• Support PF! Buy your school textbooks, materials and every day products Here!

A dam, torque problem

  • Thread starter Swatch
  • Start date
  • #1
89
0
The uppper edge af a gate in a dam runs along the water surface. The gate is 2.00 m high and 4.00 m wide and is hinged along a horizontal line trough its center. I have to calculate the torque about the hinge arising from the force due to the water. What I have done so far is to consider the top of the gate as point y=0 , the center as y=1 and the bottom y=2.
Then I calculate the torque of a thin horizontal strip at a depth y and integrate that over the gate.

Total torque from the top to the center would be

Torque=intergrate(r*g* dy *dA ) or 4*r*g*integrate(y^3)
(where r=density of water or 1.00*10^3 kg/m^3)

from this I get torque to be 9800 N*m

Then I do the same for the gate from height y=1 to y=2
and get -147000

The net torque I get is not the right answer, which is by the way2.61*10^4 N*m

Could someone please give me a hint to what I am doing wrong
 

Answers and Replies

  • #2
Doc Al
Mentor
44,904
1,169
It's not clear to me what you are integrating. The force on a thin horizontal strip is [itex]dF = 4 \rho g y dy[/itex]. For the top half of the gate, the torque on that piece (about the center) is [itex]d\tau = (1-y) 4 \rho g y dy[/itex]. Now integrate from y = 0 to 1 to find the net torque on the top half.

Write a similar expression for the torque on the bottom half.
 
  • #3
89
0
Did what you told me Doc Al, and got the right answer. Just one thing I'm not understanding. You say that the force on a thin horizontal strip is dF=4*rho*g dy

The way I see the force is dF=dp*dA
since dp=rho*g dy and dA=4*dy
shouldn't the integration contain y^2
 
  • #4
Doc Al
Mentor
44,904
1,169
Swatch said:
Just one thing I'm not understanding. You say that the force on a thin horizontal strip is dF=4*rho*g dy
No, I said: [itex]dF = 4 \rho g y dy[/itex] (Don't forget the y.)


The way I see the force is dF=dp*dA
since dp=rho*g dy and dA=4*dy
shouldn't the integration contain y^2
[itex]dF = P dA[/itex], not [itex]dP dA[/itex]. You need the actual pressure at the depth y: [itex]P = \rho g y[/itex], not [itex]\rho g dy[/itex].
 
  • #5
89
0
O.K.

So at this little horizontal strip or dA I consider the pressure as constant. Is that the correct way to think of it?
 
  • #6
Doc Al
Mentor
44,904
1,169
Swatch said:
So at this little horizontal strip or dA I consider the pressure as constant. Is that the correct way to think of it?
Right. For a small horizontal strip you can consider the pressure to be uniform, just like you can consider it to be at a single depth.

(FYI: Realize that the pressure varies from P to P + dP across the thickness of the strip. So, the average pressure is P + dP/2. So the average force would be (P + dp/2)(dA) = PdA + dp dA/2. Speaking loosely, that last term is a higher order infinitesimal and can be ignored compared to PdA.)
 

Related Threads on A dam, torque problem

  • Last Post
Replies
1
Views
3K
  • Last Post
Replies
3
Views
2K
Replies
1
Views
5K
  • Last Post
Replies
18
Views
20K
  • Last Post
Replies
1
Views
4K
  • Last Post
Replies
4
Views
6K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
8
Views
4K
Replies
3
Views
5K
Replies
3
Views
3K
Top