Hydrostatic Problem: Hinged water gate at the bottom of a reservoir

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CWatters
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Homework Statement:
See Below
Relevant Equations:
F = ρgAd
Question.jpg

Diagram.jpg

My son and I are having a problem with part c that suggests our answer for part a and b are wrong..

Part a). If we've understood the teaching notes correctly the force on the gate acting through its centroid perpendicular to the gate and is given by

F = ρgAd

where
ρ is the density of water 997 kg/m^3
g is 9.81m/s^2
A is the area of the gate
d is the depth of the centroid of the gate

The area of the gate A is

A = 1.1 * 0.8/cos(45)
= 1.24 sqm

The centroid is half way up the gate so at a depth d below the surface of

d = 2-0.4
= 1.6m

So the Force F acting through the centroid is

F = 997 * 9.81 * 1.24 * 1.6
= 19,405 N

Part b) The centroid of the gate is at..

y = 0.5 * 0.8/cos(45)

so the moment M acting on the gate due to water pressure is

M = 19,405 * 0.5 * 0.8/cos(45)
= 10,977 Nm

Part c) If the gate isn't accelerating the net moment must be zero.

The clockwise moment due to the mass Q is

= 0.8*Q*g

The anti clockwise moment due to water pressure is 10,977 Nm from above.

Equating these gives

0.8*Q*g = 10,977

and

Q = 10,977/(0.8*g)
= 1,398 kg

So we have failed to prove mass Q needs to be 1,527 kg

Where are we going wrong?
 
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Answers and Replies

  • #2
BvU
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[edit]misfired :)
 
  • #3
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I think part b may not have been done right. If the pressure at y is p(y). the moment of the pressure force is $$M=\int{wyp(y)dy}$$This would not be equal to the force times the distance to the centroid.
 
  • #4
CWatters
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Thanks Chester! That was the problem. Looking back at his notes confirms you need to integrate over the length of the gate to get the net force and moment. He's written it up in the attached MSword doc.
 

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