# Dam problem!

## Homework Statement

Consider a simple model of a dam. A reservoir of water behind the dam is filled to a height h. Assume that the width of the dam is L. As the reservoir behind a dam is filled with water, the pressure that the water exerts on the dam increases. Eventually, the force on the dam becomes substantial, and it could cause the dam to collapse.

There are two significant issues to be considered: First, the base of the dam should be able to withstand the pressure. This means that the material of which the dam is made needs to be strong enough so that it doesn't crack (compressive strength).

(a) What is the total force that the water in the reservoir exerts on the dam?

The second issue has to do with the strength of the foundation of the dam. The force of
the water produces a torque on the dam. In a simple model, if the torque due to the water
were enough to cause the dam to break free from its foundation, the dam would pivot
about its base (point P). The foundation of the dam should be strong enough so that the
dam does not topple, which means that the material has to be strong enough that the dam
does not snap (shear strength).

(b) What is the magnitude of the torque about the point P due to the water in the reservoir?

F= P X A

## The Attempt at a Solution

For part a)
break dam into little strips, each strip has a dif force...
dF = PdA = density(g)(L)(h)dh <-- integrate this?

= 1/2 (density *g * L)(h^2)???

for part b)
to find torque, sum up the little pieces?
Integration of d(Torque) = integration of dFrmoment arms = integration of dFy

from here I'm not sure where to go??? any help is appreciated :)

## Answers and Replies

Related Introductory Physics Homework Help News on Phys.org
Doc Al
Mentor
For part a)
break dam into little strips, each strip has a dif force...
dF = PdA = density(g)(L)(h)dh <-- integrate this?

= 1/2 (density *g * L)(h^2)???
OK. But when you are setting up your little strips, don't use 'h' as your variable, since h is a constant equal to the height of the water. (Don't forget about atmospheric pressure.)

for part b)
to find torque, sum up the little pieces?
Integration of d(Torque) = integration of dFrmoment arms = integration of dFy
Yes. Use the same little strips as before, but this time find their torque. Set up an expression for dT similar to what you did for dF. (Heed my advice about changing your variable of integration.)

OK. But when you are setting up your little strips, don't use 'h' as your variable, since h is a constant equal to the height of the water. (Don't forget about atmospheric pressure.)

Yes. Use the same little strips as before, but this time find their torque. Set up an expression for dT similar to what you did for dF. (Heed my advice about changing your variable of integration.)
Alright so if I used y instead of h, that is okay?
so For part a)
break dam into little strips, each strip has a dif force...
dF = PdA = density(g)(L)(y)dy <-- integrate this?

= 1/2 (density *g * L)(y^2)???

and that seems about right? I think I can forget about atmospheric pressure because its the same on either side of the dam correct?

moving on...

Integration of d(Torque) = integration of dFrmoment arms = integration of dFy = integration of (density * g * L)y dy * y? = integration of (density)(g)(L)y^2dy
= 1/3(density*g*L)y^3??

Thanks for the help, correct me if I'm wrong (sure I did something wrong!)

Doc Al
Mentor
Alright so if I used y instead of h, that is okay?
so For part a)
break dam into little strips, each strip has a dif force...
dF = PdA = density(g)(L)(y)dy <-- integrate this?
Perfect.

= 1/2 (density *g * L)(y^2)???
Once you put in the limits of integration, you'll get it in terms of 'h'. (Not y--y is just a variable.)

and that seems about right? I think I can forget about atmospheric pressure because its the same on either side of the dam correct?
Depends what they are looking for. For compression, I'd think that you'd need to consider it.

moving on...

Integration of d(Torque) = integration of dFrmoment arms = integration of dFy = integration of (density * g * L)y dy * y? = integration of (density)(g)(L)y^2dy
= 1/3(density*g*L)y^3??
Almost. Hint: Where is y measured from? (Where is y = 0?)

Perfect.

Once you put in the limits of integration, you'll get it in terms of 'h'. (Not y--y is just a variable.)

Depends what they are looking for. For compression, I'd think that you'd need to consider it.

Almost. Hint: Where is y measured from? (Where is y = 0?)
Oh so you're saying that h and 0 are the limits of the integration so when its integrated, I should get '= 1/2 (density *g * L)(h^2-0)??? I think that makes sense

In regards to where y is measured from, I guess y is zero at the bottom? Hmm I don't really know where y is measured from

Doc Al
Mentor
Oh so you're saying that h and 0 are the limits of the integration so when its integrated, I should get '= 1/2 (density *g * L)(h^2-0)???
Yes. (But you don't need to write the '-0' part.)

