In this screencast we will go through an example focused on determining the hydro static forces
on an inclined plane surface.
Specifically let's look at this case where we have a rotating gate that is braced to
prevent water from spilling out of a reservoir.
The brace is between the gate and a movable stop such that the angle which we will designate
as theta can be varied.
Let's find the force the brace exerts on the gate as a function of theta, the gate length
which we will designate as "lg" and the brace length which we will designate as "lb".
With any problem the first place to start is by drawing a picture.
You can see we have a pretty good picture of the system already here so since we want
to determine the forces we should start with a free body diagram on the gate.
We will start by drawing our gate and out brace acting against that gate.
We will label some kind of origin.
We will start down here at the pivot and say that in the up direction we have our positive
y and in the horizontal we have our positive x.
At this point we are going to call this the force from the brace, we have a weight force
acting down through the center of the gate and we have the force acting from the water
on the gate which we will designate as "Fr" for the resultant force from the water acting
on the gate.
To keep going from here we need forces in the horizontal and vertical axis so we need
to label another angle in this picture which we will call phi.
This allows us to write the force from the brace in the vertical axis as Fb*sin(phi)
and in the horizontal axis as Fb*cos(phi).
We will need some more lengths to accurately represent this system.
The horizontal distance, origin to where weight acts through the gate is going to be Lg divided
by 2 times cos(theta).
The total distance from the origin to where the brace acts on the gate is going to be
Lg cos(theta).
The resultant force on the water acting on the gate acts through the center of pressure
which for a gate like this would be 2/3 from the top or 1/3 from the origin.
We would designate this distance as Lg/3.
The two vertical distances that we care about are from the origin to the center of the gate
where the weight acts.
This is going to be Lg/2*sin(theta) and from the origin to where the brace acts which is
the top of the gate.
This distance is going to be Lg*sin(theta).
At this point we can write our force balance on the gate.
We know that the sum of all forces must be 0.
This means that the force acting in the clockwise direction must be balanced with that acting
in the counterclockwise direction.
When we do our force balance the force of the water acting on the gate is the resultant
force, Fr, times the distance at which it is acting.
We write Lg/3.
This is going to be plus the weight force of the gate which is the distance times the
origin that that force acts which is Lg/2*cos(theta).
This is going to equal the counterclockwise direction.
The force from the brace in the horizontal direction times cos(phi) times its distance
from the origin, Lg sin(theta) plus the brace force acting in the vertical direction would
be sin(phi) times the distance from the origin which will be Lg*cos(theta).
We know that the resultant force, Fr, can be rewritten as the specific weight times
the height to the center times the area at which it is acting.
This vertical distance hc we know has the height to the center so we can rewrite this
as a specific weight of water times Lg/2*sin(theta)* the area Lg times some width that is going
into the paper for the gate as w.
At this point we are plugging the resultant force into this first equation and rewrite
it so that it looks like the following.
At this point we can divide the whole equation by the length of the gate and solve for Fb.
We need a relationship between theta and phi.
If we draw our diagram again we know that this angle is theta, This angle is phi.
We could draw two triangle where we know this distance to be the length of the gate and
this distance to be the length of the brace.
We could say that the length of the gate times sin(theta) is equal to the length of the brace
times sin(phi) since each equals this common length inside of the triangle.
Therefore we could say that theta is equal to the inverse sin(lb/lg*sin(phi)).
As much as plugging this back into this equation would be fun I would suggest using some kind
of solver using both equations so that you can plug in values for the weight of the gate,
the specific weight of the water, the length of the gate, as well as the length of the
brace.
Then given a certain value of theta you could therefore solve the force of the brace acting
on the gate.