# Fluid Force problem

1. Jan 30, 2007

### dean_travers

I would appreciate any help I can get w/the following problem:
A square plate of side 'a' ft is dipped in a liquid of weight density 'p' lb/ft^3. Find the fluid force on the plate if a vertex is at the surface and a diagonal is perpendicular to the surface.

I understand the surface has a diamond shape--I have tried dividing it into two triangles and finding the fluid force for each and adding them but have been unable to obtain the textbook's answer.

The two integrals I get are:
fluid force 1 = integral (limits: a*sqrt2/2 and 0) p * 2h^2 * dh
fluid force 2 = integral (limits: a*sqrt2 and a*sqrt2/2) p * ( h + (a*sqrt2)/2) * (2h) * dh

Thanks,
Dean

2. Jan 31, 2007

### HallsofIvy

Staff Emeritus
Since this is clearly homework, I am moving it.

Yes, dividing into two triangles is probably the best way to go. If we take a coordinate system with (0,0) at the upper corner, positive y-axis along the vertical diagonal, then the sides of the upper triangle are given by y= x (x> 0) and y= -x (x< 0). The horizontal diagonal is at y= $a\sqrt{2}/2$, half the length of the diagonal. A thin (thickness dy) horizontal rectangle, at given y will have length y-(-y)= 2y and so area 2ydy. The depth is y so the pressure at that depth is py. The total force on that thin rectangle is py2dy. Integrate that from y= 0 to $a\sqrt{2}/2$ to get the total force on the upper triangle.

The lower triangle can be done in a similar way. Now, let the origin, (0, 0) be at the lower vertex, the positive y axis upward. The calculation for the area is exactly the same: 2ydy. However, the depth of that vertex at (0,0) is now $a\sqrt{2}$ and the depth of the horizontal line at height y above the vertex is $a\sqrt{2}-y$ so you must integrate $2py(a\sqrt{2}-y)$ from 0 to $a\sqrt{2}/2$.
Hmmm- if you are clever, the hard part of those two integrals will cancel!