Hydrostatic force


by stunner5000pt
Tags: force, hydrostatic
stunner5000pt
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#1
Feb27-13, 11:43 PM
P: 1,445
1. The problem statement, all variables and given/known data
A triangular plate with height 6 ft and a base of 8 ft is submerged vertically in water so that the top is 3 ft below the surface. Express the hydrostatic force against one side of the plate as an integral and evaluate it. (Recall that the weight density of water is 62.5 lb/ft3.)


2. Relevant equations

hydrostatic pressure is [itex] p = \rho g h[/itex]
3. The attempt at a solution
The hydrostatic force is the pressure times the area of the surface that is submerged

the width of the triangle is solved using the following ratio:
[tex] \frac{w}{8}=\frac{9-x}{6} [/tex]

[tex] w = \frac{8}{6} (9-x) [/tex]

The force is then given by:

[tex] \int_{3}^{9} \rho g \frac{8}{6} (9-x) [/tex]

Are the upper and lower bounds of the integral correct?
In this question the weight density is given in lb/ft^3 - does this value replace the value of rho in the above equation or does it represent rho times g?

Thanks for your help!
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SteamKing
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#2
Feb27-13, 11:58 PM
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The weight density of fresh water is 62.5 lb force / ft^3. This value represents rho*g.

You don't explicitly define x. Your integral has an expression for the width of the triangle (which I would check again), but it lacks dx.

I would check to see if your relation for the width gives zero width for depths from 0 to 3 feet. If this relation is in error, I suggest a change in the limits of integration to match the depths of the base and apex of the triangle.
SteamKing
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#3
Feb28-13, 12:51 AM
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Your integral is also missing h to convert rho*g to a pressure.

stunner5000pt
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#4
Feb28-13, 11:25 AM
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Hydrostatic force


Quote Quote by SteamKing View Post
The weight density of fresh water is 62.5 lb force / ft^3. This value represents rho*g.

You don't explicitly define x. Your integral has an expression for the width of the triangle (which I would check again), but it lacks dx.

I would check to see if your relation for the width gives zero width for depths from 0 to 3 feet. If this relation is in error, I suggest a change in the limits of integration to match the depths of the base and apex of the triangle.
Quote Quote by SteamKing View Post
Your integral is also missing h to convert rho*g to a pressure.
Thank you for your help. Based on what you've told, I correct the integral to this:

[tex] F = \int_{3}^{9} \rho g \frac{8}{6}\left( 9 - x \right) x dx [/tex]

How does this look?
SteamKing
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#5
Feb28-13, 02:06 PM
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Since the apex of the triangle is submerged, there can be no additional area from a depth of 3 feet to 0 feet. Your expression for the width of the triangle needs some adjustment to account for this.
stunner5000pt
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#6
Mar1-13, 07:51 AM
P: 1,445
Quote Quote by SteamKing View Post
Since the apex of the triangle is submerged, there can be no additional area from a depth of 3 feet to 0 feet. Your expression for the width of the triangle needs some adjustment to account for this.
Does this mean that the 9 - x should be actually 6 - x?

From the question, the triangle is inverted and submerged 3 ft under water... wouldn't the force be from 0 depth to 9 ddepth??
SteamKing
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#7
Mar1-13, 09:16 AM
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I think you need to take a moment and make a sketch of the problem.

The bottom of the triangle is 9 feet below the surface, and the top is 3 feet below.
From 3 feet below to the surface, there is no triangle, so the width is zero.

If x is going to be your overall depth measured from the surface, then the vertical position relative to the base of the triangle must be the depth minus 3 feet. The depth is required to determine hydrostatic pressure, but the depth coordinate must be modified when determining position between the base and the top of the triangle.


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