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

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The discussion centers on calculating the total force exerted on a vertical dam by water, specifically a rectangular dam with a width of 25m and a depth of 15m. The pressure exerted by the water varies with depth, leading to confusion in initial calculations. To find the average force, it's suggested to use the average depth (7.5m) in the formula F = ρgA, resulting in an approximate force of 27,582 kN. A discrepancy arises when the correct answer is noted as 27,590 kN, attributed to using different values for gravitational acceleration. The conversation emphasizes the importance of understanding pressure variation and using precise values in calculations.
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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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