
#1
Feb2713, 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{9x}{6} [/tex] [tex] w = \frac{8}{6} (9x) [/tex] The force is then given by: [tex] \int_{3}^{9} \rho g \frac{8}{6} (9x) [/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! 



#2
Feb2713, 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. 



#3
Feb2813, 12:51 AM

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




#4
Feb2813, 11:25 AM

P: 1,445

Hydrostatic force[tex] F = \int_{3}^{9} \rho g \frac{8}{6}\left( 9  x \right) x dx [/tex] How does this look? 



#5
Feb2813, 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.




#6
Mar113, 07:51 AM

P: 1,445

From the question, the triangle is inverted and submerged 3 ft under water... wouldn't the force be from 0 depth to 9 ddepth?? 



#7
Mar113, 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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