Hydrostatic Force on a Plate: Solving with Reimann Sums and Integrals

[redjohncorn]

Homework Statement

Find the hydrostatic force against one side of the plate by Reimann sum. Then express the force as an integral and evaluate it.

Homework Equations

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

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)

 

[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"?

 

[redjohncorn]

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, as a function of depth more or less.

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

 

[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

 

[redjohncorn]

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

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

 

[QuarkCharmer]

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

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?

 

[redjohncorn]

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?

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.

 

[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?

 

[redjohncorn]

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?

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

 

[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.