Regarding r and sign convention for Position vectors for moments

AI Thread Summary
The discussion centers on understanding the position vectors r1, r2, and r3 in relation to torque calculations. The participant struggles with the notation, specifically why r1 is represented as -1.5j and r2 and r3 are zero. Clarification is provided that r2 and r3 equal zero because they intersect at points A and B. The conversation also touches on the definition of torque, emphasizing that it is calculated using the cross product of the position vector and the force vector. The distinction between torque about a point and about a fixed axis is highlighted, noting that the parallel component of the moment remains constant regardless of the point chosen on the axis.
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Homework Statement



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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.
 
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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.
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ehild
 

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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.
 
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