What will be the value of r here.

  • Context: Undergrad 
  • Thread starter Thread starter dE_logics
  • Start date Start date
  • Tags Tags
    Value
Click For Summary

Discussion Overview

The discussion revolves around the concept of torque and the vector nature of distance (r) and force (F) in a rotational system. Participants explore the implications of defining the origin and the signs associated with r and F, particularly in the context of maintaining equilibrium and understanding torque calculations.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants assert that if the origin is taken as the fulcrum, r will be positive, but question the sign of r if the origin is elsewhere along r.
  • Others argue that r is a vector and does not inherently have a sign, suggesting that torque, being a vector, results from vector operations.
  • One participant proposes that the reactive force on the fulcrum must equal the applied force for translational equilibrium, introducing conditions based on the mass of the bar.
  • There is a discussion about how the direction of F may change during rotation, leading to a proposed need for r's sign to compensate for this change to maintain consistent torque direction.
  • Some participants emphasize that vectors have magnitude and direction, and question the notion of "sign flipping" in relation to force and displacement vectors.
  • A later reply suggests that while the direction of F and r matters, the concept of signs may be irrelevant as long as the vectors are treated correctly in calculations.

Areas of Agreement / Disagreement

Participants express differing views on the nature of r and F as vectors, the relevance of signs, and the implications for torque calculations. There is no consensus on how to interpret the signs of r and F in the context of the original question.

Contextual Notes

Participants reference earlier discussions and diagrams to clarify their points, indicating that the understanding of vectors and their operations is complex and may depend on specific definitions and contexts.

Who May Find This Useful

This discussion may be of interest to those studying mechanics, particularly in understanding torque, vector operations, and the implications of coordinate systems in physics.

dE_logics
Messages
742
Reaction score
0
Suppose I'm given a simple task to taking out the moment about a point -


o-------------------F

o is the fulcrum...and F is the place on which force applies (of course in the perpendicular direction) and r is the distance between the fulcrum and force.

If we take o as the origin, then r will be positive.

But what if we take the origin as somewhere on r?...what will be its sign then?
 
Physics news on Phys.org
Hi dE_logics! :smile:
dE_logics said:
If we take o as the origin, then r will be positive.

No, that doesn't make sense …

r is a vector (and so doesn't really have a sign), and so is the https://www.physicsforums.com/library.php?do=view_item&itemid=175", which is r x F :wink:
But what if we take the origin as somewhere on r?...what will be its sign then?

(see above, but it's worth adding …)

You would always take https://www.physicsforums.com/library.php?do=view_item&itemid=64" (torques) about the fulcrum anyway, because there's a reaction force there, of unknown size and direction,

and the only way of keeping that unknown force out of the equations is by taking moments about the fulcrum! :biggrin:
 
Last edited by a moderator:
That reactive force on that fulcrum should be equal to the force applied...cause if this is not so, the body will not be in translational equilibrium.Initially I was thinking there would be no reactive force on the fulcrum.

But there are 2 cases...if the bar is mass less, then there has to be a normal reaction, cause as compared to a real body having mass, the normal reaction is given by its inertia...so there should be no application for force on the fulcrum, cause the normal reaction is given by the inertia.

In a mass less body, there is no inertia, so the normal reaction will be given by the fulcrum.

r is a vector (and so doesn't really have a sign), and so is the torque, which is r x F

But we came to that conclusion here -

https://www.physicsforums.com/showthread.php?p=2216723#post2216723

If this is not so, the F taken here should be its modulus.
 
No, again, the torque is a vector and is the result of vectors and their operations.
 
dE_logics said:
If this is not so, the F taken here should be its modulus.

Sorry, dE_logics, I've no idea what you're talking about. :confused:

r x F is a vector expression (ok ok :rolleyes: … technically, it's two vectors making a psuedovector :wink:).
 
Point is r and F should have their respective signs.

Cause...if suppose in a rotating bar, the direction (and so sign) of F will flip (as it rotates over time), but the torque and its sign remains the same (i.e clockwise or anticlockwise), so to balance the sign flip of F, r's sign should also be flipped...that is the only way, the torque's sign can be maintained.

