Comparing two dams (fluid mechanics question)

[ChiralSuperfields]

Homework Statement

Please see below

Relevant Equations

Please see below

For this problem,

The solution is,

However, I though dams were built to withstand the average force exerted by the water. Therefore, from $PA = F$ a larger width dam would have less force exerted on it due to the greater area so would not have to be as thick (less mass, and thus inertia). Can someone please give me some guidance to come to their way of thinking?

Many thanks!

 

[Lnewqban]

The pressure that the water exert on the wall is the same for both dams in the problem and also for a wider dam containing the same depth of water.
Therefore, the rate F/A remains the same for all three dams.
As much more force will be supported by the wider dam, as wider than the other two it is.

That increased force only need to be accounted for when calculating its resistance to bending (aerial view), but not for overturning of its cross-section (side view).

 

[jbriggs444]

However, I though dams were built to withstand the average force exerted by the water. Therefore, from $PA = F$ a larger width dam would have less force exerted on it due to the greater area

In the case at hand, we are told that both dams have the same width. So this reasoning is not relevant.

As @Lnewqban points out, a hypothetical pair of dams where the widths were different would be subject to the same pressure (same depth of water). But the wider one would have more submerged surface area and hence be subject to more total force from the water.

 

[Lnewqban]

Looking from above, the bending load that your dams are resisting is comparable to the bending load that a fixed ends beam with uniformly distributed load is resisting.

Please, see how the internal stresses (moments and shear) in the walls of the dams, as well as in our imaginary beam, are estimated:
https://www.engineeringtoolbox.com/beams-fixed-both-ends-support-loads-deflection-d_809.html

Note that the internal moment depends on the square of the length of the wall (width of the dam), reason for which walls of big dams are made forming a horizontal arc rather than straight.

 

[haruspex]

Homework Statement: Please see below
Relevant Equations: Please see below

though dams were built to withstand the average force exerted by the water

No, each part of the dam needs to withstand the forces and torques exerted on it. Typically, the dam is thicker at the base to withstand the greater pressure there.

 

[ChiralSuperfields]

No, each part of the dam needs to withstand the forces and torques exerted on it. Typically, the dam is thicker at the base to withstand the greater pressure there.

Thank you for your help @haruspex!

 

[ChiralSuperfields]

Thank you for your replies @Lnewqban and @jbriggs444!

Sorry I though I had already thanked you.

Many thanks!

 

[ChiralSuperfields]

No, each part of the dam needs to withstand the forces and torques exerted on it. Typically, the dam is thicker at the base to withstand the greater pressure there.

Do you please know whether that is from the equation that pressure at a point is $P = \frac{dF}{dA}$ where dA is the area of a small point at the dam

Many thanks!

 

[haruspex]

Do you please know whether that is from the equation that pressure at a point is $P = \frac{dF}{dA}$ where dA is the area of a small point at the dam

Many thanks!

No, it's the other way around. The pressure is greater at the base because of the greater depth of water. That means there is a greater force per unit area.
I should also have mentioned that the torque about an axis across the dam is also greatest at the base; at height h from the base of the dam that torque is $\int _{y=h}^H\rho g(H-y)(y-h)dy=\frac 16\rho g(H-h)^3$, where H is the height of water in the dam.

 

[ChiralSuperfields]

No, it's the other way around. The pressure is greater at the base because of the greater depth of water. That means there is a greater force per unit area.
I should also have mentioned that the torque about an axis across the dam is also greatest at the base; at height h from the base of the dam that torque is $\int _{y=h}^H\rho g(H-y)(y-h)dy=\frac 16\rho g(H-h)^3$, where H is the height of water in the dam.

Thank you for your help @haruspex!