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Consider a square plate rotated 45 degrees and submerged in water (diagonal of square is perpendicular to surface of water). To find force on one side of square plate, I assume an $x$-$y$ axis passing through the center of the square, with the $y$-axis perpendicular to the water surface.

At a height of $y$ from the center, I assume a rectangular strip of breadth $dy$. The length of this strip is $2(\frac{a}{2^{0.5}}-y)$. Let $\delta$ be the weight density of water. Therefore, the force on one side of the rectangle will be integral of $f(y)$ as $y$ goes from $-a/2^{0.5}$ to $a/2^{0.5}$, where

$$f(y)=2*\delta*(a/2^{0.5} - y)^{2}.$$

The final answer comes out to be: $a^3*\delta*\frac{2^{5/2}}{3}$.

I am told that this answer is wrong. Where am I going wrong in this one?

  • $\begingroup$ You should consider area strip on the plate, which means it has width $dy~\cos\theta$, where $\theta$ is angle of plate with vertical. $\endgroup$
    – Deep
    Oct 8, 2016 at 5:39
  • $\begingroup$ I don't understand. Can you elaborate on that? $\endgroup$
    – Sat D
    Oct 8, 2016 at 5:49
  • $\begingroup$ Work with coordinate system lying in the plane of the plate. In your case y-axis is sticking out of the plate because you have set it up perpendicular to water surface. Refer Fluid Mechanics by F.M. White. $\endgroup$
    – Deep
    Oct 8, 2016 at 5:53
  • $\begingroup$ Oh no, my y-axis is perpendicular to water surface. Essentially it is concurrent with the diagonal. $\endgroup$
    – Sat D
    Oct 8, 2016 at 5:55
  • $\begingroup$ Please add a figure. $\endgroup$
    – Deep
    Oct 8, 2016 at 6:05

1 Answer 1


Have a look at your length of strip formula $2(\frac{a}{2^{0.5}}-y)$ when $y = -\frac{a}{2^{0.5}}$.
It does not give the required value of zero.
You probably need to split the integration into two terms.


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