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Fluid Submersion Question:

  1. Jun 11, 2005 #1

    DDS

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    A 2.09 cm thick bar of soap is floating on a water surface so that 1.50 cm of the bar is underwater. Bath oil of specific gravity 0.639 is poured into the water and floats on top of the water. What is the depth of the oil layer when the top of the soap is just level with the upper surface of the oil?

    Given the thicknesses and distances of the materials i have found that the distance between the waters surface and that of the highest point of the bar of soap is :

    2.09-1.50=5.9e-3m

    since the bath oil floats on the water i figured that the depth of the bath oil will be the distance of the part of soap that is petruding out of the water. However this is not the case.

    I also know that the depth or volume is not the petruding distance multiled of divided by the density of the liquid.

    Any help??
     
  2. jcsd
  3. Jun 11, 2005 #2

    DDS

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    does anyone have any suggestions for this one??
     
  4. Jun 11, 2005 #3

    OlderDan

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    Both liquids will provide a buoyant force. The net buoyant force will equal the weight of the soap, and the total volume displaced (two different liquids) will equal the volume of the soap. You can't just use the exposed height of the soap before the oil is added because adding the oil will raise the level of the soap relative to the water surface. You have to look at the system as a whole.
     
  5. Jun 11, 2005 #4

    DDS

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    I understand your analysis its just im still confused as to how to apply this in a matematical sense. Can you help me with that
     
  6. Jun 11, 2005 #5

    OlderDan

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    Think of a bar of soap as having perfectly uniform thickness of h = 2.09 cm. It does not matter what volume the soap has. Assume the soap has volume V = A*h, where A is a cross-sectional area. The volume of the water displaced intially is 1.5cm*A. Use the density of water to find the weight of water displaced. It will be in terms of A. That is also the weight of the soap.

    Let x = thickness of oil layer, and y = the depth the soap sinks into the water. You can easily write y in terms of x. The total weight of liquid displaced will be the sum of two terms: the weight of water displaced, which depends on the water density and its volume plus the weight of oil, which depends on the density of oil and its volume. All volumes and weights will be proportional to A, so A will divide out in the end. The only variable in the equation will be x, and you can solve for that.

    That is a verbal description of all the relationships you need. See if you can turn that into the equation you need to solve.
     
  7. Jun 11, 2005 #6

    DDS

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    what do i assume my density of water is?
     
  8. Jun 11, 2005 #7

    DDS

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    jsut 1000 kg/m^3
     
  9. Jun 11, 2005 #8

    DDS

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    this is what i have come up with so far:

    H=0.0209 m
    Volume of H20 displaced=0.015 m
    Exposed portion= 5.9e-3 m

    Denstity=D=0.639

    Weight of H20 Displaced:

    w=(1000)(0.015)(9.81)
    w=147.15 i havea feeling thas wrong because its not interms of A as you said

    what i think it could be but doesnt make sense to me is that if the volume of soap is V=A*H
    then u sub that V into the equation for weight to get it in terms of A. That doesnt make sense to me becuase
    the relation V=A*H describes the volume of soap so why would the volume of soap go into the position of the volume of fluid in Archimedes Principle.

    Nevertheless could it be:

    w=(1000)(A*0.0209)(9.81)
    w=205.029 A
     
  10. Jun 12, 2005 #9

    DDS

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    am i on the right track
     
  11. Jun 12, 2005 #10

    DDS

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    should it be interms of A or can i use my first answer
     
  12. Jun 12, 2005 #11

    DDS

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    can anyone let me know if im on the right track....anyone??
     
  13. Jun 12, 2005 #12

    OlderDan

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    I think it will help you to keep track of dimensions of quantities. How can volume have units of meters? Volume is three dimensional, and must have dimensions of length cubed.

    If the volume of the soap is V=AH, where H is .0209 m, and the soap sinks into the water a depth of .015m, then the volume of the water displaced is V_w = A(.015m). The density of water is 1000 kg/m^3 as you said, so you can use volume and density to find the weight of displaced water = weight of soap. When the oil is added, the volume of oil displaced is A times the depth of oil, and the volume of water displaced is A times the depth the soap sinks into the water. Every quantity that depends on volume in this problem will have an A in it, and the As will all divide out in the end.

    In the final configuaration, the volume of fluid is the voilume of the soap because the soap floats with its upper surface aligned with the top surface of fluid. The soap is displacing its entire volume.
     
  14. Jun 12, 2005 #13

    DDS

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    so where did i go wrong originally and which equation do i use, the one with the A or the first one without the A
     
  15. Jun 12, 2005 #14

    OlderDan

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    I have told you where you went wrong, and I have told you a complete outline of how to do the problem. Pay attention to the dimensions of the quantities involved and you will see that you need the area for intermediate steps to follow the logic that follows from Archimedes Principle. You need to write the equations yourself based on the information already given. It's not enough to plug numbers into equations that someone else writes for you.
     
  16. Jun 12, 2005 #15

    DDS

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    This is where i get to and then i get confused by your explanation:

    Volume of soap:
    Vs=AH
    Vs=0.0209A

    Volume of H20 Displaced:
    Vw=0.015A

    Weight of H20:
    W=Dw*Vw*g
    w=1000*0.015A*9.81
    =147.15A

    Weight of soap:
    147.15A
     
  17. Jun 13, 2005 #16

    OlderDan

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    Now add the oil until the top of the oil is even with the top of the soap. When that is done the thickness of the oil will be some value not yet introduced (except in my earlier post) Let's call it x. Then how deep does the soap sink into the water expressed in terms of x?

    Now, in terms of x calculate the volume of oil displaced, and the new volume of water displaced (it is going to change when the oil is added). Use these two volumes to calculate the weight of oil displaced and the weight of water displaced, using the fact that their sum must be the weight of the floating soap. All of this will still involve A, but it will divide out in the end.
     
  18. Jun 13, 2005 #17

    DDS

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    Then how deep does the soap sink into the water expressed in terms of x?

    thats what confuses me, i know that the soap will be added to the exposed area which is 5.9e-3 but thats all i know . How do i calculate how deep the soap sinks. Is it just equalt to the exposed area?

    and when i calculate the volume of oil displaced what weight do i use, the weight of the soap?
     
  19. Jun 13, 2005 #18

    OlderDan

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    Draw yourself a picture of the soap. Its thickness is 2.09cm: given. Extend the top line in both directions. That line is the top of the oil layer. Draw a line parallel to the top and bottom of the soap, through the soap, somewhere near the middle of the soap. That line represents the bottom of the oil and the top of the water. Call the distance from the top of the soap to the oil-water line x. How far is it from this line to the bottom of the soap?
     
  20. Jun 13, 2005 #19

    DDS

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    5.9e-3m

    because if your given the thickness to be 2.09 and the submerged distance is 1.50 then the oil occupies the remaing space which is 0.59 cm
     

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  21. Jun 13, 2005 #20

    OlderDan

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    No, it's not. I told you that when you add the oil the distance the soap sinks into the water is going to change. The buoyant force of the oil is going to raise the soap in relation to the water line. Call the distance from the top of the oil to the water line x. What is the distance from the the water line to the bottom of the soap in terms of x?
     
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