whenever they ask me to find the pressure of a book acting on the table, since the object in question is actually the table so is the force the reaction force on the table by the book and not actually weight since weight is an internal force?

    Again, with the hydraulic system information in my textbook a small force acting on a small piston gives rise to a large force in the large piston as the pressure is the same. But I'm quite confused as they say that a forfeit is exerted on the piston. I thought that the object that we have to take into perspective is the liquid? Also since my force is acting the piston that force's area has to be the same as the piston's or else the pressure won't be the same since P=F/A. I'm not very sure about this since all they say is the force on the piston so I don't quite understand it in detail.

    Lastly, pressure in a liquid acts at all directions. What does this mean? Does it mean that when I use P=hpg to get the pressure in the liquid (taking that there is no atmospheric pressure above) that pressure acts at all directions? But won't that mean that the force is infinite since the pressure acts at all direction so P=F/A is infinite as at one point upwards pressure is same as downwards.

    Thanks guys for the help! Hope to hear (or read) from you guys soon! :smile:
     
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  3. Apr 13, 2012 #2

    tiny-tim

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    hi sgstudent! :smile:
    the pressure of one body on another is always reaction force / area

    (btw, weight is an external force … it comes from the Earth)
    i don't understand your question :confused:

    the smaller piston moves a longer distance, so it can do the same work with less force

    so it's like a lever, where again the forces are inversely proportional to the distance moved
    sorry, i don't understand :confused:
     
  4. Apr 13, 2012 #3
    If you calculate or measure the pressure at a depth h, yes the pressure will be the same in all directions. If you can imagine lowering a very small cube down to the depth h then the pressure on all faces of this small cube is the same ie front, back, left, right, top and bottom. Since the force on all faces of the cube would be also the same. No matter which way you orient the small cube the pressure will be the same on all of its faces - there is no preferred direction as far as pressure is concerned.

    I am puzzled how you managed introduce infinity.
     
  5. Apr 13, 2012 #4
    Hi tiny Tim :smile: the part I'm confused about is that since we have to use reaction force/area then what does it mean by we apply a force onto a piston? Since the force that we exert is smaller than the surface area of the piston so I don't understand how can we calculate the pressure acting on the piston. Also, I thought the object that we should focus on is the water so why isn't it the force acting on the water?

    Also for the other part, I meant that in a single point underwater pressure is outwards in all direction at that point right? So at the centre of it, from the left right up down the pressure is the same right? So won't the force be infinite?

    Thanks for the help guys!
     
  6. Apr 14, 2012 #5

    haruspex

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    How can a force be less than an area?
    Perhaps you mean the force you exert on the first piston is exerted over a smaller area than the area of the piston against the fluid? Irrelevant.
    You somehow exert a force F1 on the first piston.
    That piston spreads the force over an area A of fluid, generating pressure P = F1/A.
    The pressure is transferred throughout the fluid.
    At the second piston the pressure acts over area B, generating force F2 = P*B = F1*B/A.

    Regarding pressure at depth, consider a small cube, area C on each face.
    Force on each face F = C*P.
    As C tends to zero, F tends to 0, not infinity.
     
  7. Apr 14, 2012 #6

    tiny-tim

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    hi sgstudent! :smile:
    as haruspex :smile: says, how can a force be less than an area? :confused:

    the force on "our" end of the piston (where we're pushing it) is equal and opposite to the force on the water end of the piston

    the pressure is that force divided by the area of the piston …
    … the force on "our" end of the piston is the same as the force on the water
    if you have a sphere, the force will be the same all over, and will be proportional to the surface area of the sphere

    as you reduce the sphere to a point, the force reduces to zero (because the area reduces to zero, and the force is proportional to the area)

    btw, it you consider the force of the water on just one half of the sphere, that's calculated in exactly the same way as the force of your body on the spherical top of your femur …
    http://www.arthursclipart.org/skeletons/skeletons/hip%20joint.gif
    (http://www.arthursclipart.org/skeletons/skeletons/hip joint.gif)
     
  8. Apr 14, 2012 #7
    Oh, so the force acting on the piston is equal to the force acting on the water? Why is this so, since if I have a 10N force on the water there will be a reaction force of 10N on the piston so won't I have to apply 20N of force rather than 10N so I'm confused about this part.

    Thanks for the help guys!

