Understanding Moment of Force: Exploring its Concept and Applications

• dorothy
In summary: There are four forces acting on the plank. The condition or relation that must be obeyed by these four forces is that they are all located at the same distance from a pivot point.
dorothy
Homework Statement
I want to know whether I draw the free body diagram correctly? Thank you.

T = tension
f = friction
W = gravitational force
Relevant Equations
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The forces that you have drawn look good. But you have left out a force. See if you can spot the missing force.

Also, you will want to explain from your diagram why there must be a friction force at point A.

TSny said:
The forces that you have drawn look good. But you have left out a force. See if you can spot the missing force.

Also, you will want to explain from your diagram why there must be a friction force at point A.
Is it the normal reaction missing?

dorothy said:
Is it the normal reaction missing?
Yes. To be sure that you are thinking correctly, I would need to see where you place the force and the direction of the force.

TSny said:
Yes. To be sure that you are thinking correctly, I would need to see where you place the force and the direction of the force.

R stands for the normal reaction. Is it correct? also, I’ve added a new T for tension. Is the T in red needed？

The reaction force R should be a normal force acting normal to the wall. I assume there is a wall.

The red tension force is not a force acting on the plank. So, it would not be included in a free-body diagram of the plank.

If this is a graded assignment, can you give us assurance that you are allowed to get external help? (I should have asked this at the beginning )

TSny said:
The reaction force R should be a normal force.

The red tension force is not a force acting on the plank. So, it would not be included in a free-body diagram of the plank.

If this is a graded assignment, can you give us assurance that you are allowed to get external help? (I should have asked this at the beginning )
It’s not a graded assignment. It’s just a simple exercise because now I’m in vacation so I would like to do some more practice to train myself

TSny
TSny said:
The reaction force R should be a normal force acting normal to the wall. I assume there is a wall.

The red tension force is not a force acting on the plank. So, it would not be included in a free-body diagram of the plank.

If this is a graded assignment, can you give us assurance that you are allowed to get external help? (I should have asked this at the beginning )

So I do draw the R correctly?

dorothy said:
So I do draw the R correctly?
No.
My interpretation of the diagram is that the left end of the plank is up against something like a wall. The wall exerts a force against the left end of the plank. This force from the wall has a component parallel to the wall (this is the friction force) and a component normal to the wall.

Looks very good. Can you use this diagram to explain why the friction force cannot be zero and why it must be in the upward direction?

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TSny said:
Looks very good. Can you use this diagram to explain why the friction force cannot be zero and why it must be in the upward direction?
Is it because ‘friction has to be there to balance the gravitational force W ’?

dorothy said:
Is it because ‘friction has to be there to balance the reaction force in order to stay at rest’?
No. The friction force is vertical while the reaction force is horizontal. Since they are in different directions they cannot balance each other.

What are the basic conditions that the forces and the torques of the forces must obey in order for the plank to be in equilibrium?

dorothy said:
Is it because ‘friction has to be there to balance the gravitational force W ’?
I‘ve just change my thought , is it correct in this case?

TSny said:
What are the basic conditions that the forces and the torques of the forces must obey in order for the plank to be in equilibrium?
The two forces should be parallel and with equal magnitude but opposite in direction

dorothy said:
The two forces should be parallel and with equal magnitude but opposite in direction
There are four forces acting on the plank. What condition or relation must be obeyed by these four forces if the plank is in equilibrium?

Is there also a condition involving torques?

Take a look at your FBD and see whether or not each force has a counterpart which prevents the plank from moving in any direction, as it actually happens.
Remember, summation of forces along x-axis and y-axis must be zero, if balance exists.

If any of those forces is located at certain distance from a pivot point, it could make the body rotate about that point (not a balance condition); therefore, something must prevent that from happening.

TSny said:
There are four forces acting on the plank. What condition or relation must be obeyed by these four forces if the plank is in equilibrium?

Is there also a condition involving torques?
Can I answer this question in this way?

TSny said:
There are four forces acting on the plank. What condition or relation must be obeyed by these four forces if the plank is in equilibrium?

Is there also a condition involving torques?
Net force=0?

TSny said:
There are four forces acting on the plank. What condition or relation must be obeyed by these four forces if the plank is in equilibrium?

Is there also a condition involving torques?
Torque can be calculated by 5f?

dorothy said:
Can I answer this question in this way?View attachment 299700
No.
The tension force, T, has a vertical component. Perhaps this vertical component of T happens to cancel W. If so, the friction would not be needed. However, there is a way to show that f needed. Think about torques.

dorothy said:
Torque can be calculated by 5f?
I think we need to remember the basic conditions required for equilibrium:

(1) The vector sum of all of the forces acting on the plank must be zero.

(2) The sum of all of the torques acting on the plank must be zero for any choice of the location of the origin for calculating the torques.

Condition (1) can be restated as saying two things:
(1a) The sum of the horizontal components of the forces must add to zero.
(1b) The sum of the vertical components of the forces must add to zero.

