Question on Moment of a force

  • Thread starter dorothy
  • Start date
  • #1
dorothy
36
1
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:
/
9C7E83E8-BF8B-495B-8654-F10F0BEB15CC.jpeg
 

Answers and Replies

  • #2
dorothy
36
1
I think there’s something missing, please help.
 
  • #3
TSny
Homework Helper
Gold Member
13,831
4,009
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.
 
  • #4
dorothy
36
1
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?
 
  • #5
TSny
Homework Helper
Gold Member
13,831
4,009
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.
 
  • #6
dorothy
36
1
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.
DB145291-2BFA-4AB6-917D-39E6301A4AF1.jpeg

R stands for the normal reaction. Is it correct? also, I’ve added a new T for tension. Is the T in red needed?
 
  • #7
TSny
Homework Helper
Gold Member
13,831
4,009
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 :oldsmile: )
 
  • #8
dorothy
36
1
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 :oldsmile: )
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 😅
 
  • #9
dorothy
36
1
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 :oldsmile: )

So I do draw the R correctly?
 
  • #10
TSny
Homework Helper
Gold Member
13,831
4,009
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.
 
  • #11
TSny
Homework Helper
Gold Member
13,831
4,009
1649616247061.png


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?
 
Last edited:
  • #12
dorothy
36
1
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 ’?
 
  • #13
TSny
Homework Helper
Gold Member
13,831
4,009
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?
 
  • #14
dorothy
36
1
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?
 
  • #15
dorothy
36
1
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
 
  • #16
TSny
Homework Helper
Gold Member
13,831
4,009
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?
 
  • #17
Lnewqban
Homework Helper
Gold Member
2,610
1,411
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.
 
  • #18
dorothy
36
1
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?
06605A14-C3E9-4EF4-914F-8D939766A005.jpeg
 
  • #19
dorothy
36
1
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?
 
  • #20
dorothy
36
1
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?
 
  • #21
TSny
Homework Helper
Gold Member
13,831
4,009
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.
 
  • #22
TSny
Homework Helper
Gold Member
13,831
4,009
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.
 
  • #23
dorothy
36
1
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 🥲
 
  • #24
TSny
Homework Helper
Gold Member
13,831
4,009
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?
 
  • #25
dorothy
36
1
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?
 
  • #26
TSny
Homework Helper
Gold Member
13,831
4,009
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 :oldsmile:

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.
 
  • #27
dorothy
36
1
You have the right idea :oldsmile:

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?
 
  • #28
TSny
Homework Helper
Gold Member
13,831
4,009
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.
 
  • #29
dorothy
36
1
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.
 
  • #30
TSny
Homework Helper
Gold Member
13,831
4,009
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.
 
  • #31
jbriggs444
Science Advisor
Homework Helper
11,569
6,215
(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.
 
  • Like
Likes hutchphd and TSny

Suggested for: Question on Moment of a force

  • Last Post
Replies
9
Views
312
Replies
10
Views
189
  • Last Post
Replies
4
Views
223
  • Last Post
Replies
9
Views
256
Replies
9
Views
572
Replies
9
Views
440
Replies
3
Views
324
Replies
13
Views
350
Top