    BTW tiny Tim, you seem to be able to pick out the bits that I'm confused about really quickly, that's amazing! Thanks for the help! :smile:
     
  9. Apr 14, 2012 #8

    tiny-tim

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    are we talking about a particular device? :confused:

    usually there's two pistons, we push one, and the other pushes a wheel etc

    if we push the first piston with 10 N, the water pushes back on the first piston with 10 N

    at the other end, the same water pushes on the second piston with 20 N, and the water pushes back on the second piston with 20 N

    (so it's geared: we push the first piston, and the second piston moves half as far with twice the force)
     
  10. Apr 14, 2012 #9
    Huh, I don't understand the 'I push with 10N so the water push back with 10N part.' Why would the water push back with an equal but opposite force? Thanks for the help!
     
  11. Apr 14, 2012 #10

    tiny-tim

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    newton's third law? :confused:
     
  12. Apr 14, 2012 #11
  13. Apr 14, 2012 #12

    tiny-tim

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    to make things easy, let's assume that you put your whole weight on the piston

    so the force from you on the top of the piston is mg

    and the force from the piston on you is also mg

    the force from the bottom of the piston on the water is mg

    and the force from the water on the piston is mg

    everything is mg :smile: (until you get to the other side of the water, where the diameter is different)​
     
  14. Apr 14, 2012 #13
    Oh so the piston has no net force? Actually are you able to deduce these stuff just like this because previously in most dynamics question we treat two bodies differently like when I push a box across a table my hand is one object and the book as well so if I push 10N on the book there will be a reaction force of 10N so just from this information I don't really know about the other forces acting on my hand to balance the reaction force which is opposing me. So I'm pretty confused about this part here. Thanks for the help tiny tim! :smile:
     
  15. Apr 14, 2012 #14

    tiny-tim

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    if the piston is moving steadily, its acceleration is zero, so the net force must be zero
    i'm not following you :confused:

    there's a 10 N force all the way along, finishing where you, or something, is pushing against the Earth
     
  16. Apr 14, 2012 #15
    Oh sorry for being vague. I mean like when I push a box with 10N, my hand experiences a 10N force too which is the N3L reaction force. But I'm not sure if the force I apply to my hand will be 10N or greater than 10N. So that's why I'm confused about why I can assume all the forces in the piston and the water like the weight example you gave just now. Thanks for the help! :smile:
     
  17. Apr 14, 2012 #16

    tiny-tim

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    the horizontal component of the force you apply to your hand will be 10 N :smile:

    (this is because your hand is not accelerating, so if there's 10N on one side, there must be 10 N on the other side, net horizontal force 0)
     
  18. Apr 14, 2012 #17
    But how do I know if my hand is accelerating since I might apply even more force to counteract the reaction force such that my hand will accelerate at the same rate as the object in question. So I'm guessing that there is no one answer for this and it depends on the person? But then if this is so then how can the pressure in the piston like the wright example you gave be justified? Thanks for the help tiny tim!
     
  19. Apr 14, 2012 #18

    tiny-tim

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    your hand will always have the same acceleration as the object it is in contact with

    (as i said, the forces on either side of your hand will only be equal if there is no accceleration)
    the piston will on average have zero acceleration

    (and in a water system like this, the acceleration would be very small anyway)
     
  20. Apr 14, 2012 #19
    Oh okay thanks for the help! But will questions be ambiguous by saying something like a hammer head has a surface area of 0.01m^2 a man applies a force of 100N on it to hit a nail. What is the pressure exerted on the nail? since what's exerted on my hammer might he different from what's exerted on the nail. or is there a connection between them that I'm not seeing. thanks for the help Tim.:smile:
     
  21. Apr 15, 2012 #20

    haruspex

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    First, you don't normally apply a force with a hammer. You swing the hammer to strike the nail. This provides an impulse (momentum).
    But say you instead place the hammer head carefully on the nail and then push with a force of 100N. The area of contact between hammer and nail will be the area of the head of the nail (unless it's a very thin hammer). The pressure at the interface is therefore 100N/area of nail head. The hammer and nail exert that pressure on each other, over the same area.

    This is why thumbtacks have wide heads. The pressure between thumb and tack is the force / area of tack head. Pressing directly on the much smaller area of a nail head would create a much greater pressure on a small patch of your skin, potentially piercing it. Correspondingly, the tack or nail has a fine point in order to create a huge pressure on a tiny area of substrate.
     
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