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Aquarium Pressure Problem

  1. Apr 27, 2008 #1
    1. The problem statement, all variables and given/known data
    An aquarium tank is 100 cm long, 35 cm wide, and 40 cm deep. It is filled to the top

    a)What is the force of the water on the bottom (100 cm x 35 cm) of the tank?

    b) What is the force of the water on the front window (100 cm x 40 cm) of the tank?


    2. Relevant equations



    3. The attempt at a solution

    For part a, can I just use I find the mass of the water and use mg? Actually I think I might have to use integration, but could someone explain it to me?
     
    Last edited: Apr 27, 2008
  2. jcsd
  3. Apr 27, 2008 #2

    Dick

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    Yes, for part a you can just use mg. Why? Think of a force diagram. For part b you have to integrate pressure over area.
     
  4. Apr 27, 2008 #3
    For part b, I'm pretty sure I have to use integration.
    P=Po + ro*g*y where y is the depth
    P= (101300Pa) + (1000kg/m^3)(9.80)d

    Say I cut the front surface into little rectangles--each with width .35 and height dy
    So would I multiply the above by .35dy and take the integral evaluated from 0 to 40 ? Please help me set up this integral. Thanks
     
  5. Apr 27, 2008 #4

    Dick

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    You don't cut it into little rectangles, it's one big rectangle. Just integrate. P= (101300Pa) + (1000kg/m^3)(9.80)y. Where y is depth and x is the distance coordinate across the front. You'll want to integrate this dx*dy. What's the range of x and y? BTW it only asks for the pressure of the water, so you might want to drop the 101300Pa. That's from the air. The question is a little ambiguous about that. Try it both ways to make sure.
     
    Last edited: Apr 27, 2008
  6. Apr 28, 2008 #5
    I don't understand what you mean by integrate this dx*dy.

    Would I take the integral from 0 to .40 of (1000kg/m^3)(9.80)y * (1.0m) dy?
     
  7. Apr 28, 2008 #6

    Dick

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    That works also. If you haven't done double integration don't worry about it.
     
  8. Apr 28, 2008 #7
    Ok so is this right: Integral of 1000*9.80*y dy

    = 4900 y^2 evaluated from 0 to .4

    4900* (.4)^2 = 784 N

    Also why is it that I drop the pressure from the air--is it because the force from the air is down, but the force of the water on the front of the glass is horizontal?
     
  9. Apr 28, 2008 #8

    Dick

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    Yes. That's right. But try to understand why you would or would not keep the air pressure. Pressure doesn't have a direction so it's not because of direction. The TOTAL force on the inside of the front of the aquarium is the sum of the force due to air above PLUS the force due to the water inside. If that's what you want, you include it. If you don't want it then don't include it. You have to read the question and decide. A reasonable reason to not include it is that there is also air pressure pushing into the tank from the outside air. Which approximately cancels the contribution of the air to the inside pressure. So the net force on the front of the tank is just that due to the water.
     
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