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Moment of forces about axes

1. Homework Statement

Shafts A and B connect the gear box to the wheel assemblies
of a tractor, and shaft C connects it to the engine. Shafts A and
B lie in the vertical yz plane, while shaft C is directed along
the x axis. Replace the couples applied to the shafts with a
single equivalent couple, specifying its magnitude and the
direction of its axis.



3. The Attempt at a Solution

We have to resolve each moment into its components Mx, My, Mz

I am having trouble determining the signs of each... How is this exactly done?

I have:
Ma = (1600 Nm)(sin 20) j + (1600 N m)(cos 20) k
Mb = (1200 Nm)(sin 20) j + (1200 N m)(cos 20) k

I'm being told that the y component of Ma has a NEGATIVE SIGN... Why?

Thanks in advance
 

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tiny-tim

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hi aero_zeppelin! :smile:
I'm being told that the y component of Ma has a NEGATIVE SIGN... Why?
the direction of the torque is the same as the direction of the axis

the axis for MA is positive along the z direction, but negative along the y direction :wink:

(unlike the axis for MB, which is positive along both)
 
Ok, so I can see that Shaft A is pointing towards the POSITIVE Z and the NEGATIVE Y

But I see NEGATIVE Z and the NEGATIVE Y for Shaft B (pointing down).

How am I supposed to be looking at this? lol

Thanks!
 

haruspex

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Imagine the arcing arrow projected onto the xz plane, and view it looking along the y axis, with positive y away from you. For A, it will curve anticlockwise, making it a negative torque. For B, it will appear clockwise, making it a positive torque. Similarly, project onto the xy plane and look out along the increasingly positive z axis. Both should now go clockwise, making them positive.
 
But isn't the convention CCW = Positive and CW = Negative ? Or is this only in North America?

I'm not sure if I'm imagining this correctly... My book mentions something about"an observer" at the tip of the moment vector for determining the sense of rotation. Is this what you meant? In this case, I should stand at A or B and have the y axis as a pole in front of me?
 

haruspex

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But isn't the convention CCW = Positive and CW = Negative ? Or is this only in North America?
I never remember what the convention is, but it makes no difference as long as you are consistent.
 
Ok, consistency.

And am I interpreting what you said correctly? About the arcing arrows..
 

haruspex

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Ok, consistency.

And am I interpreting what you said correctly? About the arcing arrows..
Again, there are different ways of going about this, which should all lead to the same answer. If what your book says is something you can remember and work with, do that.
 

tiny-tim

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hi aero_zeppelin! :smile:

(just got up :zzz:)

are you ok about this now?​
 
hi aero_zeppelin! :smile:

(just got up :zzz:)

are you ok about this now?​


I don't see it quite well yet...

I don't understand why for Ma we have -y and +z and for Mb we have both positive. Are the shafts the moment vectors per se?

Thanks for asking!
 

tiny-tim

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Are the shafts the moment vectors per se?
yes, the vector of the moment is a multiple of the vector of the shaft :smile:

(and since one shaft is sloping / and the other is sloping \ , one is ++ and the other is +-)
 
So the fact that shaft B is "pointing" in the - z direction (or lies in the -z side) has nothing to do?
 

tiny-tim

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ah, i see what you mean …

no, the side of the box doesn't matter, it would be the same if shaft B went right through the box and came out the other side

all that matters is which way round that circular arrow is going

curl the fingers of your right hand to match the circular arrow, and your thumb (in the thumbs-up position!) points along the moment vector :wink:
 

haruspex

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So the fact that shaft B is "pointing" in the - z direction (or lies in the -z side) has nothing to do?
The shaft, as such, doesn't point in a particular one of the two directions. If you were concerned with, say, the bending moment from a laterally applied load on the shaft then you would associate a position vector with the shaft, and you would have a basis for deciding which is to be the origin. Likewise, in the present problem, it doesn't become a vector, and thus acquire a direction, until you consider its rotation.
 
all that matters is which way round that circular arrow is going

curl the fingers of your right hand to match the circular arrow, and your thumb (in the thumbs-up position!) points along the moment vector :wink:


Ok so Shaft B is spinning.... Clockwise? And Shaft A Counterclockwise?
I could take them the other way also right? For consistency
 

tiny-tim

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from where i'm looking, they're both rotating the same way (anti-clockwise)

(of course, if you stand "behind" the shaft, so that you're looking along it towards the gearbox, then one is clockwise and the other is anti-clockwise, but that is not a good way to look at it, the position of the gearbox doesn't matter!)
 

haruspex

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Ok so Shaft B is spinning.... Clockwise? And Shaft A Counterclockwise?
I could take them the other way also right? For consistency
Yes. Clockwise and anticlockwise depend on where you stand.
I recommend sticking to the right-hand rule convention tiny-tim quoted. That translates the direction of rotation into a vector direction within your coordinate system.
 
It's clear now, thanks a lot guys!
 

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