Moment library article - question

[lxman]

Hi all,

I am reading the moment library article, and in particular pertaining to this section

Every force F has a moment about any point P.

To find the moment, draw R, the point of application of the force, and L, the line of the force, and draw the perpendicular line PQ from P to L (so both Q and R lie on L).

For the Moment of a velocity, R is the position of the centre of mass, and L is the line of the velocity.

Then the moment of F about P is the vector written "r x F" (pronounced "r cross F"), where r is the position vector PR. Its direction is perpendicular to both L and the line PR (and PQ). And its magnitude is PQ times F.

PQ is sometimes called the lever arm.

Note that if P is on the line L, then P = Q, so PQ = 0, so the moment of the force is 0.

I am attempting to visualize what is explained.

I have created the following diagram (click for a larger image):

http://www.freeimagehosting.net/uploads/th.6994c04925.png

I believe that I have diagrammed correctly. The point where I get a bit confused is:

Then the moment of F about P is the vector written "r x F" (pronounced "r cross F"), where r is the position vector PR. Its direction is perpendicular to both L and the line PR (and PQ). And its magnitude is PQ times F.

PR should be the vector extending from P to R2 along the Z axis (as I have drawn). Is this correct? Or would it be the vector from P to the existing point of application of F - R1?

Then the magnitude of the vector would be the cross product of vector PQ and the force vector? Or would it simply be the magnitude of PQ (a scalar) times the magnitude of the force vector (another scalar) to arrive at a scalar?

Thank you for your helping me clear this up.

 

[tiny-tim]

hi lxman!

PR should be the vector extending from P to R2 along the Z axis (as I have drawn). Is this correct? Or would it be the vector from P to the existing point of application of F - R1?

i don't understand …

why have you drawn two Rs? what are they supposed to be?

if your vector from P to R₂ is along the z-axis (ie, vertically out of the page), then it can't possibly be PR

the cross product is of F (along L) and any vector PS where S is any point on L (including R and Q) … you need to prove for yourself that the result is the same whichever S is chosen

Then the magnitude of the vector would be the cross product of vector PQ and the force vector?

a magnitude is not a cross product …

a magnitude is a scalar, and a cross product is a vector

do some reading about cross products, and the article will become clear

Or would it simply be the magnitude of PQ (a scalar) times the magnitude of the force vector (another scalar) to arrive at a scalar?

yes … "And its magnitude is PQ times F."

 

[lxman]

Hi tiny-tim

i don't understand …

Then we are in agreement. That seems to be my main point of confusion I think. From reading the article, I get the impression that, in the end, my vector of the moment of force should be in the Z axis (coming out of the page towards me). If I draw my vector from point P to point R, the point of application of the force, I have a vector in the XY axis.

Hmm, re-reading, I think I understand the source of my confusion. I believe that I was confused by the use of the pronoun "Its":

Then the moment of F about P is the vector written "r x F" (pronounced "r cross F"), where r is the position vector PR. Its direction is perpendicular to both L and the line PR (and PQ). And its magnitude is PQ times F.

As I am reading along, I see that "it is the vector PR." Subsequent to this, I am told that "Its direction is perpendicular to . . ." With the successive use of the two pronouns I was assuming that the second "Its" also referred to the position vector PR. But obviously my assumption was incorrect, now that I look back. (Yes, I know what happens whence we assume things :shy:.)

So, the correct interpretation, I believe, is that "It is the vector PR" refers to the vector PR in the XY plane. Whereas "Its direction is perpendicular to . . ." refers to "r x F" which extends along the Z axis, as I have drawn. I just need to relabel my original R1 as R and then come up with some other point as the termination of my vector "r x F".

Then the magnitude of my resultant vector "r x F" will be a result of the magnitude of my force vector (in the XY plane) and the magnitude of the vector PQ (also in the XY plane) to give me the magnitude of my resultant vector in the YZ plane.

I know, this is a long-winded way of looking at it, but I was apparently confused . Do I have it (!) right now?

 

[K^2]

Here is what the picture is supposed to look like. The force F is applied at point R. The moment arm around point P is line PQ. The magnitude of the moment is length of PQ times magnitude of force F. The vector representing the moment is perpendicular to PQ and F, so it points directly towards you out of the screen (right hand rule to determine direction), and is therefore, not represented on the picture.

 

[tiny-tim]

hi lxman!

Do I have it (!) right now?

yes