1. ## Fluid Pressures/Statics

THE QUESTION IS POSTED ON THE ATTACHMENT FILE BELOW

1. The problem statement, all variables and given/known data
density of the gates is 4000 kg/m^3
width of the gate is 6m

2. Relevant equations
Static Equilibrium : every equation that sums up forces or moments equals zero

You also must find the force of the weight going down F=mg
but first you have to find the mass which is the volume multiplied by the density.

3. The attempt at a solution

I know the force that is produced by the pressure which is 265 KN that is obvious from that fact that Ax=265 KN (different direction)as well .

but if Ay=377 KN[up] and Cy=1389 KN[up]

that means force of the gate is 1012 KN(down)

now my main problem is

1. I've tried manipulating the height, base and width of the triangle to find the volumes for a hour now and I can't find any way you could get the right volume that leads to the forces of 1021 KN

-yes I know you have to multiply the mass by the 9.81 m/s^2 to get the force

-whats driving me crazy is the whole volume issue

2. If the force is 1012KN like they say how did they get the moment/torque arm from the force to point A to be 4.4 m (in order to solve for Cy)???
-I know you have to find the centroid, but I kind of keep getting stuck after that

thanks...

2. Maybe I'm being a bit dense, but I am confused about the diagram. Are we looking down on this? Are we looking at a horizontal cross section? Where is the water?

-Dan

3. the gate is the shaded triangle.

we are looking from the side, so you cannot see the 6m width of the gate.

the water is one the left side of the diagram, you could see the line that determines the surface.

the gate opens when there is a rotation at point (A) causing the the water to be released.

hope this clears it up...

4. Your logic chain is essentially correct. The problem is not only in getting the volume, and hence mass, of the gate, but you have to have the weight of the gate in the correct spot as well. It looks to me like the volume of the gate is going to be
$\displaystyle V = (6)A = 6 \cdot \frac{1}{2}ab~sin(\theta)$
where a = 3 m, b is the length of the line CA, and $\displaystyle \theta$ is the angle ACB.

$\displaystyle V = 6 \cdot 3 \cdot 5\sqrt{2}~cos(45^o)$

As to the location of the weight you need to find the CM of the triangle. You might find this site to be helpful.

-Dan