# Hydrostatic force problem

1. Nov 15, 2011

### redjohncorn

1. The problem statement, all variables and given/known data
Find the hydrostatic force against one side of the plate by Reimann sum. Then express the force as an integral and evaluate it.

2. Relevant equations
pg do not need to be solved. they can be left like that.
F=force
p= density
g=gravity
a= area
d= distance

3. The attempt at a solution
so far i have the equation
in the attachment. However, I do not know where to put the 1 ft that lies above the triangle.
i have the area as (4/3)(3-y)

#### Attached Files:

• ###### hydrostatic.png
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Last edited: Nov 15, 2011
2. Nov 15, 2011

### QuarkCharmer

That one foot needs to be depth as a function of y. For instance, if you put the top of the triangle as y=0, increasing positively downward, then d(y) = (y+1), because at y=0, the depth will then be 1, as expected. At y=3, the depth will be 4, and so on.

You are really computing the volume of water above a particular area (more specifically, the force it causes), as a function of depth more or less. To solve these problems you can always think "How can I express depth/length/etc as a function of my independent variable"?

3. Nov 15, 2011

### redjohncorn

would this integral be correct? i can never understand the whole concept about which way y is

#### Attached Files:

• ###### hydrostatic.png
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4. Nov 15, 2011

### QuarkCharmer

Where are you calling zero? Can you label your axis.

Just think of it like this:
$$F_{H} = δ \int_{a}^{b}L(x) D(x) dx$$

5. Nov 15, 2011

### redjohncorn

y= 0 starts at the very bottom of the triangle for this integral.

6. Nov 15, 2011

### QuarkCharmer

Okay, so you have (4-y) for your D(y), when y = 1, is the depth at that point 3? Can you test the length function L(y) in the same manner against the diagram?

7. Nov 15, 2011

### redjohncorn

if by L(x) you are talking about the area that needs to be integrated, then the area is (4/3)(3-y) since y is that unknown length above the dy section.

8. Nov 15, 2011

### QuarkCharmer

Yes, the area. L(y) dy is the area of the strip/rectangle. Using your axis, where y=0 is at the bottom of the triangle. When y increases (going closer to the surface), does your function 4/3(3-y) represent the length of the strip at that point? When y = 0, that L(y) function evaluates to 4/3(3-0) = 4, does that sound right to you?

9. Nov 15, 2011

### redjohncorn

no. it doesn't sound right.
ok i changed the area formula to A(y)= (4/3)y
now i think it's correct. thank you

10. Nov 15, 2011

### QuarkCharmer

That sounds right!

Usually, an easier way to do these problems is to put y=0 at the surface of the water, that way your depth function D(x) is just x. Then you simply need to find a way to get out the Length at some particular x value. This can greatly simplify your integrals in most cases that I have seen. If the object is circular, putting the y=0 at the center of the circle can do the same thing.