What is the Total Force Exerted on a Vertical Dam by Water?

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Homework Help Overview

The problem involves calculating the total force exerted on a vertical rectangular dam by water, given specific dimensions and depth of the water. The subject area pertains to fluid mechanics and hydrostatic pressure.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of pressure at varying depths and the implications of using average pressure versus point pressure. There is also consideration of the correct value for gravitational acceleration.

Discussion Status

The discussion is active with participants exploring different methods to calculate the force on the dam. Some guidance has been provided regarding the variability of pressure with depth and the suggestion to consider average pressure for simplification.

Contextual Notes

Participants note discrepancies in the gravitational constant used in calculations, which may affect the final result. There is an acknowledgment of the provided answer differing slightly from calculations based on different values of g.

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Homework Statement



A vertical rectangular dam of width 25m contains water in a reservoir of depth 15m. What is the overall force exerted on the dam by the water?

Homework Equations



P=F/A ... F=PA
P = pgh

The Attempt at a Solution



I'm really not sure about this one to be honest.

I tried finding the Pressure (1000*9.807*15) and then multiplying that by area (15*25) but that doesn't give the right answer. Any tips would be much appreciated.
 
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The reason that F = \rho\cdot A does not give the correct answer is that the pressure varies with depth.

Does that help nudge you in the right direction?
 
Thanks for the reply, so I need to find the Pressure at each point along the Dam?

Which would be (1000 * 9.807 * 1) + (1000 * 9.807 * 2) ... up to 15m?
 
bcfcdan said:
Thanks for the reply, so I need to find the Pressure at each point along the Dam?

Which would be (1000 * 9.807 * 1) + (1000 * 9.807 * 2) ... up to 15m?
You could do that, but it would be easier (and sufficient in this case) to find the average force on the damn.
 
Would that be (1000 * 9.807 * 7.5) * (15 * 25)?

Which gives 27,582kN
 
bcfcdan said:
Would that be (1000 * 9.807 * 7.5) * (15 * 25)?

Which gives 27,582kN

Yup :biggrin:
 
Thanks for your help. Just looked at the answer that we were given for this question and it was 27,590kN.
 
bcfcdan said:
Thanks for your help. Just looked at the answer that we were given for this question and it was 27,590kN.
No problem.

You would have got exactly that answer if you'd have used g=9.81\;\text{m}\cdot\text{s}^{-2}.
 
Ah yes, I thought that might be it. Bit odd considering we get told to use g = 9.807 and then the correct answer is through using it as 9.81!
 

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