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Fluid Force Problem- Caclulus II

  1. Oct 10, 2007 #1
    A porthole on a vertical side of a submarine submerged in sea water has a diameter of 2r feet. Sea water has a weight density of 64 pounds per cubic foot. Find the fluid force on the porthole, assuming that the center of the porthole is d feet below the surface.

    Relevant Equations:
    fluid pressure= weight * height aka p=wh
    fluid force= p*A, A being the area the force is exerted on.

    Attempt at Solution:

    I know its going to be the integral of p*A bound from zero to some number. I thought that p might equal 64y (the weight times the height), but that seems too simple and I think "d" has to be incorporated into that part of the integral somehow. For A, I used the equation of the circle, x^2 + y^2 = r^2, A=y= square root (r^2-x^2). That's where I stopped, because none of that seems right, I feel like I need to have everything in terms of one or two variables in order to integrate and actually find the fluid force. P

    Please help, I am so stuck and frustrated, I drew a diagram but it's not helping much! I desperately need to figure out the correct answer to this problem!!!
    Last edited: Oct 11, 2007
  2. jcsd
  3. Oct 11, 2007 #2


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    "A porthole on a vertical side". The porthole is vertical and different horizontal lines drawn across the porthole have different depths. Your diagram should be a circle with at least one horizontal line across it representing a section of the porthole at a constant depth. Since the center of the porthole is at d feet, and has radius r feet, the depth of each part of the porthole ranges from d-r feet (top) to d+r feet (bottom). Those should be the limits of integration. The equation of a circle of radius r with center at (0, d) is, of course, [itex]x^2+ (y-d)^2= r^2[/itex] or [itex]x= \sqrt{r^2- (y-d)^2}[/itex]. A horizontal strip at depth y, thickness dy, will have area [itex]\sqrt{r^2- (y-d)^2}dy[/itex]. The pressure at depth y is 64y so the total force on that strip is [itex]64y\sqrt{r^2- (y-d)^2}dy[/itex]. That's what you need to integrate.
  4. Oct 11, 2007 #3
    thank you VERY much, but how does one go about integrating that? help, i'm very stuck, i tried using u-substitution for the square root, but i feel that somehow there should be an easy way to integrate the square root function, because its a circle...
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