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Freeboard of a Soup Can in a Pool

  1. Apr 13, 2010 #1
    Willie took a soup can from the kitchen trash, added a handful of rusty ball bearings from the bottom of his Dad's junk drawer, and is now about to put it into the deep end of the family pool that Dad has just finished cleaning. Dad is not amused. "Don't worry," Willie says, "It'll float." "It had better," his Dad says, "or you'll be going in after it."

    The can, plus bearings, weights exactly 0.5 pounds (force). Measurements are shown on the right. Will the can float? If so, how much freeboard will it have? (Freeboard, a nautical term, is distance between the waterline of a ship, and the ship's deck. In this case, it's the distance from the water level to the top of the can.) Show your calculations. (Hint: Convert pounds (force) to Newton’s and work in the MKS system.)



    2. Relevant equations

    Archimedes Principle?

    lbs to Newtons

    Vol=Pi*h*r^2

    Freeboard = distance from the water level to the top of the can

    3. The attempt at a solution

    Ok so here is where I get confused. I believe the can will float because the can is less than 1 kg/l which is the density of water. So far here is what I have....any suggestions on where to start???

    Weight = .5
    Converted to Newtons
    .5lbs*4.4482216 = 2.224 N

    Calculated Volume of Soup Can

    3.14159*10cm*32
    V = 282.74
     
  2. jcsd
  3. Apr 13, 2010 #2
    Here is the image. BTW this is an old homework assignment that I never could figure out....any help is much appreciated. Thanks.
     

    Attached Files:

  4. Jul 7, 2011 #3
    OP did you ever figure this out... I have this exact same problem and can't figure it out!

    As a matter of fact I got the exact same answers as you did above with the add info below:

    density of can (pc)=0.805cm3

    after that I get just as stuck you were.
     
    Last edited: Jul 7, 2011
  5. Jul 7, 2011 #4

    gneill

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    Staff: Mentor

    It's an old post; the OP may no longer be around or be interested in this particular question. Why don't you show your own work on it?

    Hint: How about if you were to consider how much of the can, with the given base area, would have to be submerged in order to displace the can's weight in water?
     
  6. Jul 7, 2011 #5
    ok, well after looking at this for a VERY long time and using a friends help this is what I have come up with...

    Vc=282.74cm3
    Mc=0.227kg
    Finding the specific gravity of the can (pc):
    pc=m/V
    pc=0.227/0.2827
    pc=0.805m3

    Taking this and the mass of the can I *think* that I can find the volume of displaced water

    Displacement(diss)=mass/pc
    (diss)=227/0.805
    (diss)=281.99

    I then took the volume of the can (282.74cm3) and subtracted the displacement (281.99cm3) which game me 0.75cm3

    Knowing this I reversed the area equation used to find the volume of the can Vc=[itex]\pi[/itex]r2h [itex]\rightarrow[/itex] h = [itex]\pi[/itex]r2/Vc to find the height of the can out of water
    which I found to be .0283cm [itex]\rightarrow[/itex] 28.3[itex]\mu[/itex]m

    If this is wrong SOMEONE please point this out to me!!! I have no clue if I did this right or not.
     
  7. Jul 8, 2011 #6

    gneill

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    Staff: Mentor

    You've lost some accuracy here, possibly due to rounding and/or using a not too accurate value for g. The number should be closer to .80213 g/cm3 (or perhaps kg/L, since you seem to have converted the volume units to liters for your calculation).

    That would be the volume of displaced water if the WHOLE CAN was submerged. Effectively you've done the calculation:
    [tex] p_c = m_c/v_c [/tex]
    [tex] diss = m_c/p_c = v_c [/tex]
    Which doesn't produce any new information...

    The difference in value between this result and the volume that you calculated previously for the volume of the can is due to the rounding error mentioned above.
    I'm not sure why you would want to calculate the specific gravity (actually the density) of the can, except as a check to see if it is less than the density of water and thus can float. Once that is ascertained you don't really need the can density.

    Instead, use the density of the water (the thing that's being displaced) to determine what volume of water needs to be displaced in order to equal the weight of the can. Then look at how much of the can needs to be submerged in order to displace that volume of water.
     
  8. Jul 8, 2011 #7
    I see where I might have gone wrong, but...

    so should I use this formula to find the water displaced?
    [tex] diss = m_c/p_w = v_w [/tex]

    If so that doesn't seem right to me... would I take this new displacement and try to convert it back to Newtons so that I can find the weight difference? After that... what?

    and how would I find how much needs to be submerged??? The previous attempt was the best that I had...


    I'm so lost...
     
  9. Jul 8, 2011 #8

    gneill

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    Staff: Mentor

    Yes. This is the amount of water that the can must displace if it is to float; the can floats when it displaces a volume of water whose weight (or mass) is equal to the weight (or mass) of the can.
    This "new displacement", as you put it, is the amount of the can that needs to be submerged; the portion of the can that is underwater and displacing water.
    You now have the volume of the portion of the can that needs to be submerged. You should be able to work out the length of the can cylinder that corresponds to that volume. What's the formula for the volume of a cylinder?
     
  10. Jul 11, 2011 #9
    OK So I took your equation and I think that I did the math correct, but I think that some of my conversions are incorrect in places.

    Anyway: here is what I got

    [tex] diss = m_c/p_w = v_w [/tex]
    [tex] diss = 227/1 = 227cm^3 [/tex]

    Taking that and subtracting the original volume of the can:
    [tex]New Vol = 282.74-227 [/tex]
    [tex]New Vol = 55.74 cm^3 [/tex]

    Now taking this and plugging it back into the volume eqation:
    [tex] V = \pi r^2h[/tex]
    [tex] H = \pi r^2/V[/tex]
    [tex] H = \pi 3^2/ 55.74[/tex]
    [tex] H = 1.97cm[/tex]

    So I *think* that my answer should be 1.97cm of freeboard?

    Like I said I know that somewhere in that my conversions went a little wacky but I think that my answer is right this time.
     
  11. Jul 11, 2011 #10

    gneill

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    Staff: Mentor

    Looks good to me! :smile:
     
  12. Jul 12, 2011 #11
    Thank you so much for all your help!!!
     
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