Regarding r and sign convention for Position vectors for moments

Click For Summary

Homework Help Overview

The discussion revolves around understanding position vectors and their implications for torque calculations in a physics context. The original poster expresses confusion regarding specific position vector values and their significance in relation to moments.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the definition of torque and its relationship to position vectors. Questions arise about the visibility of referenced diagrams and the implications of certain position vector values being zero. There is also a discussion about the conditions under which the moment of force is zero.

Discussion Status

Participants are actively engaging with the original poster's confusion, offering clarifications about torque and position vectors. Some are questioning assumptions and definitions, while others provide insights into the nature of moments about fixed axes. There is a lack of consensus on the clarity of the original poster's understanding.

Contextual Notes

There is mention of a diagram that may not be visible to all participants, which could be affecting the discussion. The original poster's specific values for position vectors and their implications are under scrutiny, indicating potential gaps in information or understanding.

7Lions
Messages
9
Reaction score
0

Homework Statement



34qqy3b.png




The Attempt at a Solution



I'm having trouble understanding how r1 = -1.5j and r2= r3 = 0.

Can anyone make this a little clearer for me? I've spent quite a while trying to wrap my head around it but to no avail.
 
Physics news on Phys.org
The torque with respect to a point A is \vec \tau = \vec r \times \vec F, where \vec r is the position of the point of application of the force with respect to A. A has to be along the axis of rotation. The vector r1 (red in the picture) points from A to the application point of the forces.
attachment.php?attachmentid=66707&d=1392656774.jpg

ehild
 

Attachments

  • torquer.JPG
    torquer.JPG
    11.3 KB · Views: 628
Last edited by a moderator:
Red in the picture? Did you post a picture soewhere? I'm unable to see it
 
Could it be that the picture in post 1 continues a bit more? r2 = r3 = 0 because those lines intersect AB at B and at A. At the lower edge I see a u##_{AB}## pop up and it might well continue with a calculation of the distance between r##_1## and u##_{AB}##
 
7lions: Do you have a definition for the moment of the force about an axis?
Do you know in what conditions this will be zero?
 
You have ##\vec \tau = \vec r \times \vec F## for the ##moment## about the axis prependicular to ##\vec r## and ##\vec F##. The axis "goes trhough" the origin of ##\vec r##.
 
Ah, I'm' "helping" nose, which is unnecessary, sorry. It looks as if the lions are asleep in the jungle tonight. I turn in too... :smile:
 
BvU said:
You have ##\vec \tau = \vec r \times \vec F## for the ##moment## about the axis prependicular to ##\vec r## and ##\vec F##. The axis "goes trhough" the origin of ##\vec r##.

Well, this is the torque (moment of the force) about a point.
The moment about a fixed axis is less frequently mentioned.
This is the reason I asked about it.

The definition I know is that you take any point on the axis and calculate the moment about that point. And then you find the component of the moment parallel to the axis. It turns out that no matter what point you choose on the axis, the parallel component is the same.
And this is the moment about the axis. The perpendicular component will try to rotate the axis. But as the axis is "fixed", this has no effect.

I just wanted to make sure that this is the definition the OP knows about.
 

Similar threads

  • · Replies 9 ·
Replies
9
Views
13K
Replies
4
Views
2K
  • · Replies 30 ·
2
Replies
30
Views
2K
  • · Replies 6 ·
Replies
6
Views
5K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 20 ·
Replies
20
Views
3K
  • · Replies 17 ·
Replies
17
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
5
Views
9K
  • · Replies 5 ·
Replies
5
Views
2K