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Archemedies principle

  1. Nov 17, 2012 #1
    1. The problem statement, all variables and given/known data

    A circular cylinder, of cross sectional radius r = (2 + A) m, f
    oats in equilibrium on the
    horizontal
    flat surface of a liquid. The horizontal plane through the axis of the cylinder is
    parallel to the fl
    at liquid surface, which is very large in extent. A cartesian Oxyz coordinate
    system is constructed such that the axis of the cylinder is aligned with the y coordinate axis and
    the z coordinate axis is vertically upwards. The origin O is located on the axis of the cylinder.
    The region above the liquid surface is filled with air at atmospheric pressure patm = (110) kPa.
    The density of the material making up the cylinder is dc = (4)x10^2 kg/m3
    and
    the density of the liquid is dl = (30)x10^2 kg/m3
    The only body force is gravity, which is constant and acts in the negative z direction. The acceleration due to
    gravity is g = 9:81 m/s^2
    . You should assume that the cylinder is long and that you may neglect end effects.
    If the interface between the air and the
    liquid is determined by the plane z = r cos (beta), determine
    the value of (beta) in degrees and give an indication of the numerical accuracy of your result.


    2. Relevant equations



    3. The attempt at a solution

    i have attempted this a few different ways which involve solving the final equation with the bisection method. i have taken the centroid of the cylinder where z = 0, i have then used the equal pressure equations for the points of the bottom of the cylinder directly below the centroid and a place on the surface of the water, the following equation is what i believe to be correct however the answer given tell me i may have gone wrong (since altering the density of the materials does not change the angle found to what i think sufficiently)

    f_value = (dl*r*g - dl*g*(r*cosd(x))) * ( (pi*r^2/2) + (pi/2 * (r^2 - ((r*sind(x))^2))) - (pi*r^2*dc*g))

    my question is have i gone about this correctly or is there another way im not seeing?

    it may take me a little while to get back i work part time on weekend so may struggle to find time to reply thanks for any help given :)
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 17, 2012 #2

    haruspex

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    Let's take this in stages.
    What does r = (2+A) mean in the problem statement?
    Here's your equation, cleaned up a bit:
    f = (dl r g - dl g r cosd(x)) ( πr2/2 + π/2 (r2 - (r sind(x))2) - πr2dcg)
    = dl r g(1 - cosd(x)) πr2(1 - (sind(x))2/2 - dcg)
    What are cosd(x), sind(x)?
    Instead of leaping straight to that equation, what formula do you have for the cross-sectional area below the liquid level?
    Labelling that area As and the whole cross-section At (=πr2), say, what equation do you have for the balance of forces?
     
  4. Nov 18, 2012 #3
    hi sorry, A is a constant in this case 5 (our numbers change depending on our student numbers)

    as for sind(x) and cosd(x) they are just the matlab programming equivalent of sine and cosine but for angles in degree.

    for below the liquid area i have assumed that the water level will be above the centre point, and then used

    πr^2 - π/2 (r sin (x))^2 : where the second part of the equation is the segment area above the water level

    since we don't know the level at which the cylinder will fall into equilibrium the x value is a constant to be found which for the formula ive stated should be between [0 : 90]

    i have then used the equal pressure equation

    Pa + Da g Za = Pb + Db g Zb << where a and b denote subscripts of the point a and the lowest level below the water surface in contact with the cylinder and b at the water surface level, Z = 0 is taken at the centroid so for point a Z would be -7 and point b Z would be r cos(x)

    i have taken both Pa and PB to be atmospheric pressure (which is where i may have gone wrong but don't know what else to do)

    substituting this in i derived the equation for upthrust of

    [pl r g + pl (r cos(x)) g ] * area submerged

    i have then equalled this to the weight of the cylinder πr^2 * pc * g

    and then final came to my original equation
     
    Last edited: Nov 18, 2012
  5. Nov 18, 2012 #4

    gneill

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    Can you justify the above assumption? It seems to me that the density of the liquid is much greater than that of the floating object (3000 versus 400 kg/m3), so a check on this assumption would be in order.

    Suppose that the cylinder was held so that exactly half of it was submerged. What would be the force of buoyancy? How would it compare to the cylinder weight? (Note, you can use Archimedes' principle easily here since you can calculate the volume of displaced liquid).
     
  6. Nov 18, 2012 #5
    Hi I've just had a look I can see what you mean the density of my cylinder would mean it sitting much higher on the surface.. I tried with a density of cylinder much closer to the density of water so it would sit below half way but the results it gives me never differ from around 30

    I'm writing this reply on my phone.. Sorry if it doesn't make much sence
     
  7. Nov 18, 2012 #6

    haruspex

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    That doesn't look right to me. Where are you getting that equation from? It ought to have a mix of linear and trig functions of x.
     
  8. Nov 19, 2012 #7
    okay, so ive taken everything into account, and reworked the equations, however the answer im getting range from 0-45 degrees, between variations of density of the cylinder ranging from zero to half the liquid density, but from whats been said i believe this to be half of what i should be getting?, since if the density is half that of the water then the circle cross sectional face should be split in half?

    the equation that this happens for is now

    pl*g*( r - r*cos(x) ) * r^2/2 * (sin(2x) + (2*pi*x)/180)
     
  9. Nov 19, 2012 #8

    haruspex

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    Yes.
    That's an expression, not an equation. Please show your working.
    (I believe you should end up with an equation of the form dc/dl = f(β), with no mention of r, g or air pressure.)
     
  10. Nov 20, 2012 #9
    well .. i think i may have completely over thought this problem.. surly any floating object with a given weight has to displace the amount of water equal to that of its weight?

    on this basis i started again and turned out a very simple formula i assumed gauge pressure and then quite simply solved the following equation using a my bisection program


    f_value = pc*pi*r^2 - (pl * ((r^2/2)*(-sin(x) + ((pi*x)/180))))

    where ((r^2/2)*(-sin(x) + ((pi*x)/180)) is the segment area of the submerged area


    unsure as to whether this is correct but ive spent too long on the problem and the deadline is tomorrow, thanks for every ones help kyle
     
  11. Nov 20, 2012 #10

    haruspex

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    I think it should be sin(2x), as you had previously.
     
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