Fluid Pressure and Fluid Force in Calculus II

In summary, to set up the integral for finding the fluid force on a half-full cylindrical gasoline tank with a diameter of 3 feet and gasoline weighing 42 pounds per cubic foot, one must use integral calculus and strips of area to account for the varying pressure and non-uniform area. This involves finding the expressions for the depth of the fluid and horizontal length at a given depth, as well as using the density of the fluid in the integral expression.
  • #1
Shay10825
338
0
Hi. I need help on how to set up the integral for these problems.

1. A cylindrical gasoline tank is placed so that the axis of the cylinder is horizontal. Find the fluid force on a circular end of the tank if the tank is half full, assuming that the diameter is 3 feet and the gasoline weighs 42 pounds per cubic foot.
Answer: 94.5 lbs

Thanks
 
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  • #2
1. You have to find the force on the bottom half of a circular area. In effect, you will be finding the force (local pressure at the depth * element area) on an elemental strip (horizontal) and summing them (i.e. calculus integration). Have you done any of this yet ? i.e making up horizontal strips.
 
  • #3
Fermat said:
Have you done any of this yet ? i.e making up horizontal strips.

No i have not. Why do you make horizontal strips? How would you set up the integral?
 
  • #4
2. It just looks like you should take the average/effective pressure acting on the porthole as the pressure of seawater at a depth of 15 ft.
And multiply that be the area of the porthole.
 
  • #5
yeah I got number 2 a couple of seconds ago
 
  • #6
How do I set up number 1?
 
  • #7
Shay10825 said:


No i have not. Why do you make horizontal strips? ...
That's what you do when doing integral calculus from basics.

Have you not done any of that stuff before? You should have, if you've been given that question.
 
  • #8
Fermat said:
That's what you do when doing integral calculus from basics.

Have you not done any of that stuff before? You should have, if you've been given that question.

THe last thing we did was Centroids and for all of the problems I just use the formula in the book. But for this problem i keep using the book's formula but I can't get the correct answer. I don't make strips.
 
  • #9
You're not getting the right answer because the pressure varies over the depth of the fluid, and you can't use an average pressure because the area varies non-uniformly over the depth. That's why you have to use integral calculus and strips of area to work it out.

Have you never worked out the area under a curve using cakculus , where you use strips of small areas?
 
  • #10
I used the formula:

F= w integral[ h(y)*L(y) dy
 
  • #11
Could you tell me what h(y) and L(y) are?

That integral looks like it will solve some problems - but what area is involved? square, circular?
 
  • #12
h(y) = the depth of the fluid

L(y) = the horizontal length of the region y
 
  • #13
Ok.

What YOU have to do now is work out the correct expressions for h(y) and L(y).

I'll start you off. h(y) is h = y, where y is the depth of fluid below the half-way line.

L(y) is the length of the horizontal line at the depth y. i.e L is a chord of the circle, the mid-point of which is at a distance y from its centre.

The w in your integral expression should be the pressure at the depth y - is that correct ?

Edit: no - now I think w is the density of the fluid.
 
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1. What is fluid pressure?

Fluid pressure is the force exerted by a fluid per unit area. It is caused by the constant random motion of molecules in a fluid and is measured in units of force per unit area, such as pounds per square inch (psi) or pascals (Pa).

2. How is fluid pressure calculated?

Fluid pressure is calculated using the equation P = F/A, where P is the pressure, F is the force exerted by the fluid, and A is the area over which the force is applied. In calculus II, this equation is used to find the pressure at a specific point within a fluid, taking into account the density and velocity of the fluid at that point.

3. What is the difference between static and dynamic fluid pressure?

Static fluid pressure is the pressure exerted by a fluid at rest, while dynamic fluid pressure is the pressure exerted by a fluid in motion. In calculus II, both types of pressure are considered and can be calculated using different equations and principles.

4. How does fluid force relate to fluid pressure?

Fluid force is the total force exerted by a fluid on a surface, and it is directly related to fluid pressure. In calculus II, the concept of fluid force is used to calculate the net force exerted on a submerged object, taking into account the varying fluid pressure at different points on the object's surface.

5. What are some real-life applications of fluid pressure and fluid force?

Fluid pressure and fluid force have many practical applications, such as in hydraulics, aerodynamics, and hydrodynamics. They are used in the design of hydraulic systems, aircraft wings, and ship hulls, as well as in studying the motion of fluids in pipes, pumps, and turbines. Understanding these concepts is also crucial in fields like meteorology and oceanography.

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