We've already seen that (1b) doesn't give us enough information to decide whether or not the friction force is required. And (1a) won't be of any help since it deals with horizontal forces while f is vertical.

Lnewqban
TSny said:
I think we need to remember the basic conditions required for equilibrium:

(1) The vector sum of all of the forces acting on the plank must be zero.

(2) The sum of all of the torques acting on the plank must be zero for any choice of the location of the origin for calculating the torques.

Condition (1) can be restated as saying two things:
(1a) The sum of the horizontal components of the forces must add to zero.
(1b) The sum of the vertical components of the forces must add to zero.

We've already seen that (1b) doesn't give us enough information to decide whether or not the friction force is required. And (1a) won't be of any help since it deals with horizontal forces while f is vertical.
I get the concept but I don’t know how to explain it in Q1

dorothy said:
I get the concept but I don’t know how to explain it in Q1
Try picking point B as an origin for the torques.

(1) Does the tension force T produce any torque about B? If so, is the torque clockwise or counterclockwise?

(2) Does the weight W produce any torque about B? If so, is the torque clockwise or counterclockwise?

(3) Does the reaction force R produce any torque about B? If so, is the torque clockwise or counterclockwise?

Based on your answers to these three questions, can you argue that there must be a friction force at A and that the friction force must be upward?

TSny said:
Try picking point B as an origin for the torques.

(1) Does the tension force T produce any torque about B? If so, is the torque clockwise or counterclockwise?

(2) Does the weight W produce any torque about B? If so, is the torque clockwise or counterclockwise?

(3) Does the reaction force R produce any torque about B? If so, is the torque clockwise or counterclockwise?

Based on your answers to these three questions, can you argue that there must be a friction force at A and that the friction force must be upward?
1. Yes, clockwise
2. Yes, anti-clockwise
3. No
Ohhh is it because we need the moment due to f to balance with the moment due to W in order to reach the equilibrium?

dorothy said:
1. Yes, clockwise
2. Yes, anti-clockwise
3. No
Ohhh is it because we need the moment due to f to balance with the moment due to W in order to reach the equilibrium?
You have the right idea

But, the tension force T does not produce any torque about the origin at B. Can you see why?
This is important in order to be able to logically deduce that there must be an upward friction force.

TSny said:
You have the right idea

But, the tension force T does not produce any torque about the origin at B. Can you see why?
This is important in order to be able to logically deduce that there must be an upward friction force.
because the force T is pointing upward and it doesn’t “push” the plank?

dorothy said:
because the force T is pointing upward and it doesn’t “push” the plank?
No.

Have you covered the concept of "line of action of a force"? What can you say about the torque of a force about some origin if the line of action of the force passes through the origin?

I'm curious as to your reason for stating (correctly!) that the torque due to R about point B is zero.

TSny said:
No.

Have you covered the concept of "line of action of a force"? What can you say about the torque of a force about some origin if the line of action of the force passes through the origin?

I'm curious as to your reason for stating (correctly!) that the torque due to R about point B is zero.
oh because I think that T is pointing upward so it won’t pass B. But for R, it is pointing horizontally and will pass B so its torque is 0.

dorothy said:
oh because I think that T is pointing upward so it won’t pass B. But for R, it is pointing horizontally and will pass B so its torque is 0.
Good reasoning for R, but not for T. If you drew a line through T, the line would pass through B. So, just like R, the line of action passes through B. So, the torque about B due to T is zero.

It should be clear that: any force that acts at the origin would not produce any torque about that origin since the line of action of the force clearly passes through the origin.

TSny said:
(2) The sum of all of the torques acting on the plank must be zero for any choice of the location of the origin for calculating the torques.
If the sum of the forces is zero and if the sum of the torques about a chosen origin is zero then the sum of the torques about any other choice of origin will also be zero.

hutchphd and TSny

1. What is a moment of force?

A moment of force, also known as torque, is the measure of the rotational force applied to an object. It is calculated by multiplying the magnitude of the force by the distance from the axis of rotation to the point of application of the force.

2. How is moment of force different from regular force?

Moment of force is a type of force that causes an object to rotate around an axis, while regular force causes an object to move in a straight line. Moment of force also takes into account the distance from the axis of rotation, while regular force does not.

3. What are some real-life applications of moment of force?

Moment of force is used in many everyday activities, such as opening a door, using a wrench to tighten a bolt, and throwing a ball. It is also essential in more complex tasks, such as driving a car, operating machinery, and performing sports movements like swinging a golf club or throwing a discus.

4. How is moment of force calculated?

The formula for calculating moment of force is: Moment of force = force x distance. The force is measured in Newtons (N) and the distance is measured in meters (m).

5. How does the concept of moment of force relate to stability?

Moment of force plays a crucial role in determining the stability of an object. The greater the moment of force applied to an object, the more likely it is to rotate and become unstable. This is why objects with a lower center of mass are more stable, as the distance from the axis of rotation is smaller, resulting in a smaller moment of force.

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