# Regarding r and sign convention for Position vectors for moments

1. Feb 16, 2014

### 7Lions

1. The problem statement, all variables and given/known data

3. 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.

2. Feb 17, 2014

### ehild

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.

ehild

#### Attached Files:

• ###### torquer.JPG
File size:
11.3 KB
Views:
136
Last edited by a moderator: Feb 17, 2014
3. Feb 17, 2014

### 7Lions

Red in the picture? Did you post a picture soewhere? I'm unable to see it

4. Feb 17, 2014

### BvU

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$_{AB}$

5. Feb 17, 2014

### nasu

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?

6. Feb 17, 2014

### BvU

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 $\vec r$.

7. Feb 17, 2014

### BvU

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....

8. Feb 18, 2014

### nasu

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