1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Fluid mechanics problem

  1. Dec 4, 2008 #1
    1. The problem statement, all variables and given/known data
    A cube of ice floats partly in water and partly in kerosene oil. Find the ratio of the volume of ice immersed in water to that in kerosene oil. Specific gravity of kerosene oil is 0.8 and that of ice is 0.9


    2. Relevant equations

    Upthrust = V(imm)*d*g

    3. The attempt at a solution

    I am confused. The forces acting on the block will be mg downwards, upthrust by the water upwards. In what direction will the thrust due to K.oil act??? upward or downwards??
     

    Attached Files:

  2. jcsd
  3. Dec 4, 2008 #2

    alphysicist

    User Avatar
    Homework Helper

    Hi iitjee10,

    Your attachments have not yet been approved, so I cannot see them. However, the buoyant force is the vector sum of the forces on the top and bottom of the object (due to the pressure of the fluid), and so the buoyant force will be upwards for both kerosene and water (since the pressure is greater the farther down you go).
     
  4. Dec 5, 2008 #3
    how can the kerosene oil apply upward buoyant force when it is completely above the block?
     
  5. Dec 5, 2008 #4

    alphysicist

    User Avatar
    Homework Helper

    The buoyant force is the difference between the force due to the pressure on the bottom and top of the block. The bottom pressure will be larger, so the buoyant force will be upwards.

    If you look more closely, the actual physical upwards force is the water force on the bottom, and it is true the the actual force from the kerosene will be downwards. But we don't look at those two forces individually; the buoyant force takes into account both of those. The water pressure is increased due to the kerosene above it, so the total buoyant effect due to the presence of the kerosene is upwards.

    If this does not make sense, you can calculate the forces directly. Just let the pressure at the top of the block be P0, and let the cube have sides of length x and faces with area A. Find the pressure at the bottom of the block, and then calculate the forces and find their vector sum. You will get back the buoyant force formula with both kersone and water terms.
     
  6. Dec 6, 2008 #5
    can u please do it and include a diagram so that i can understand better?

    Can u also list a website that can explain this concept a bit more clearly??
     
  7. Dec 6, 2008 #6

    alphysicist

    User Avatar
    Homework Helper

    Did you mean the idea in my last post, or the problem in your original post?

    If you meant my last post, call the pressure at the top of the cube P0 as in this figure:

    http://img372.imageshack.us/img372/2446/buoyantcm3.jpg

    Assuming the cube floats "upright", and if the area of the top face is A, then the force downwards from the kerosene on the top face is:

    [tex]
    F_{\rm down} = P_0 A
    [/tex]

    Now find the pressure P1 on the bottom face, using the textbook formula:

    [tex]
    P=P_0 + \rho g h
    [/tex]

    Since there are two different fluids there will be two extra terms, and so the pressure at the bottom of the cube will be:

    [tex]
    P_1 = P_0 + \rho_k g x_1 + \rho_w g x_2
    [/tex]

    where x1 and x2 are given in the figure. The the upwards force on the bottom face is:

    [tex]
    F_{\rm up} =P_1 A = (P_0 + \rho_k g x_1 + \rho_w g x_2) A
    [/tex]
    This is the crucial point to answer your question, because the upward effect of the kerosene shows up in this term. The upward force is only exerted by the water as you pointed out, but that force is increased because of the presence of the kerosene causing the water pressure to be greater. (And the downward force of the kerosene will be cancelled out as you can see next.)

    Finally the buoyant force is the vector sum of those forces:

    [tex]
    \begin{align}
    B &= F_{\rm up} - F_{\rm down}\nonumber\\
    & = (P_0 + \rho_k g x_1 + \rho_w g x_2) A - P_0 A\nonumber\\
    &= ( \rho_k g x_1 + \rho_w g x_2) A\nonumber\\
    &= \rho_k g V_k + \rho_w g V_w\nonumber
    \end{align}
    [/tex]

    In the last step Vk= x1A is the volume of kerosene displaced, and the same for the water volume displaced. These are the two buoyant force terms and they are both upwards.




    For the problem in your original post, I would begin by labeling the volumes V1 and V2 for the parts of the cube in kersone and water. From your force diagram, use Newton's law to write down a force equation, and then solve for V2/V1. What do you get?
     
  8. Dec 9, 2008 #7
    if the kerosene is completely above the block, then there is no kerosene below the block to apply the necessary upward force.
     
  9. Dec 9, 2008 #8

    alphysicist

    User Avatar
    Homework Helper


    That does not matter. As I said in my last post:

    The upward force is only exerted by the water as you pointed out, but that force is increased because of the presence of the kerosene causing the water pressure to be greater.


    The buoyant force is not a single exerted force, it is the total effect of the pressure pushing down on the top and the pressure pushing up on the bottom. And here the pressure pushing up on the bottom of the object is increased due to the presence of the kerosene. So when you consider the total buoyant effect of the kerosene, it is upwards.


    Which means the answer to the question you ask in your original post is that to draw your force diagram for the ice cube, the buoyant forces from the water and the kerosene will both be upwards.
     
  10. Dec 10, 2008 #9
    thanx i have understood.
     
  11. Dec 10, 2008 #10
    the answer is 1:1 right??
     
  12. Dec 10, 2008 #11

    alphysicist

    User Avatar
    Homework Helper

    That looks right to me.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Fluid mechanics problem
Loading...