# What will be the value of r here.

1. Jun 1, 2009

### dE_logics

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?

2. Jun 1, 2009

### tiny-tim

Hi dE_logics!
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
(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!

Last edited by a moderator: Apr 24, 2017
3. Jun 2, 2009

### dE_logics

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.

But we came to that conclusion here -

If this is not so, the F taken here should be its modulus.

4. Jun 2, 2009

### Born2bwire

No, again, the torque is a vector and is the result of vectors and their operations.

5. Jun 2, 2009

### tiny-tim

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

r x F is a vector expression (ok ok … technically, it's two vectors making a psuedovector ).

6. Jun 2, 2009

### dE_logics

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 dont know what is a psuedovector.

7. Jun 2, 2009

### dE_logics

Yeah...I agree with that, but that section of coordinate system is causing a problem.

8. Jun 2, 2009

### tiny-tim

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

what do you mean by "flip"? what is this F that's flipping?

9. Jun 2, 2009

### Born2bwire

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. Jun 2, 2009

### dE_logics

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 -

and -

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 -

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

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11. Jun 2, 2009

### dE_logics

Ok...thanks!

12. Jun 2, 2009

### tiny-tim

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.

13. Jun 2, 2009

### dE_logics

It all appears extremely confusing...need to work on vectors.