Results 1 to 4 of 4

Math Help - Hydrostatic force

  1. #1
    Newbie
    Joined
    Jul 2008
    Posts
    1

    Hydrostatic force



    Find the hydrostatic force exerted on one end of the plate.

    Help~~
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    If we let the surface of the water be the x-axis and bisect it with the y-axis,

    then perhaps we copuld find the equation of the line which represents the

    side of the plate.

    The bottom right corner will have coordinates (10,-8) and the top right corner will have coordinates (6,0)

    Using these two points to find our line equation we get y=-2x+12 or x=6-\frac{y}{2}

    The depth is -y and the length of the region at y is 2x = 2\left(6-\frac{y}{2}\right)=12-y

    So, we can set up the integral:

    62.4\int_{-8}^{0}(y^{2}-12y)dy=34611.2

    You could also try similar triangles and see if you get the same thing.
    Last edited by galactus; July 9th 2008 at 06:30 PM. Reason: I did err. Used wrong coordinates. Fixed now.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Apr 2005
    Posts
    1,631
    Quote Originally Posted by porknbeans View Post


    Find the hydrostatic force exerted on one end of the plate.

    Help~~
    Hydrostatic force F on a submerged area A is F = \rho*A
    where \rho = hydrostatic pressure at the centroid of A,
    and A is perpendicular to \rho

    Since the given A is in feet or English, then the unit weight of water is 62.4 lbs per cubic ft., so,
    \rho = (62.4*y) lbs/sq.ft ....where y is the depth from water surface.

    The given figure of A can be subdivided into two ploygons:
    >>>a rectangle that is 12ft long by 8ft high
    >>>and an isosceles triangle that has a base of 8ft and an altitude of 8ft.

    For the rectangle:
    We make the water surface as the x-axis, and going down is positive.
    dA1 = 12*dy ....the element of A1.
    F1 = INT(0 to 8)[(62.4(y) *12dy]
    F1 = (12*62.4)INT(0 to 8)[y]dy
    F1 = 748.8[(y^2)/2](0 to 8)
    F1 = 748.8[64/2 -0]
    F1 = 23.962 kips

    For the isosceles triangle:
    We put the (0,0) at the apex, and going down is positive also.
    We get the relationship between x and y for the right hypotenuse,
    Let dA3 be anywhere between y = 0 and y = 8, so
    base for dA3 = x
    depth at dA3 = y
    >>>dA3 is element for A3. A3 = area of one right triangle whose base is 4 and height is 8.
    By proportion,
    y/x = 8/4
    4y = 8x
    x = y/2
    dA3 = x*dy = (y/2)dy
    So, dA2 = (y/2 dy)*2 = y dy2 ....there are 2 right triangles in the isosceles
    Then,
    F2 = INT(0 to 8)[(62.4*y)(y dy]
    F2 = (62.4)INT(0 to 8)[y^2]dy
    F2 = (62.4 /3)[y^3](0 to 8)
    F2 = 20.8[(8)^3 -0]
    F2 = 10.65 kips

    Therefore, F = F1 +F2 = 23.962 +10.65 = 34.612 kips ----answer.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    I made a mistake by using the wrong coordinates for my line equation on the last post, but I fixed it. Thanks for the heads up ticbol.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. hydrostatic force
    Posted in the Calculus Forum
    Replies: 0
    Last Post: March 24th 2009, 01:32 PM
  2. hydrostatic force
    Posted in the Calculus Forum
    Replies: 6
    Last Post: December 9th 2008, 09:24 PM
  3. Triangle hydrostatic force!
    Posted in the Calculus Forum
    Replies: 4
    Last Post: June 24th 2008, 03:36 PM
  4. Hydrostatic Force Problem!!!
    Posted in the Calculus Forum
    Replies: 7
    Last Post: June 23rd 2008, 07:01 PM
  5. Hydrostatic Force
    Posted in the Calculus Forum
    Replies: 1
    Last Post: June 20th 2008, 03:49 PM

Search Tags


/mathhelpforum @mathhelpforum