How Do You Calculate Fluid Force on a Rotated Square Tank?

[tnutty]

A metal oil tank has cross-section that is a square rotated 45 degrees as shown in the figure above. Its height is 12 ft and width is 12 feet. The oil in the tank has a weight-density of 57 lb/ft3.
Write a definite integral that expresses the fluid force on the end of the tank when the oil is 4 feet deep?

Here is a link to the photo :
http://img17.imageshack.us/img17/2294/76984288.jpg

hopefully you can make it out.

I am not sure how to set up the definite integral.

 

[minger]

Well, I think you can figure out out that
[tex] f = pA[/tex]
Right?

Area is a function that you have. Pressure is a function as well. What are these functions?

 

[nlightner]

I would approach this problem by first understanding the concept of fluid force. Fluid force is the force exerted by a fluid on an object placed in it, and it is directly proportional to the depth of the fluid and the area of the object. In this case, the object is the end of the metal oil tank and the fluid is the oil inside the tank.

To calculate the fluid force on the end of the tank, we need to first determine the pressure exerted by the oil at a depth of 4 feet. This can be calculated using the formula P = ρgh, where P is the pressure, ρ is the weight-density of the oil (57 lb/ft3), g is the acceleration due to gravity (32.2 ft/s2), and h is the depth of the fluid (4 ft). This gives us a pressure of 732.8 lb/ft2.

Next, we need to calculate the area of the end of the tank. Since the cross-section of the tank is a square rotated 45 degrees, we can use the formula A = (s^2)/2, where s is the side length of the square. In this case, s = 12 ft, so the area of the end of the tank is 72 ft2.

Now, we can use the formula for fluid force, F = PA, where F is the fluid force, P is the pressure, and A is the area. Plugging in the values we calculated, we get F = (732.8 lb/ft2)(72 ft2) = 52,761.6 lb.

To express this as a definite integral, we can use the formula ∫F(x)dx, where F(x) is the function representing the fluid force at a given depth x. In this case, x represents the depth of the oil in the tank, and F(x) = (ρgx)/2, since the pressure and area both vary with depth. So our definite integral becomes ∫(57(32.2)x/2)dx, with the limits of integration from 0 to 4 (since we are calculating the fluid force at a depth of 4 ft). Solving this integral gives us the same result of 52,761.6 lb.

In conclusion, the definite integral that expresses the fluid force on the end of the tank when the oil is 4 feet deep is ∫(57