Forces Due to Static Fluids

In summary,-Calculate the height (h) of the water level which can result in failure of the dam by tipping it over due to water pressure at U/S of the dam-Draw the free body diagram for the water pressure-Integrate the moment of the water pressure from top to bottom about the tipping axis A and equate that to the moment generated about the same axis by gravity.
  • #1
CGI
74
1

Homework Statement


Find height (h) of the water level which can result in failure of dam by tipping it over due to water pressure at U/S of dam. Also draw the free body diagram.

Density of water to be = 1000 kg/m^3 and its specific weight=62.4 lb/ft^3
Hints: The specific weight of concrete is = 150 lb/ft^3.
Width of dam = 1ft.

IMG_1005.JPG


Homework Equations



Fr = (Specific Weight)(Height to Centroid)(Area)

The Attempt at a Solution



I've thought about rearranging the Fr equation so that I get
h = Fr/(Specific Weight * Area)
But the part where I'm confused is where to get the Fr from.

I thought that I could get Fr for the concrete by using
150 lb/ft^3 for the specific weight
(12ft * 1 ft*) As the Area
If I were to go about it this way(which I'm not even sure if it is right) how would I go about finding the height
to the centroid?

I tried to draw the free body diagram and I know that the pressure due from the water is in a prism-like shape as shown, but this is all I have. The resultant force from the water is also shown on there.
IMG_1006.JPG


I hope I used the template right! If anyone could help me on this, I would really appreciate it!
 
Physics news on Phys.org
  • #2
I think you'd be allowed to consult tables to get the centroid of that shape. Either the trapezium directly or work it out as the centroid of two combined simple shapes using a formula you can look up.
 
  • #3
Okay, so once I find the centroid, I use that to calculate Fr? And then from there I can use that to find the height to the center of pressure?
 
  • #4
The centroid of the concrete's cross-section shape is where you can assume all its weight is concentrated, for the purpose of determining moments, etc.
 
  • #5
CGI said:
Okay, so once I find the centroid, I use that to calculate Fr? And then from there I can use that to find the height to the center of pressure?
I would integrate the moment of the water pressure from top to bottom about the tipping axis A and equate that to the moment generated about the same axis by gravity. That is safer than assuming some sort of "center of pressure" since the moment arm of the water force is a function of depth, not only in length but also in direction.
 

What is a static fluid?

A static fluid is a fluid in which there is no net movement or flow. This means that the fluid is at rest and all forces acting on it are balanced.

What are the different types of forces due to static fluids?

The two main types of forces due to static fluids are pressure forces and buoyant forces. Pressure forces are caused by the weight of the fluid pressing on an object, while buoyant forces are caused by the difference in pressure between the top and bottom of an object submerged in a fluid.

How do pressure forces affect objects in a fluid?

Pressure forces exerted by a fluid on an object will depend on the density of the fluid and the depth at which the object is submerged. Objects will experience a greater pressure force at deeper depths due to the weight of the fluid above it.

What is the equation for calculating pressure forces?

The equation for calculating pressure forces is F = PA, where F is the force, P is the pressure, and A is the area of the surface the pressure is acting on. This equation is known as Pascal's Law.

How does the shape of an object affect buoyant forces?

The shape of an object affects buoyant forces because it determines the volume of fluid displaced by the object. The larger the volume of fluid displaced, the greater the buoyant force will be. This is why objects with a larger surface area or a hollow shape tend to experience greater buoyant forces.

Similar threads

  • Introductory Physics Homework Help
Replies
9
Views
958
  • Introductory Physics Homework Help
Replies
7
Views
1K
  • Introductory Physics Homework Help
Replies
10
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
14
Views
642
  • Introductory Physics Homework Help
Replies
12
Views
2K
  • Introductory Physics Homework Help
Replies
9
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
5K
  • Introductory Physics Homework Help
Replies
6
Views
5K
Back
Top