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Overall Force Acting on a Submerged Cube (in water)

  1. Aug 14, 2014 #1
    This is not homework or coursework. I have an exam coming up and this question is from a past paper and I am unsure why but it is confusing me a bit.

    1. The problem statement, all variables and given/known data
    A cube of side 0.3m which has an average density of 4500 kg/m^3 . Determine the overall force acting on the cube when it is submerged in water.

    Density of water is given to be 1000 kg/m^3


    2. Relevant equations
    [itex]
    p=h \rho g \\
    [/itex]
    Upthrust = [itex] \rho g V = \rho g l^3 [/itex]



    3. The attempt at a solution
    The main reason I am confused is because I thought one would need to know the depth, h, the cube is submerged at?

    I understand that one only needs to consider the forces acting on the top face and bottom face because the others cancel out but again, I have no idea how I am supposed to do that without knowing the depth.

    Is it simply that the total force is equal to the upthrust? Because that's the only equation I can think of that does not include the depth.

    If so then it is simply
    [itex] \rho g l^3 = 4500 \times 9.81 \times 0.3^3 = 1192 N [/itex]

    Also, is there a symbol used to represent upthrust or is it simply something like F with a subscript?

    Any help appreciated. Thanks :)
     
  2. jcsd
  3. Aug 14, 2014 #2

    HallsofIvy

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    I'm no sure what you mean by "submerged under water". Since an ice cube normally floats if left to itself, there would be a net upward force causing it to rise to the surface. If the ice cube is being held at a specific depth then obviously the net force is 0.

    If the ice cube is not being held at a specific depth then use Archimede's principle: Subtract the mass of the ice cube from the mass of water it displaces. That, times "g", gives the upward force on the ice cube.
     
  4. Aug 14, 2014 #3

    Orodruin

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    No, you do not need to know the depth. Pressure on top of the cube is going to push the cube down. Pressure on the bottom of the cube is going to push it up. The net buoyancy force is going to be the difference between those forces and as the top and bottom faces have the same area, it follows that the difference is proportional to the pressure difference between the top and bottom of the cube, which depends on the height of the cube - not on the overall pressure.
     
  5. Aug 14, 2014 #4

    Orodruin

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    The cube has a specific gravity of 4.5, it is not made of ice ... :smile:
     
  6. Aug 14, 2014 #5
    I don't know sorry, that is the exact wording on the exam paper.

    Isn't the equation for upthrust I used the result of Archimede's principle?

    Right OK, that makes it a little clearer for me thanks. So is the answer I got correct do you know?

    Thanks
     
  7. Aug 14, 2014 #6

    Orodruin

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    Your answer would be the force from the pressure of the water on the cube. Is there perhaps another force which is also acting on the cube?

    Regarding Archimedes' principle: Your formula is correct. Just remember that when an object is only partially submerged, the volume that should be used is the volume of the part of the object that is below the surface.
     
  8. Aug 14, 2014 #7
    Regarding Archimedes' principle: Thanks, I assumed in this question that is was completely submerged as it did not say otherwise.

    Reading the question, that is what I took it to mean, the force of the water acting on the cube. I cannot think of any other force, apart from gravity.
     
  9. Aug 14, 2014 #8

    Orodruin

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    The question asks for the "overall force". I would take this to mean the net force on the cube, which definitely should include gravity. If only the force from the water was asked for, it would most likely say something to the effect of "force of the water on the cube" or "buoyancy".
     
  10. Aug 18, 2014 #9
    Ok fair enough. Isn't gravity already taken into account though? Since [itex]mg=\rho V g [/itex]?

    The main reason I thought that the wording meant only the forces of water on the cube is because on the exam paper this is part iiib of one question all about buoyancy and the Archimedes principle.
     
    Last edited: Aug 18, 2014
  11. Aug 18, 2014 #10

    Orodruin

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    There are two forces acting on the cube, one is the pressure from the water, which is in the up direction and is given by the weight of the displaced water through Archimedes' principle (positive direction up)
    $$
    F_{\rm pressure} = \rho_{\rm H_2O} Vg.
    $$
    The other is the gravitational force on the cube due to its actual mass, which is given by (again, positive direction up)
    $$
    F_{\rm gravity} = -m_{\rm cube} g = -\rho_{\rm cube} V g.
    $$
    The net force on the cube is the sum of these
    $$
    F_{\rm net} = V g (\rho_{\rm H_2O} - \rho_{\rm cube}).
    $$
    Among other things, this tells us that the cube would be floating freely around without any net force acting upon it (if water drag is neglected) if it had the same density as water.
     
  12. Aug 18, 2014 #11
    Right ok thanks, thats cleared it up a little. So my original answer is that the force of the water on the cube or gravity because I used the density of the cube rather than the density of water.

    Thanks,
     
  13. Aug 18, 2014 #12

    Orodruin

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    Your original numerical answer would be the gravitational force on the cube and does not include the force from the water on the cube.
     
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