In this problem we have a gate or a dam that supports a channel of water and we are trying
to figure out the moment acting at the hinge at the bottom of this gate to support it.
So naturally there is a pressure force of the water acting on the gate in the clockwise
direction. So we want to calculate the moment acting about the hinge in the counterclockwise
direction. And the gate extends a width of b into the screen and the height of the gate
is the same as the height of the water, height a. Let's define a coordinate axis, y=0 at
the top, it extends a length y below the surface of the water. The x-axis extends into the
screen. So here is my picture prism, a pressure of 0 at the top and it increases linearly
with depth, or linearly with y. And because of this effect, instead of acting right at
the centroid, the resultant force of the pressure acts a distance 2/3 of he way down from the
top of the water. So if the total height of the gate is a, then the resultant force, we
could call this distance yR, yR is equal to 2/3a. And this means that this distance above
the hinge is equal to a-2/3a or simply 1/3a. So calculate the moment about the hinge, we
need both the lever arm, in this case 1/3a, and the magnitude of the resultant force.
If I look pack at the pressure prism the magnitude the the resultant force FR is equal to the
average pressure times the area of the gate. And because it is a linear pressure profile
with the pressure of 0 at the top, the average pressure is the pressure acting right at the
centroid. And in this case of a rectangular gate the centroid is a value of a over 2.
We will call this hc the depth of the centroid, is 1/2a. And the average pressure in that
case is simply its density times h. So here I have got rho of the fluid times hc, times
g, times the cross sectional area of the gate. And the cross sectional area of the gate is
simply the height of it times the depth into the screen, or in this case a times b. And
if I pug in hc is equal to 1/2a I have got 1/2 rho and g a^2 b is the magnitude of the
resultant force. And finally to calculate the moment, the moment is simply the magnitude
of the resultant force times this distance from the hinge which is 1/3 a. And with the
right hand rule a negative sign is used to indicate a clockwise moment of the fluid acting
on the gate. So then substituting in for FR I get -1/6 rho g a^3 b. I can also calculate
the moment by integrating directly. So if I say the moment is, I'll use again the negative
sign for a clockwise moment, negative times some distance a-y. So if I have a force acting
a distance y from the surface of the water I have got a distance a-y above the hinge.
I will say the integral of a-y times differential force and this differential force, if I continue
this, a and a-y is simply P times the differential area. And continuing with this I have got
negative the area, times again a-y, the lever arm, times rho g y dA. Where rho g y is the
static pressure at some value y below the surface. And if I pull the negative rho g,
I will integrate from a distance 0 to b into the screen and zero to a above the gate, times
again a-y times y dy dx. And I will let you evaluate this integral on your own but when
you go through it you get a value of -1/6 rho g a^3 b, just like we did using the rule
of thumb. The third way to get the same answers, is if I set it up again with the resultant
force acting some distance yR from the surface of the water. And I use this expression, where
yR is equal to the second moment of the area, Ixc over yc times A plus yc. For a rectangular
gate I know that Ixc is equal to 1/12 ba^3. And this is an expression that is tabulated
and it depends on the geometry that you are looking for. And in this case we are looking
at a rectangular gate so this geometry is given to us. In the centroid of a rectangular
object x, right at the geometric center of it. So we have got yc, this distance right
here, is equal to a/2. So this comes again, half the height of the gate below the surface
of the water. So if I plug in, I have got yR is equal to 1/12 ba^3 and it is divided
by 1/2a, y at the centroid, times the area, which is a times b plus 1/2a and if I simplify
this the b's cancel out and I have got a in the numerator and this equals 1/6 times a
plus 1/2 a. And that is equal to 2/3a. So again the moment is equal to negative FR times
a-yR and like we have shown before, FR is negative 1/2 rho g a^2 b. And a-yR now is
simply a-2/3 a. And again what we come up with is negative 1/6 times rho g a^3 b.