# Hydrostatic force

by stunner5000pt
Tags: force, hydrostatic
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 P: 1,439 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 $p = \rho g h$ 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: $$\frac{w}{8}=\frac{9-x}{6}$$ $$w = \frac{8}{6} (9-x)$$ 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!
 Emeritus Sci Advisor HW Helper Thanks PF Gold P: 6,471 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.
 Emeritus Sci Advisor HW Helper Thanks PF Gold P: 6,471 Your integral is also missing h to convert rho*g to a pressure.
P: 1,439
Hydrostatic force

 Quote by 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.
 Quote by SteamKing 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?
 Emeritus Sci Advisor HW Helper Thanks PF Gold P: 6,471 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.
P: 1,439
 Quote by 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.
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??
 Emeritus Sci Advisor HW Helper Thanks PF Gold P: 6,471 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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