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Force of the water on the plug

  1. Nov 28, 2011 #1
    1. The problem statement, all variables and given/known data

    The plug in a bathtub is 10cm2 in area and is 0.8m below the surface of the water. What is the force of the water on the plug?

    Given Data: Area: 10cm^2
    h: 0.8m


    too lazy to post attempts. sorry.
     
  2. jcsd
  3. Nov 28, 2011 #2
    If you are too lazy to post attempts, then we are too lazy to help you.
     
  4. Nov 28, 2011 #3
    Okayy fine. I get lost on what formula i should use.
     
  5. Nov 28, 2011 #4
    Let's see the formulas you are contemplating using.
     
  6. Nov 28, 2011 #5
    Hmm..
    Formula for Pressure P=F/A ,
    or Pascal's principle?
     
  7. Nov 28, 2011 #6
    OK, we now have a beginning. How is pressure computed? How is force computed from pressure.
     
  8. Nov 28, 2011 #7
    By multiplying Area and Pressure.
     
  9. Nov 28, 2011 #8
    Pressure increases the deeper you go below the surface of a fluid. Does that give you an idea of how to compute pressure?
     
  10. Nov 28, 2011 #9
    Yes, that will give you force. Now you need to compute pressure. How?
     
  11. Nov 28, 2011 #10
    Wait I found another formula, I think this one could be possible for solving the unknown.

    F=(density)(Area)(Height)(Gravity)
     
  12. Nov 28, 2011 #11
    This formula provides the weight of an object.

    We need the pressure where the plug is located because then you can determine the force on the plug by multiplying it by area.

    With formulas, units are very important. Look at pressure P. Its units are force per unit area. Recalling what I said earlier about the pressure depending on how far one is below the surface, what two things when multiplied together provide you with force per unit area?
     
  13. Nov 28, 2011 #12
    Hint: What is the density of water?
     
  14. Nov 28, 2011 #13
    density, gravity and height? Coz' height cancels out the m^3?

    Sorry for messing things up.
     
  15. Nov 28, 2011 #14
    You've got it. Pressure equals depth times distance under surface if the density is expressed in units of weight per unit volume. If density is mass per units volume, then you have to multiply by gravity as you cite above.

    So now you can compute the force on the plug.
     
  16. Nov 28, 2011 #15
    9800N/m^3 then what?
     
  17. Nov 28, 2011 #16
    9800N/m2*

    Sorry, been following the thread.
     
  18. Nov 28, 2011 #17
    How can that be, sir?

    Density * Gravity = 1.0x103kg/m3(9.81m/s2)

    = 9800N/m3

    What made the cubic meter turn to a meter squared?
     
  19. Nov 28, 2011 #18
    Density x gravity x height, as you stated earlier.

    (kg/m3) x (m/s2) x m

    = N/m2
     
  20. Nov 28, 2011 #19
    I'm totally aware of that. But I was only at the first two.

    With multiplying with the height the answer would be "7840N/m^2"
     
  21. Nov 28, 2011 #20
    The choices that are given in the book is
    7.38N
    7.62N
    7.75N
    7.84N
     
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