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Homework Help: Bouyancy problem

  1. Feb 9, 2008 #1
    [SOLVED] bouyancy problem

    1. The problem statement, all variables and given/known data

    If a steel ax and a aluminum piston have the same apparent weights in the water which can be written as
    (Pax)(Vax) - (Pw)(Vax) = (Pp)(Vp) - (Pw)(Vp)
    where P=rho(density), p=piston, V=volume, and w=water
    Find the ratio (Wax/Wp) of their weights in the air. (Neglect the buoyant force in the air.) Assume densities Pax=7.7g/cm^3 and Pp=2.5g/cm^3

    2. Relevant equations


    3. The attempt at a solution

    I do not even know how to start it so any help would be greatly appreciated.
  2. jcsd
  3. Feb 9, 2008 #2
    well, if the apparent weight is the same, that means the masses of the 2 will be different because the densities of each are different. Look up the densities for steel and aluminum (should be given) and solve for mass. Look up your equations for density and volume. solve for the unknowns. it should get things rolling.

  4. Feb 10, 2008 #3
    With problems like these, I always find it helpful to draw a free-body diagram for each substance I am investigating.

    In this case you should, have weight pointing straight down, Bouyant force pointing straight up AND Apparent weight acting upward. I always include W_app as a force on my FBD as this will help me determine my masses. I think of as there must be something measuring the W_app such as a spring scale.

    I don't know if this is the customary way to account for W_app, but it has always worked for me.
  5. Feb 10, 2008 #4

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    Write down Archimedes' Principle and we can proceed from there. Try to think how you can apply that in this problem.
  6. Feb 10, 2008 #5
    Ok, so if (Pax)(Vax) - (Pw)(Vax) = (Pp)(Vp) - (Pw)(Vp)
    then (770000)Vax - (1000)Vax = (250000)Vp - (1000)Vp
    and (769000)Vax = (249000)Vp
    right? The only problem is that I don't know how to find the volume now.
  7. Feb 10, 2008 #6
    Oh is Archimedes Principle the one that says that the buoyant force is equal to the weight of the fluid displaced?
  8. Feb 10, 2008 #7
    You have 2 variables in this problem. Solve for one variable and then plug it back into the orginal equation leaving you only 1 variable to solve for. i.e. Vp = 769000Vax/ 249000. Now plug that in for Vp and solve for Vax.

  9. Feb 10, 2008 #8
    Ok, thanks :)
  10. Feb 10, 2008 #9
    but if (76900)Vax = (249000)(769000Vax/24900) then wouldn't Vax get eliminated because the 249000s cross each other out?
  11. Feb 10, 2008 #10
  12. Feb 10, 2008 #11
    Sorry, I think I was thinking a little unclear. Its been a while. Ive done a problem like this in the past but just dont remember how. There should be someone here who knows, but ill look into it tomorrow.

  13. Feb 10, 2008 #12
    Oh, ok thanks
  14. Feb 10, 2008 #13

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    The eqn has been given to you.

    (Pax)(Vax) - (Pw)(Vax) = (Pp)(Vp) - (Pw)(Vp), which means (if multiplied with g),

    vol*density*g - vol*(density of water)*g is equal for both. Also,

    Apparent wt in water = Actual wt - buoyant force. So you can find the actual weights.
  15. Feb 11, 2008 #14
    Oh! Duh, sorry that makes sense, thanks. :)
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