# Hydrostatic force

by stunner5000pt
Tags: force, hydrostatic
 P: 1,443 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!
 HW Helper Thanks P: 5,152 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.
 HW Helper Thanks P: 5,152 Your integral is also missing h to convert rho*g to a pressure.
P: 1,443

## 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?
 HW Helper Thanks P: 5,152 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: 1,443
 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??
 HW Helper Thanks P: 5,152 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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