In regards to where y is measured from, I guess y is zero at the bottom? Hmm I don't really know where y is measured from
Then you'd better figure it out. Hint: How did you determine the water pressure of each strip?

Yes. (But you don't need to write the '-0' part.)

Then you'd better figure it out. Hint: How did you determine the water pressure of each strip?
Ohh hold on. So y would be zero at the surface of the water (y is the thickness) and it gets larger as you go down. Maybe this has to do with h? so maybe I'd have to convert this to h again like in the last problem?
so 1/3(density*g*L)h^3??

Doc Al
Mentor
Ohh hold on. So y would be zero at the surface of the water (y is the thickness) and it gets larger as you go down.
Right!

Maybe this has to do with h? so maybe I'd have to convert this to h again like in the last problem?
so 1/3(density*g*L)h^3??
Not exactly. Yes, you'll need to end up with 'h', not 'y', just like before. But that's not the correct answer. Hint: What's the torque from each strip? (Note that you are finding torque with respect to the bottom.)

Right!

Not exactly. Yes, you'll need to end up with 'h', not 'y', just like before. But that's not the correct answer. Hint: What's the torque from each strip? (Note that you are finding torque with respect to the bottom.)
I thought I already posted the torque from each strip? 1/3(density*g*L)y^3?? (at least thats what I guess) If I'm finding torque with respect to the bottom, then that is the highest positive y value. I guess I'm confused as to where you are taking me here?? Should I not post that integration anymore? Should I take it back before it was integrated?

Doc Al
Mentor
Should I not post that integration anymore? Should I take it back before it was integrated?
Yes. Write the torque (dT) of each small strip as a function of y and h. Once you have that correct, then you can integrate.

Yes. Write the torque (dT) of each small strip as a function of y and h. Once you have that correct, then you can integrate.
I worked with someone from my class on this problem and they said they ended up with this:

1/2L*density*g*h^3 - 1/3L*density*h^3 = 1/6L*density*g*h^3

They're pretty positive its right and I see where they get what is subtracted but how did they find 1/2L*density*g*h^3??

Doc Al
Mentor
I worked with someone from my class on this problem and they said they ended up with this:

1/2L*density*g*h^3 - 1/3L*density*h^3 = 1/6L*density*g*h^3

They're pretty positive its right and I see where they get what is subtracted but how did they find 1/2L*density*g*h^3??
If you do as I suggested in my last post and set up the correct expression for dT, then just by doing the integration you'll get both terms. Get the expression for torque correct--the rest is just turning the crank (integrating).

If you do as I suggested in my last post and set up the correct expression for dT, then just by doing the integration you'll get both terms. Get the expression for torque correct--the rest is just turning the crank (integrating).
Integration of torque = integration of dF * moment arms

so 1/2 * integration of (density *g * L)(y)dy * y??

thats the best I got, I hope thats (finally) right.

end up with 1/6 density * g * L *h^3

Doc Al
Mentor
Integration of torque = integration of dF * moment arms

so 1/2 * integration of (density *g * L)(y)dy * y??

thats the best I got, I hope thats (finally) right.
No, that's not quite right. Since 'y' is zero at the top of the water, it cannot be the moment arm.

No, that's not quite right. Since 'y' is zero at the top of the water, it cannot be the moment arm.
if y is zero, then should I use h instead?

Doc Al
Mentor
if y is zero, then should I use h instead?
Express the moment arm in terms of y and h. (Remember that h is a constant.)

Express the moment arm in terms of y and h. (Remember that h is a constant.)
moment arm = h-y

1/2 * integration of (density *g * L)(y)dy * (h-y)

so integration of density * g * l * y dy * h - integration of density * g * l * y dy * -y

so 1/2 density * g * l * h^3 - 1/3 density * g * l * y * h^3

= 1/6 density g * L * h^3!

Doc Al
Mentor
moment arm = h-y
Yes!

1/2 * integration of (density *g * L)(y)dy * (h-y)
Good! (Except for that factor of 1/2.)

so integration of density * g * l * y dy * h - integration of density * g * l * y dy * -y

so 1/2 density * g * l * h^3 - 1/3 density * g * l * y * h^3

= 1/6 density g * L * h^3!
Good!

Yes!

Good! (Except for that factor of 1/2.)

Good!
Thanks for stickin with me!

:surprised