Three equal forces applied to a rectangle, find net torque direction?

In summary, the forces at points A, B, and C produce equal and opposite torques, but the force at point F3 does not produce any torque.
  • #1
paulimerci
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47
Homework Statement
Attached below.
Relevant Equations
T = F.d
This is how I interpreted the problem,
a) The net torque about point A is zero. This is because the forces F1 and F2 are equal and opposite, and they act at the same distance from point A. Therefore, they produce torques that cancel each other out..
The force F3 doesn’t does not produce any torque because it acts along the line of action of F1 and F2 and it passes through the COM.
b) The net torque about point B is counterclockwise because the forces F1 and F2 produce torques that are both CCW, while the force F3 doesn’t produce a torque.
c) The net torque at point c is zero, because it's moment is zero.

I'm not sure with the answers I gave, please point out what mistakes I did. Thank you!
 

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  • #2
Part (a)
F1 and F2 are not equal and opposite. Their sum is not zero. Also, when calculating torques be sure you multiply the magnitude of the force by the lever arm. What are the lever arms of F1 and F2?

Part (b)
You are correct that F3 produces no torque. Are you sure that F1 and F2 both produce ccw torques about point B? Check again.

Part (c) is correct.
 
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  • #3
kuruman said:
Part (a)
F1 and F2 are not equal and opposite. Their sum is not zero. Also, when calculating torques be sure you multiply the magnitude of the force by the lever arm. What are the lever arms of F1 and F2?

Part (b)
You are correct that F3 produces no torque. Are you sure that F1 and F2 both produce ccw torques about point B? Check again.

Part (c) is correct.
Thank you, for Part a) I know how to find a lever arm for forces but in this case I don't know how to do it. If I take c as the axis of rotation, the line of action for force F1 is crossing the point c in that case it has no moment. Which point should I take, should I take A the axis of rotation?
 
  • #4
paulimerci said:
Thank you, for Part a) I know how to find a lever arm for forces but in this case I don't know how to do it. If I take c as the axis of rotation, the line of action for force F1 is crossing the point c in that case it has no moment. Which point should I take, should I take A the axis of rotation?
Yes, for part (a) you need to take point A as the axis of rotation because that is what the problem is asking you to do.
 
  • #5
kuruman said:
Yes, for part (a) you need to take point A as the axis of rotation because that is what the problem is asking you to do.
Thank you, If A is the axis of rotation, then F1 produces a CCW torque, and F2 is parallel to the moment arm, so it doesn’t produce any torque. where F3 will exert a CW torque. Correct?
 
  • #6
The only statement that is correct is that F2 produces no torque. Also, it is not enough to say what the direction of the torques for F1 and F3 are. If they are in opposite directions, you need to find which is larger which means you need to find their moment arms in order to calculate the net torque.
 
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  • #7
kuruman said:
The only statement that is correct is that F2 produces no torque. Also, it is not enough to say what the direction of the torques for F1 and F3 are. If they are in opposite directions, you need to find which is larger which means you need to find their moment arms in order to calculate the net torque.
The torque of F3 must be greater than F1 because the lever arm from point A is perpendicular to F1. Whereas the lever arm makes obtuse to the force F3.
 
  • #8
Hi,

Are you aware you can shift a force along its line of action ? If you shift ##F_3## to point ##B## and ##F_1## to ##C##, which of the two has a longer 'arm'?

##\ ##
 
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  • #9
BvU said:
Hi,

Are you aware you can shift a force along its line of action ? If you shift ##F_3## to point ##B## and ##F_1## to ##C##, which of the two has a longer 'arm'?

##\ ##
I understand what you say but am unable to verbalize it. I think F1 has a longer arm.
 
  • #10
paulimerci said:
I understand what you say but am unable to verbalize it. I think F1 has a longer arm.
Since F1 has a longer arm than F3, it makes more torque than F3. F1 is CW, and F3 is CCW. Right?
 
  • #11
paulimerci said:
Since F1 has a longer arm than F3, it makes more torque than F3. F1 is CW, and F3 is CCW. Right?
Part b): The forces F1 and F2 produce equal and opposite torques about point c, while the force F3 doesn't exert any torque. Hence, the net torque at B is zero. correct?
 
  • #12
paulimerci said:
Part b): The forces F1 and F2 produce equal and opposite torques about point c, while the force F3 doesn't exert any torque. Hence, the net torque at B is zero. correct?
Yes.
Note that, because A is in the line of action of F2, you can instead treat F2 as being applied at A. This produces a symmetry.
 
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  • #13
Can't resist... does this help ?
1679997145504.png
##\qquad##
1679997176672.png


1679997192338.png
##\qquad\qquad##
1679997209691.png


[edit] spot the mistake: top red arrow in B is r2 , not r1
 
Last edited:
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  • #14
haruspex said:
Yes.
Note that, because A is in the line of action of F2, you can instead treat F2 as being applied at A. This produces a symmetry.
Thank you!
 
  • #16
BvU said:
Can't resist... does this help ?
View attachment 324141##\qquad##View attachment 324142
For part A) The moment arm r1 is greater than r3. Therefore, the F1 exerts more torque than the F3. F1 is a CW torque, and F3 is CCW. The answer for part a is CW. Right?
 
  • #17
Right.
 
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  • #18
If you still have to ask, we haven't succeeded in explaining adequately ... ?

On the other hand, if you were in error, there would have been a 'gentle' nudge in the right direction :smile:
 
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  • #19
BvU said:
If you still have to ask, we haven't succeeded in explaining adequately ... ?

