Fluid force on a dam just trying to reason with the problem

In summary, the conversation discusses the calculation of total pressure on a dam shaped like a right triangle with a height of 4 and a base of 6. The right angle is at the top right when looking at the face of the dam. The density is 64 lb/cubic foot and the pressure increases with depth. The solution involves flipping the dam over and using similar triangles to calculate the width of each horizontal slice. The total force is found by integrating over the full height of the dam.
  • #1
1MileCrash
1,342
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Homework Statement



A dam is shaped like a right triangle, 4 height and 6 across.

The right triangle is oriented so that the right angle is at the TOP RIGHT. (we are looking at the face of the dam)

Density is a weight density, and is 64 lb/cubic foot

Homework Equations





The Attempt at a Solution



OK, to work this out, am I allowed to just flip the dam over? If the pressure increases with depth, it decreases with height, and if I flip it over, it increases with height on the plane.

So I flipped it over.

Now, the area of a rectangular slice of the triangle (which must be horizontal slices so that tiny slices have constant pressure) have an area of delta y X width.

To find the width, I invoked the wrath of similar triangles. Since width is always 3/2 of height, and the height of a "triangle" made from the larger one is given by 4-y (because partitioning a triangle at y=4 would be nothing, while one at y=0 is the whole thing) so the width of the triangle is (3/2)(4-y)delta y

Since I flipped the dam, pressure increases with depth according to p = 64y

And since force = area X pressure,

64y(3/2)(4-y)delta y is the force on any slice of the dam.

Integrating over the full height of the dam, from 0 to 4, should give me total force.

Is

[itex]\int^{4}_{0} 64y(1.3(4-y)) dy[/itex]

The total pressure on the dam? Do I roughly know what I'm doing?

Just seeing if I intuitively grasp it before actually reviewing it for final. Always helps me.
 
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  • #2
For some reason the edit button is missing... integral should have 1.5 in it, not 1.3.
 

Related to Fluid force on a dam just trying to reason with the problem

1. What is fluid force?

Fluid force is the force exerted by a fluid (such as water or air) on an object. It is a result of the pressure and velocity of the fluid acting on the surface of the object.

2. How does fluid force affect a dam?

Fluid force can cause significant stress on a dam, especially during events like floods or high water levels. The force of the water pushing against the dam can potentially cause damage or failure if the dam is not designed to withstand it.

3. How is fluid force calculated on a dam?

Fluid force on a dam is calculated using the formula F = ρghA, where ρ is the density of the fluid, g is the acceleration due to gravity, h is the height of the fluid above the dam, and A is the surface area of the dam. This formula takes into account the pressure and velocity of the fluid acting on the dam.

4. How can fluid force on a dam be controlled or mitigated?

There are several ways to control or mitigate fluid force on a dam. One method is to increase the height and width of the dam to better withstand the force of the fluid. Another option is to incorporate spillways or other structures that can release excess fluid and reduce the pressure on the dam. Regular maintenance and inspection of the dam can also help identify and address any potential issues before they become major problems.

5. What are the potential consequences of not considering fluid force on a dam?

Ignoring fluid force on a dam can have serious consequences, including dam failure and flooding. This can result in damage to surrounding areas, loss of property and life, and significant financial costs. It is important for engineers and scientists to carefully consider fluid force when designing and maintaining dams to ensure their safety and stability.

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