Rotational balance problem

tracker890 Source h
Homework Statement:
To determine F
Relevant Equations:
moment balance equation

Please help me to understand why it is wrong to take moment for point ## A’ ## , because I think static equilibrium should be static equilibrium for any point in space.

Method 1：

$$\sum{M_A=0:}$$

$$F\cdot R=\left( F_p \right) _x\cdot \left( R-y_p \right) +\left( F_p \right) _y\left( x_p \right)$$

$$F=\frac{1}{R}\left[ \left( F_p \right) _x\cdot \left( R-y_p \right) +\left( F_p \right) _y\left( x_p \right) \right] ..........\text{(}Ans\text{)}$$

Method 2：

$$\sum{M_A’=0:}$$

$$F\cdot R=\left( F_p \right) _x\cdot \left( R-y_p \right)$$

$$F=\frac{1}{R}\left[ \left( F_p \right) _x\cdot \left( R-y_p \right) \right] ...........\left( wrong\ answer \right)$$

Gold Member
I think you forgot the torque from weight of the water acting at the centroid? Generally, that is what is needed to find the horizontal position of the center of pressure.

Lnewqban
tracker890 Source h
I think you forgot the torque from weight of the water.
The force of water acting on the Gate: ##\left( F_p \right) _x\text{、}\left( F_p \right) _y##
reference.

Gold Member
The force of water acting on the Gate: ##\left( F_p \right) _x\text{、}\left( F_p \right) _y##
reference.

So you are saying that ##F_p## is the force of weight?

Never mind, I think you are saying that ##F_p## is the resultant force acting on the gate. How did you figure out where the ##x## coordinate of ##A'## is?

Also, and maybe this is the problem but there has to be a vertical reaction force acting on the gate? If ##F_p## is the resultant, from what I'm seeing you have shown no vertical reaction force that could possibly balance the vertical component of ##F_p##?

Last edited:
tracker890 Source h
So you are saying that ##F_p## is the force of weight?

Never mind, I think you are saying that ##F_p## is the resultant force acting on the gate. How did you figure out where the ##x## coordinate of ##A'## is?

Also, and maybe this is the problem but there has to be a vertical reaction force acting on the gate? If ##F_p## is the resultant, from what I'm seeing you have shown no vertical reaction force that could possibly balance the vertical component of ##F_p##?

Gold Member
Yeah, I get that.

Where is the vertical reaction force at ##A## that is necessary to balance the vertical component of ##F_p##? You have to satisfy two relationships.

##\sum F = 0 ##

##\sum M = 0 ##

tracker890 Source h
tracker890 Source h
Yeah, I get that.

Where is the vertical reaction force at ##A## that is necessary to balance the vertical component of ##F_p##? You have to satisfy two relationships.

##\sum F = 0 ##

##\sum M = 0 ##
Thank you!
I think the free body diagram should be changed as follows:

Therefore, it is more convenient to take the moment at point A.

erobz
Gold Member

Last edited:
Steve4Physics and tracker890 Source h
tracker890 Source h

erobz
Homework Helper
Gold Member
2022 Award
Just to add to what @erobz has said…

The reaction force of the hinge on the door (a point A) has an unknown magnitude and direction. This reaction force produces a moment about point A'. This moment hasn’t been included in (Post #1) Method 2

(When taking moments about point A, as in (Post #1) Method 1, the reaction can be ignored as it passes through point A.)

tracker890 Source h and erobz
Homework Helper
Gold Member
Therefore, it is more convenient to take the moment at point A.
I would say both points are equally convenient.
You know that the reaction forces at hinge A are:
Fax=F+Fpx
Fay=Fpy

Note that yp will be the location of the centroid of a triangle formed by the horizontal pressure distribution, while xp will be the location of the centroid of a quarter of circle formed by the vertical pressure distribution.

tracker890 Source h and erobz
Homework Helper
Gold Member
2022 Award
Seems to me the easiest way is to forget about centroids and centres of pressure and work from first principles.
At ##\theta## below the horizontal, the force on an element ##R\cdot d\theta## is ##R\rho g\sin(\theta)R\cdot d\theta##. Its torque about A is ##R^2\rho g\sin(\theta)R\cos(\theta)\cdot d\theta = \frac 12R^3\rho g\sin(2\theta)\cdot d\theta##. Integrate.

Gold Member
Seems to me the easiest way is to forget about centroids and centres of pressure and work from first principles.
At ##\theta## below the horizontal, the force on an element ##R\cdot d\theta## is ##R\rho g\sin(\theta)R\cdot d\theta##. Its torque about A is ##R^2\rho g\sin(\theta)R\cos(\theta)\cdot d\theta = \frac 12R^3\rho g\sin(2\theta)\cdot d\theta##. Integrate.
My fluid mechanics text (for Engineers) completely detours it in favor of the formulaic (calculus already done for you) approach. They expect less mathematical finesse of engineers!

Lnewqban
Gold Member
Seems to me the easiest way is to forget about centroids and centres of pressure and work from first principles.
At ##\theta## below the horizontal, the force on an element ##R\cdot d\theta## is ##R\rho g\sin(\theta)R\cdot d\theta##. Its torque about A is ##R^2\rho g\sin(\theta)R\cos(\theta)\cdot d\theta = \frac 12R^3\rho g\sin(2\theta)\cdot d\theta##. Integrate.
Don't we need to multiply by the width of the gate (into the page)?