On the other hand, if you were in error, there would have been a 'gentle' nudge in the right direction :smile:
Sorry for prenudging you.
 
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  • #20
BvU said:
If you still have to ask, we haven't succeeded in explaining adequately ... ?

On the other hand, if you were in error, there would have been a 'gentle' nudge in the right direction :smile:
Is there any explanation that I need to provide?
 
  • #21
paulimerci said:
Is there any explanation that I need to provide?
Even though you don't have numbers, you know that the forces are equal in magnitude. Call that ##F##. As far a distances are concerned, call the side of each square ##L##. Then each torque will be some fraction or multiple of ##FL##. Add the two ##FL##s together and see what you get. Equations provide the most convincing explanations.
 
  • #22
paulimerci said:
Is there any explanation that I need to provide?
Just a different approach:
I would vectorially add the three forces.
The resultant force should be applied at the common point C.

Imaginarily locating the pivot point at A for a), B for b) and C for c), would allow me to intuitively see how the resultant force would make the rectangle rotate.

Note that there is a non-stable balance for the case that B is the pivot point, eventually causing a rotation (CCW or CW) until reaching a stable balance.

Addition of forces.jpg
 
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  • #23
Thank you,
##F1 = F2 = F3 =F ##, ## L2=L3=L1=L4=L##
For part a)
$$ \sum \tau_A = -F L\sqrt 2 + FL$$
For part b)
$$ \sum\tau_B = F \frac {L \sqrt 2}{2} - F \frac {L \sqrt 2}{2}$$
For part c)
$$\sum\tau_C = 0 $$
Do the equations look convincing?
 
  • #24
Lnewqban said:
Just a different approach:
I would vectorially add the three forces.
The resultant force should be applied at the common point C.

Imaginarily locating the pivot point at A for a), B for b) and C for c), would allow me to intuitively see how the resultant force would make the rectangle rotate.

Note that there is a non-stable balance for the case that B is the pivot point, eventually causing a rotation (CCW or CW) until reaching a stable balance.

View attachment 324181
It looks like a good strategy! Thanks for sharing.
 
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  • #25
paulimerci said:
Thank you,
##F1 = F2 = F3 =F ##, ## L2=L3=L1=L4=L##
For part a)
$$ \sum \tau_A = -F L\sqrt 2 + FL$$
For part b)
$$ \sum\tau_B = F \frac {L \sqrt 2}{2} - F \frac {L \sqrt 2}{2}$$
For part c)
$$\sum\tau_C = 0 $$
Do the equations look convincing?
Does the equation look correct? Mistakes?
 
  • #26
paulimerci said:
Thank you,
##F1 = F2 = F3 =F ##, ## L2=L3=L1=L4=L##
For part a)
$$ \sum \tau_A = -F L\sqrt 2 + FL$$
For part b)
$$ \sum\tau_B = F \frac {L \sqrt 2}{2} - F \frac {L \sqrt 2}{2}$$
For part c)
$$\sum\tau_C = 0 $$
Do the equations look convincing?
Yes, but you have not finished part b.
 
  • #27
haruspex said:
Yes, but you have not finished part b.
Thank you; I think I've included all the forces in the equation for Part B. What is missing?
 
  • #28
paulimerci said:
Thank you; I think I've included all the forces in the equation for Part B. What is missing?
What’s missing is your answer to the question asked.
 
  • #29
haruspex said:
What’s missing is your answer to the question asked.
Part b): The force F3 acting along the point B where its moment is zero is why I didn't include it in my equation.
$$ \sum\tau_B = F \frac {L \sqrt 2}{2} - F \frac {L \sqrt 2}{2} + F (0)$$
 
Last edited:
  • #30
paulimerci said:
Part b): The force F3 acting along the point B where its moment is zero is why I didn't include it in my equation.
$$ \sum\tau_B = F \frac {L \sqrt 2}{2} - F \frac {L \sqrt 2}{2} + F (0)$$
All you have to add is "therefore, of the three options, the answer is …."
 
  • #31
haruspex said:
All you have to add is "therefore, of the three options, the answer is …."
The net torque about point B is zero.
 
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1. What is torque and how is it related to forces applied to a rectangle?

Torque is a measure of the rotational force applied to an object. In the case of a rectangle, it is the force that causes the rectangle to rotate around a fixed point, known as the pivot point or axis. The net torque is the sum of all the individual torques acting on the rectangle.

2. How do you calculate the net torque of three equal forces applied to a rectangle?

To calculate the net torque, you need to know the magnitude of each force, the distance from the pivot point to where each force is applied, and the direction of each force. The formula for net torque is T = F x d x sinθ, where T is the torque, F is the force, d is the distance, and θ is the angle between the force and the lever arm (the shortest distance from the pivot point to the line of action of the force).

3. What is the direction of the net torque when three equal forces are applied to a rectangle?

The direction of the net torque depends on the direction of the forces and their relative positions. If the three forces are applied at different angles, the net torque will be in the direction of the force with the greatest angle. If the three forces are applied at the same angle, the net torque will be zero.

4. How does the location of the pivot point affect the net torque of three equal forces applied to a rectangle?

The location of the pivot point does not affect the net torque as long as the forces are applied at the same distances from the pivot point on opposite sides of the rectangle. If the pivot point is moved closer to one of the forces, the net torque will be greater in that direction.

5. Can the net torque of three equal forces applied to a rectangle ever be zero?

Yes, the net torque can be zero if the three forces are applied at the same angle and the same distance from the pivot point on opposite sides of the rectangle. In this case, the forces will cancel each other out and there will be no rotation.

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