Why Is Taking Moments at Point A' Incorrect in Rotational Balance Problems?

In summary: No, the torque in the x-direction is the same.No, the torque in the x-direction is the same.In summary, the gate will not stay in the correct position if the moment is taken at point A'.
  • #1
tracker890 Source h
90
11
Homework Statement
To determine F
Relevant Equations
moment balance equation
1671624792635.png

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:

$$ \sum{M_A’=0:} $$

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

$$ F=\frac{1}{R}\left[ \left( F_p \right) _x\cdot \left( R-y_p \right) \right] ...........\left( wrong\ answer \right) $$
 
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  • #2
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.
 
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  • #3
erobz said:
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.
 
  • #4
tracker890 Source h said:
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##?
 
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  • #5
erobz said:
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
 
  • #6
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 ##
 
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  • #7
erobz said:
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:
1671632623632.png

Therefore, it is more convenient to take the moment at point A.
 
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  • #8
1671640909553.png
 
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  • #9
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  • #10
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.)
 
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  • #11
tracker890 Source h said:
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.
 
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  • #12
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.
 
  • #13
haruspex said:
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!
 
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  • #14
haruspex said:
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)?
 
  • #15
erobz said:
Don't we need to multiply by the width of the gate (into the page)?
I took ##\rho## as an area density,
 
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1. What is rotational balance problem?

The rotational balance problem is a physics concept that refers to the state in which an object is in equilibrium or balance while rotating around a fixed axis. This means that the object's center of mass remains at a constant distance from the axis of rotation, resulting in a stable and smooth rotation.

2. How is rotational balance measured?

Rotational balance is measured using the principle of torque, which is the product of force and distance. The object is said to be in rotational balance when the clockwise torque is equal to the counterclockwise torque, resulting in a net torque of zero. This can be calculated using the equation: Torque = Force x Distance.

3. What factors affect rotational balance?

Several factors can affect rotational balance, including the shape, mass, and distribution of weight of the object. The distance of the object's center of mass from the axis of rotation also plays a crucial role in determining its rotational balance. Other external factors such as friction and air resistance can also affect rotational balance.

4. How does rotational balance relate to everyday life?

Rotational balance is an essential concept in everyday life, especially in activities that involve rotating objects, such as bicycle wheels, ceiling fans, and car tires. It is also crucial in sports such as figure skating and gymnastics, where maintaining rotational balance is crucial for performing certain moves and routines.

5. How can rotational balance be improved?

To improve rotational balance, the object's center of mass can be adjusted by changing its shape or redistributing its weight. Reducing friction and air resistance can also help improve rotational balance. Additionally, using a more stable and balanced base for the object to rotate on can also improve its rotational balance.

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