# Fluid Pressures

1. Dec 14, 2007

### damasgate

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...

#### Attached Files:

• ###### untitled.JPG
File size:
18.8 KB
Views:
59
Last edited: Dec 14, 2007
2. Dec 14, 2007

### coomast

The force coming from the weight of the gate is as you stated nothing more than the density times the volume of the gate. The volume of the gate is simply the area times the length of it. The length is given as 6m and the area can be obtained from the calculation of elementary area's. Indeed, consider the area composed as follows, a rectangular area with width 5m and height 3m (bottom left corner is C, top right is not drawn, but in coordinates (5,3) if you consider the axes system in C) plus a triangle width 5m and height 2m on top of this rectangular area minus the triangular area width 5m and height 5m, corners in C, A and in (5,0). This area is now 7.5m². The volume and the weight is now easily obtained as 45m³ and 1766kN. The force from the water is indeed 265kN.

Now you need to determine where these forces are. The water force is at a height of 1m from the bottom pointing towards the right, the weight of the gate is at the center of gravity of the triangular shape of it. Thus pointing down at 5/3m from the Y-axes I mentioned. This weight is the sum of the two reaction forces in Y-direction, not the difference as you have used.

Can you proceed from this in solving the question? You seem to know the formula's you need.

3. Dec 14, 2007

### damasgate

thanks alot, but I still don't understand exactly the 5/3m distance from the y-axis for the force of the gate...

4. Dec 14, 2007

### coomast

The weight of the gate is situated at the center of gravity of the gate and since it is a triangle, it's center of gravity is located at 1/3 of the perpendicular height, in any of the three directions. There's more on this here (go towards the middle of the article):

http://en.wikipedia.org/wiki/Triangle

Hope this helps, if not, please post.

5. Dec 14, 2007

### coomast

Mmmm, the remark on perpendicular might be a little confusing. It is better to use the following, the center of gravity of a triangle is situated at a point which can be determined by drawing a line parallel to one of the legs at 1/3 of the distance towards the opposite corner point. If you do this for all three legs, you get a single intersection point which is the one you need.

I used this fact only to draw a parallel line with the vertical leg of the gate. This at 1/3 of the distance towards point A. Therefore the 5/3 value.

6. Dec 14, 2007

### damasgate

um...should it be 2/3 of the distance.....i.e 5x(2/3)?

7. Dec 14, 2007

### coomast

No, 1/3 of the distance between the vertical leg (Y-axes if you use the axes system I proposed) and the point A.