That's what I mean here.

I don't know what is a psuedovector.
 
Born2bwire said:
No, again, the torque is a vector and is the result of vectors and their operations.

Yeah...I agree with that, but that section of coordinate system is causing a problem.
 
dE_logics said:
… the direction (and so sign) of F will flip …

Something's obviously bothering you, but I can't work out what it is :redface: …

what do you mean by "flip"? what is this F that's flipping?
 
A vector does not have a sign, it has a magnitude and a direction. There is no sign flipping in terms of the force or displacement vectors. You need to treat these as vectors and understand how the vectors and their operations translate to scalar equivalents and not the other way around.
 
  • #10
tiny-tim said:
what do you mean by "flip"? what is this F that's flipping?

I mean the sign flip.

If you're observing a rotation, form your perspective the direction of F will change over time...since the torque or moment by force is not changing, there should be a compensation of the sign change in F...this compensation will be by r.

E.G -

attachment.php?attachmentid=19189&stc=1&d=1243940247.jpg


and -

attachment.php?attachmentid=19190&stc=1&d=1243940247.jpg


The torque exhibited by both of these is the same, but the direction force is opposite.

So the torque by one should be F*r and the other should be -F*r...i.e its direction should be opposite.

But obviously this does not happen...the r is there as the compensation, so actually this happens -

Torque by one : F*r, torque by other : -F*-r

This has been confirmed here -

https://www.physicsforums.com/showthread.php?p=2216723#post2216723

So there should be a sign of r...what will it be in the case as stated in the main question?
 

Attachments

  • Untitled 1.jpg
    Untitled 1.jpg
    1.8 KB · Views: 413
  • Untitled 2.jpg
    Untitled 2.jpg
    1.9 KB · Views: 441
  • #11
Born2bwire said:
A vector does not have a sign, it has a magnitude and a direction. There is no sign flipping in terms of the force or displacement vectors. You need to treat these as vectors and understand how the vectors and their operations translate to scalar equivalents and not the other way around.

Ok...thanks!
 
  • #12
dE_logics said:
I mean the sign flip.

If you're observing a rotation, form your perspective the direction of F will change over time...since the torque or moment by force is not changing, there should be a compensation of the sign change in F...this compensation will be by r.

E.G -
…
So there should be a sign of r...what will it be in the case as stated in the main question?

I can see from your diagrams what you mean now …

but it's not "flipping", since F (and r) rotates continuously.

Anyway, your diagram shows an arrow at the head of F, but it also needs an arrow at the head of r …

the sign of F (and r) is irrelevant, but of course the direction matters, and F is always 90º clockwise from r, so r x F is always the same.

Forget about signs … always use arrows. :smile:​
 
  • #13
It all appears extremely confusing...need to work on vectors.
 

Similar threads

Undergrad Rolling Without Slipping
  • · Replies 59 ·
2
Replies
59
Views
6K
Undergrad Arbitrary value of the moment of a force
  • · Replies 1 ·
Replies
1
Views
2K
Undergrad What is the actual definition of the position of an object?
  • · Replies 31 ·
2
Replies
31
Views
2K
High School What is the relationship between force and distance in planetary motion?
  • · Replies 4 ·
Replies
4
Views
2K
Graduate Motion involving Translation & Rotation |Kleppner and Kolenkow
  • · Replies 10 ·
Replies
10
Views
2K
High School Sign conventions in torque and non-uniform circular motion
  • · Replies 1 ·
Replies
1
Views
2K
Undergrad Visualising rotation in 3-D space
  • · Replies 9 ·
Replies
9
Views
2K
Graduate Free body diagrams, coordinate systems origin/orientation
  • · Replies 14 ·
Replies
14
Views
3K
Undergrad Solving Wheel Coming to a Stop: FBD, Friction & Hysteresis
  • · Replies 22 ·
Replies
22
Views
5K
Undergrad Misunderstanding the concept of torque
  • · Replies 24 ·
Replies
24
Views
5K