Water Pressure On Wall of Dam

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1. Dec 8, 2014

missdandy

1. The problem statement, all variables and given/known data

Consider a simple model of a free-standing dam, depicted in the diagram. Water of density ρ fills a reservoir behind the dam to a height h. Assume the width of the dam (the dimension pointing into the page) is w.

(a) Determine an equation for the pressure of the water as a function of depth in the reservoir. (Note that you will need to define your own coordinate system.)

(b) Consider a single horizontal layer of the water behind the dam wall. What is the magnitude dF of the force this layer of water exerts on the dam?

(c) The force of the water produces a torque on the dam. What is the magnitude of the torque about point P due to the water in the reservoir.

(d) Show that the torque about the base of the dam due to the force of the water can be considered to act with a lever arm equal to h/3.

2. Relevant equations
P=ρgh
P=F/A
F=PA
τ=rFsinθ
Lever arm = rsinθ

3. The attempt at a solution
a) Taking y to be the water depth measured from the reservoir floor, P=ρgy. Since ρwater≈1, P≈gy.

b) F=PA=gyA.
A=width*height of water slice = w(dy).
dF=(gyw)(dy)

c) I started with τ=rFsinθ and then tried to look at the torque applied by a single slice of water. I drew this picture:

Then I said that the length L=sqrt(p^2+y^2). Plugging that into the equation for torque along with force, I got dτ=sqrt(p^2+y^2)*(gyw)(dy)sinθ

Then integrating the above with respect to y I got τ=(1/3)gLsinθ(p^2+y^2)^(3/2)
=(1/3)gLsinθ(sqrt(p^2+y^2))(p^2+y^2)

d. Plugging my calculated value for L into the equation for the lever arm I got
Lever arm=sqrt(p^2+y^2)sinθ=h/3

Then substituting that into the equation for torque:
τ=(1/3)gp(p^2+y^2)*(h/3) = (h/9)gp(p^2+y^2)

I'm not really sure where to go from here, or even if all of my solutions to the parts are actually correct. I'd really appreciate a little bit of guidance.

2. Dec 8, 2014

haruspex

If a force magnitude F is applied at some point A, and the distance from A to B is d, the torque F has about B is not necessarily Fd. What else do you have to consider?

3. Dec 8, 2014

missdandy

Do you mean the angle between the force and the lever arm?

4. Dec 8, 2014

jbriggs444

So pressure at the bottom of the reservoir is zero and pressure at the top is ρgh?

5. Dec 8, 2014

ehild

You measure y from the reservoir floor and you state that P=ρgy that is the hydrostatic pressure is zero at the bottom of the water and highest at the surface. is it right?
ρwater≈1 is not true. The numerical value of the density depends on the chosen units. ρwater≈1 g /cm3 or 1000 kg/m3..

About the lever arm: What is r in the formula? What is the lever arm if the axis goes through point P at the bottom of the dam?

6. Dec 8, 2014

jbriggs444

So you have torque = length of lever arm times force times the sine of the angle between them. That's good. But there is some simplification that you can do. This is part of what Haruspex is trying to get at. Your formula for dτ contains a factor of L sin θ. There is a [much!] easier expression for L sin θ in terms of y.

Alternately you can arrive at the same simplification another way. Instead of taking torque as force times moment arm times the sine of the angle between them, you can view torque as force times the component of the moment arm at right angles to the force.

Don't forget to fix your coordinate system before you re-do the integration with this simplification.

7. Dec 8, 2014

ehild

No, the lever arm or moment arm contains that sin(angle) already. The lever arm is perpendicular to the force.
http://hyperphysics.phy-astr.gsu.edu/hbase/torq2.html

8. Dec 8, 2014

jbriggs444

Thank you, ehild. I should have said length of "lever" rather than "lever arm", I suppose.