# hydrostatic pressure

• March 8th 2010, 07:38 PM
cdlegendary
hydrostatic pressure
Does anyone know how to setup the integral for the question below? I'm stumped (Headbang)

http://hw.math.ucsb.edu/webwork/math...-prob8_3_7.gif

A vertical plate is submerged in water and has the shape as seen in the picture. Using the fact that the weight of water is http://hw.math.ucsb.edu/webwork/math...284282img1.gif calculate the hydrostatic force (in lbs) against the end of the tank. Give your answer correct to the nearest hundred.
• March 9th 2010, 03:07 AM
HallsofIvy
Are we to assume that the slant sides are of the same length? In that case, setting it up in a coordinate system with origin at the center of the top base, on the water line, The right side goes through the points (6, 0) and (10, -8) so it easy to see that its equation is y= -2(x- 6)= -2x+ 12. The left side goes through the points (-6, 0) and (-10, -8) and so has equation y= 2(x+ 6)= 2x+ 12.

A thin "layer of water" at constant depth will have the same amount of water above it at each point and so constant pressure. The pressure at each point is the weight of the water above, $-\gamma y$ where $\gamma$ is the density of water. (That negative sign is because y itself is negative- the depth is actually |y|.)

The total force on a "thin layer of water" at depth y, of length L and height $\Delta y$ will have total force on it of $-\gamma y L \Delta y$.

That "length", L, will be the difference of the x values at each end and, since we are going to integrate with respect to y (because we have " $\Delta y$"), we need to write x on those slant sides as functions of y.
y= -2x+ 12 on the right so 2x= -y+ 12 and x= -y/2+ 6. y= 2x+ 12 on the right so 2x= y- 12 and x= y/2- 6. At fixed y, the distance between the two sides, and the length of that "thin layer of water" is (-y/2+ 6)- (y/2- 6)= -y + 12.

The weight of that "thin layer of water" and the force on that thin section of plate is $-\gamma y (-y+ 12)\Delta y$.

The total force on the plate then, is given approximately by the Riemann sum $\sum \gamma (y^2- 12y)\Delta y$ which, in the limit, becomes $\gamma \int_{y= -8}^0 y^2- 12y dy$.