Solve Hydrostatic Force w/ Triangle Plate - Math Homework

[stunner5000pt]

Homework Statement

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.)

Homework Equations

hydrostatic pressure is p = \rho g h

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:
\frac{w}{8}=\frac{9-x}{6}

w = \frac{8}{6} (9-x)

The force is then given by:

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

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!

 

[SteamKing]

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]

Your integral is also missing h to convert rho*g to a pressure.

 

[stunner5000pt]

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.

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:

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

How does this look?

 

[SteamKing]

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]

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]

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.