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Buoyancy Problem

  1. Jul 25, 2009 #1

    jdg

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    1. The problem statement, all variables and given/known data
    A chunk of carbon steel with density, ρ = 7.84 g/cm3 , is completely submerged in fresh water. The chuck of steel weighs 30 N more in air than in water


    2. Relevant equations Fb= ρgV ?



    3. The attempt at a solution How do you find the buoyancy without being given the volume? Should I assume 1 cm^3?
     
  2. jcsd
  3. Jul 25, 2009 #2

    jgens

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    It would help if we knew the question! Assuming that the question is "what is the buoyant force acting on the chunk of carbon steel?" use the fact that the block weighs 30 N less in water than in air. Try writing out the equations for the block's weight in each case.
     
  4. Jul 25, 2009 #3

    jdg

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    I got that far, I'm not sure how to set up a problem like that. I found the specific gravity of steel in fresh water (d_steel/d_water)= (7840/1000)= 7.84
     
  5. Jul 25, 2009 #4

    jdg

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    The questions were:

    1. Find the buoyant force acting on the chunk of steel

    2. Find the the volume of the chunk of steel
     
  6. Jul 25, 2009 #5

    jgens

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    Well, the weight of the carbon steel chunck in air should be: Wair = mg

    The weight of the carbon steel chunk in water should be: Wwater = mg - ???

    And: Wair - Wwater = ???

    Edit: Thanks for the questions! If you can solve this first one, you can solve the second question really simply!
     
  7. Jul 25, 2009 #6

    jdg

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    Yea, my problem is, how do I find the mass? If I assume 1 m3 for the volume, it is 7840 kg
     
  8. Jul 25, 2009 #7

    jgens

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    Um, you're not looking at this problem the right way. If Wair is greater than Wwater by 30 N, what does that suggest about Wair - Wwater?

    Hint: To do this problem, you don't need to make any assumptions! :)
     
  9. Jul 25, 2009 #8

    jdg

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    Wair - Wwater

    Ok, so Wair - (Wair-30N)?
     
  10. Jul 25, 2009 #9

    jgens

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    Yes! So this means that: Wair - (Wair - 30 N) = ???
     
  11. Jul 25, 2009 #10

    jdg

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    mg-30?
     
  12. Jul 25, 2009 #11

    jdg

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    wait, so the Wair's cancel and you get -30N?
     
  13. Jul 25, 2009 #12

    jgens

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    No! Write out the expression for the weight of the carbon steel chunk in water and subtract it from the expression for the weight of the chunk in air.

    Hint: Your expression should include the buoyant force.

    Edit: Yes, the Wair cancels out, but: -(-30 N) != -30 N
     
  14. Jul 25, 2009 #13

    jdg

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    ok awesome, so Fb= Fair-(Fair-30N)= 30 N? Then, volume= Fb/d*g (d=density) = 30/(7840*9.81) = 3.901e-4?
     
  15. Jul 25, 2009 #14

    jgens

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    Yes! Fb = 30 N.

    That's what I get for the Volume as well, or equivalently V = 390 cm3.
     
  16. Jul 25, 2009 #15

    jdg

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    ok, so for weight in air it would be 30 N + 30 N= 60 N? And the mass would be m = Fair/g or m = (Fair-Fb)/g?
     
  17. Jul 25, 2009 #16

    jgens

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    For the mass, use the relationship: ρ = m/V

    Edit: Oops! Used the wrong density in calculating the volume, remember that the buoyant force equals the weight of the water displaced! I'm terribly sorry about that.
     
  18. Jul 25, 2009 #17

    jdg

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    thanks a lot! you have time for one or 2 more?
     
  19. Jul 25, 2009 #18

    jgens

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    Yeah, I can help on a couple more problems, but you may want to re-do a couple of the calculations. Sorry!
     
  20. Jul 25, 2009 #19
    I don't think you can find the weight in air. It's juts "30N" heavier than in the weight water.

    Following from jgens, the volume should be about 3.058e-3
     
  21. Jul 25, 2009 #20

    jdg

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    Ok, I got: Fair= mg = 30 N
    m = dV = (7840)(3.901e-4) =3.06 kg
    V = Fb/dg = (30 N) / (7840*9.81) = 3.901 m3



    Next question: A garden hose with internal diameter of 13.5 mm lies flat on a sidewalk while water is flowing in it at a speed of 6 m/s. A person happens to step on it at the very edge of the opening of the hose and decreases its internal diameter by a factor of 9

    So D (1) = 0.0135m
    r (1) = 0.00675m
    D (2) = 0.0135/9 = 0.0015m
    r (2) = 0.00075m
    A (1) = pi*r^2 = (3.14...)(0.00675^2) = 1.4134...e-4
    A (2) = pi*r^2 = (3.14...)(0.00075^2) = 1.767...e-6

    1. What was the water flow rate in the hose prior to the person stepping on it?
    - I got this part: J(1) = A(1)V(1) = 8.59 m3/s

    2. What is the flow rate of water after the person steps on it?

    3. What is the speed of the water just as it exits the hose after the person steps on it?
     
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