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：

$$ \sum{M_A’=0:} $$

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

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

 

[erobz]

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.

 

[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.

 

[erobz]

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 $F_p$?

 

[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 $F_p$?

see https://upload.cc/i1/2022/12/21/QHOYL5.jpg

 

[erobz]

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]

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]

 

[tracker890 Source h]

View attachment 319150

Thank you for your clear answer.

 

[Steve4Physics]

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.)

 

[Lnewqban]

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.

 

[haruspex]

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.

 

[erobz]

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!

 

[erobz]

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)?

 

[haruspex]

Don't we need to multiply by the width of the gate (into the page)?

I took $\rho$ as an area density,