Results 1 to 5 of 5

Math Help - Buoyancy Pressure

  1. #1
    Newbie
    Joined
    Mar 2008
    Posts
    12

    Buoyancy Pressure

    Hi,
    I am looking to work out the buoyancy pressure of an object of mass 2kg when held in a tank of tap water at a depth of 300mm (final answer in mmHg). Using Archimedes, I have that the buoyancy force is equal to:

    P*A*y*g

    Where p is the density of the water (I worked this out to be 998.23 kg/m^3 for tap water at 20 degrees, namely room temperature), A is the cross sectional area of the object (the object has dimensions 0.23m x 0.38 m x 0.15 m, the face in the water corresponds to an area of 0.23 x 0.38 = 0.0874m^2), y is the depth at which the object is submerged (0.3 metres) and g is gravity, taken as 9.81.

    Then:

    998.23*0.0874*0.3*9.81 = 256.76 Newtons.

    However, I am then unsure what to do with this. I have buoyancy force, how to I convert to buoyancy pressure?!? Do I use Pressure = Force/Area?!? If I can, which area do I consider, the cross sectional area of the object?!? If I do this, I then have:

    Pressure = 2937.757 Pa (Pascals)
    so that:

    2937.757 Pa = 22.035 Torr.

    Then, since 1 mmHg = 1 torr, we have 22.035 mmHg. Is this correct for this problem?!?


    Any help would be appreciated.
    Last edited by Abelian; June 6th 2008 at 11:20 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,888
    Thanks
    326
    Awards
    1
    Quote Originally Posted by Abelian View Post
    Hi,
    I am looking to work out the buoyancy pressure of an object of mass 2kg when held in a tank of tap water at a depth of 300mm (final answer in mmHg). Using Archimedes, I have that the buoyancy force is equal to:

    P*A*y*g

    Where p is the density of the water (I worked this out to be 998.23 kg/m^3 for tap water at 20 degrees, namely room temperature), A is the cross sectional area of the object (the object has dimensions 0.23m x 0.38 m x 0.15 m, the face in the water corresponds to an area of 0.23 x 0.38 = 0.0874m^2), y is the depth at which the object is submerged (0.3 metres) and g is gravity, taken as 9.81.

    Then:

    998.23*0.0874*0.3*9.81 = 256.76 Newtons.

    However, I am then unsure what to do with this. I have buoyancy force, how to I convert to buoyancy pressure?!? Do I use Pressure = Force/Area?!? If I can, which area do I consider, the cross sectional area of the object?!? If I do this, I then have:

    Pressure = 2937.757 Pa (Pascals)
    so that:

    2937.757 Pa = 22.035 Torr.

    Then, since 1 mmHg = 1 torr, we have 22.035 mmHg. Is this correct for this problem?!?


    Any help would be appreciated.
    I am too tired to make a full response right now but I have something to point out and then a question.

    Point: The buoyant force is equal to the weight of the water the object displaces. You cannot calculate it using the pressure equation as you have written. If the entire volume of the object has been submerged then the buoyant force on the object is B = \rho gV where V is the volume of the object and \rho is the density of water.

    Question: What are you defining buoyant pressure as? If I am not mistaken the buoyant pressure is the difference in pressure between the top and bottom of the object (if both surfaces are parallel and horizontal.) Please correct me if that isn't the definition you are using.

    -Dan
    Last edited by topsquark; June 7th 2008 at 06:56 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Mar 2008
    Posts
    12
    Hi,
    Thanks for your reply. I was using my notes from mechanics last year:

    "Ignoring friction, the only forces acting on the object are its weight due to gravity and the Archimedes buoyancy force. Archimedes principle says that the buoyancy force acting on a body is equal to the weight of the displaced fluid. The objects weight is W = mg and the buoyancy force is:

    F(buoy) = pAyg..."

    (with p, A, y and g defined as above, the 'object' was from an example given in the notes of a bottle with sand in it).

    I was using what I had above, and adapting it to this case. I couldnt understand why there should be a difference?!? Although, I have read the problem back through and have spotted the word submerged now, so once I work out the buoyancy pressure using:

    B=p*V

    with p the pressure of the water, V the volume of the object, how would I calculate the pressure at the 'top', assuming that the value of B that I have found above is for the bottom?!? Would I then get an answer in Newtons?
    Last edited by Abelian; June 6th 2008 at 03:43 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,888
    Thanks
    326
    Awards
    1
    Quote Originally Posted by Abelian View Post
    Hi,
    Thanks for your reply. I was using my notes from mechanics last year:

    "Ignoring friction, the only forces acting on the object are its weight due to gravity and the Archimedes buoyancy force. Archimedes principle says that the buoyancy force acting on a body is equal to the weight of the displaced fluid. The objects weight is W = mg and the buoyancy force is:

    F(buoy) = pAyg..."

    (with p, A, y and g defined as above, the 'object' was from an example given in the notes of a bottle with sand in it).

    I was using what I had above, and adapting it to this case. I couldnt understand why there should be a difference?!? Although, I have read the problem back through and have spotted the word submerged now, so once I work out the buoyancy pressure using:

    B=p*V

    with p the pressure of the water, V the volume of the object, how would I calculate the pressure at the 'top', assuming that the value of B that I have found above is for the bottom?!? Would I then get an answer in Newtons?
    The buoyant force is the net force on the object due to the fluid it is immersed in. If we have a simple object like a cube we can calculate it by finding the pressure on the bottom face of the object and find the upward force on it from that and subtract from that the force on the pressure from the top face. If the height of the cube is y then we get the force equation that you provided:
    F_B = \rho gy A = \rho gV
    (Sorry for the typo in my first reply!) To find the pressure at the top of the cube you use
    P = mgd
    where d is the depth of the top face. To find the pressure at the bottom of the cube, etc.

    If you are trying to find the force on the bottom of the cube with just the buoyant force and dimensions of the cube then you can't do it: all we can get is the net difference in the pressure.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Mar 2008
    Posts
    12

    Wink

    Hi,
    Right, so do I then have that if the box is held underwater at a depth of 0.3 metres:

    Pressure on top face = 2 * 9.8 * 0.3 = 5.88 Pa

    Pressure on bottom face = 2 * 9.8 * 0.45 = 8.82 Pa

    Do I then substitute both of these into the force equation for p? Or do I use

    Pressure = Force / Area

    and since I know the pressure for the top and bottom faces, and the cross sectional area, I can work out the Force's for both?!?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Calculating Buoyancy
    Posted in the Math Topics Forum
    Replies: 1
    Last Post: October 13th 2009, 09:50 PM
  2. Buoyancy Problem
    Posted in the Calculus Forum
    Replies: 4
    Last Post: March 15th 2009, 06:47 AM
  3. Buoyancy Force Question
    Posted in the Advanced Applied Math Forum
    Replies: 1
    Last Post: October 31st 2008, 08:03 AM
  4. [SOLVED] Physics questions, air buoyancy and volume
    Posted in the Math Topics Forum
    Replies: 2
    Last Post: May 20th 2008, 12:08 PM
  5. Calculating volume using buoyancy
    Posted in the Math Topics Forum
    Replies: 4
    Last Post: October 31st 2007, 03:01 AM

Search Tags


/mathhelpforum @mathhelpforum