Understanding the Moment of a Force: Statics of Rigid Bodies

In summary: Can you see why? Can you figure out the orientation (CW or CCW) of the moments produced by the rest of the forces? In summary, the conversation discusses how to identify the direction of a force in finding its moment. It mentions the concepts of magnitude of the moment, direction of the moment, moment arm, and magnitude of the force. It also brings up the use of cross-product and the right hand rule to determine the orientation of the moment. The conversation ends with a problem-solving example involving finding the moment about a point and the orientation of each individual force's moment.
  • #1
Frederica Salt
16
3
How can I identify the direction of a force in finding its moment? :rolleyes:
 
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  • #2
Do you know that magnitude of the moment? The direction of the moment? The moment arm? The magnitude of the force?
 
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  • #3
EM_Guy said:
Do you know that magnitude of the moment? The direction of the moment? The moment arm? The magnitude of the force?
I can hardly elaborate my question. :frown: I mean when you have to find its components and get the moment using its components. It's force x distance right? But the negative and positive thing there. It confuses me.
 
  • #4
Are you familiar with the cross-product and the right hand rule?

It sounds like you need to think about the basic concept of what a moment is.

Have you ever tried to push open a door by pressing on the door very near the hinge of the door? Did you find that to be difficult? What happens when you press on the door further away from the hinge?
 
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  • #5
EM_Guy said:
Are you familiar with the cross-product and the right hand rule?

It sounds like you need to think about the basic concept of what a moment is.

Have you ever tried to push open a door by pressing on the door very near the hinge of the door? Did you find that to be difficult? What happens when you press on the door further away from the hinge?
Yes we had that discussion in my physics 1 class. I have here an example. I hope this could help us understand each other.
 

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  • #6
EM_Guy said:
Are you familiar with the cross-product and the right hand rule?

It sounds like you need to think about the basic concept of what a moment is.

Have you ever tried to push open a door by pressing on the door very near the hinge of the door? Did you find that to be difficult? What happens when you press on the door further away from the hinge?
It's 2D.
 
  • #7


So, ##M = \sum(r \times F)## The object tends to rotate about the axis of the moment vector.
 
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  • #8
EM_Guy said:


So, ##M = \sum(r \times F)## The object tends to rotate about the axis of the moment vector.

Sir, thank you so much for the help.
 
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  • #9
 
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  • #10
EM_Guy said:

Thank you. :)
 
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  • #11




Hopefully these videos help.
 
  • #12
Frederica Salt said:
Yes we had that discussion in my physics 1 class. I have here an example. I hope this could help us understand each other.
That is really simple question. Firstly, you should find the moment of point "O" ,that is zero, to find Fx and Fy. Finally, F is sqrt[(Fx)^2 + (Fy)^2].
 
  • #13
Frederica Salt said:
Yes we had that discussion in my physics 1 class. I have here an example. I hope this could help us understand each other.

Do you know what the moment about O is or what the magnitude of the (3,4,5) force is? Or do you need to find the magnitude of this force so that the moment about O would be equal to 0?

The total moment about O is the sum of the moments about O for each force. The first step is to find the moment arm between O and the point at which the forces are being applied. Then you do cross-products.
 
  • #14
EM_Guy said:
Do you know what the moment about O is or what the magnitude of the (3,4,5) force is? Or do you need to find the magnitude of this force so that the moment about O would be equal to 0?

The total moment about O is the sum of the moments about O for each force. The first step is to find the moment arm between O and the point at which the forces are being applied. Then you do cross-products.
Thyphon said:
That is really simple question. Firstly, you should find the moment of point "O" ,that is zero, to find Fx and Fy. Finally, F is sqrt[(Fx)^2 + (Fy)^2].
This was it. I would always have hard time in determining the negative and positive thing when it comes to problems like this one.
 

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  • #15
Frederica Salt said:
This was it. I would always have hard time in determining the negative and positive thing when it comes to problems like this one.

Just take it one step at a time, and make sure each step makes sense.

In this problem, if you were going to find the moment about A, it would just be the sum of the moments of each force about A. The first thing to do would be to use the right hand rule to make sure you understand the orientation of the moment about A of each individual force. For example, by the right hand rule, you can tell that the force Q produces a counter-clockwise (CCW) moment about A. The upwards 80 lb force would produce a CCW moment about A, but the downwards 80 lb force would produce a CW moment about A. Also, the downwards 80 lb force would produce a moment about A of greater magnitude than the moment produced by the upwards 80 lb force. Can you see why? Can you figure out the orientation (CW or CCW) of the moments produced by the rest of the forces?
 
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  • #16
Take a look these below.

1.JPG
2.JPG
3.JPG
4.JPG
5.JPG
6.JPG
 
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  • #17
EM_Guy said:
Just take it one step at a time, and make sure each step makes sense.

In this problem, if you were going to find the moment about A, it would just be the sum of the moments of each force about A. The first thing to do would be to use the right hand rule to make sure you understand the orientation of the moment about A of each individual force. For example, by the right hand rule, you can tell that the force Q produces a counter-clockwise (CCW) moment about A. The upwards 80 lb force would produce a CCW moment about A, but the downwards 80 lb force would produce a CW moment about A. Also, the downwards 80 lb force would produce a moment about A of greater magnitude than the moment produced by the upwards 80 lb force. Can you see why? Can you figure out the orientation (CW or CCW) of the moments produced by the rest of the forces?
Thank you tho. :) We had this problem in our class for Mechanics and fortunately I was able to finally get it. :biggrin: Salamat for your help. :)
 

Related to Understanding the Moment of a Force: Statics of Rigid Bodies

1. What is the definition of a moment of a force?

The moment of a force, also known as torque, is a measure of the tendency of a force to rotate an object about an axis.

2. How is the moment of a force calculated?

The moment of a force is calculated by multiplying the magnitude of the force by the perpendicular distance from the axis of rotation to the line of action of the force.

3. What is the difference between a moment and a force?

A force is a push or pull on an object, while a moment is a rotational force that causes an object to rotate about an axis.

4. How does the moment of a force affect the equilibrium of a rigid body?

In order for a rigid body to be in equilibrium, the sum of the moments acting on it must be equal to zero. This means that the moment of a force can either create or counteract other moments in order to maintain balance.

5. What are some real-world applications of understanding the moment of a force?

Understanding the moment of a force is crucial in many engineering and physics applications, such as designing stable structures, analyzing the stability of bridges and buildings, and determining the optimal placement of weights on a lever or seesaw.

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