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Fluid Mechanics- Pressure on a gate

  • Thread starter giacomh
  • Start date
  • #1
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Homework Statement



http://imgur.com/vXkg0


Homework Equations



Fw=γAH

Ycp=bh3/(12)(h/2)b + (h/2)

The Attempt at a Solution



I found the angle of the gate from the horizontal to be 53.1 degrees. Using this, I found y to be 2.5m and H (vertical distance to centroid) to be 6 m (2.5sin(53.1)+4).

I then found Fw=9810*6*(4*5)=1177200N, and Ycp=(4*53)/(12*20*2.5) +2.5=3.33m.

By taking the moment at the ground, I found Fw*(5-3.33)-5P=0, so P= 393kN. My book says the answer should be 523 kN. Help please.
 

Answers and Replies

  • #2
rude man
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The picture does not show the angle P makes with the horizontal or vertical, so the question looks unanswerable to me. P could be an angle perpendicular to the 5m line or at some angle away from perpendicularity.
 
  • #3
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I believe that P is supposed to be perpendicular surface (opposing Fw)
 

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