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Now, Must d be the perpedicular distance from the force to the axis , or any distance?

Thanks in advance,

Aladdin

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- #1

- 45

- 0

Now, Must d be the perpedicular distance from the force to the axis , or any distance?

Thanks in advance,

Aladdin

- #2

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You have given very little information here. For example, what is "alpha"? Is this the angle between F and d?

Now, Must d be the perpedicular distance from the force to the axis , or any distance?

Thanks in advance,

Aladdin

I think there's quite a bit of confusion here. You may want to look at this and see if you've misunderstood something important.

http://hyperphysics.phy-astr.gsu.edu/hbase/woang.html#waa

Please consider formulating as clear and complete of a question next time.

Zz.

- #3

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No, Moment of force with respect to an axis,

Alpha is the angle betwen the force and d.

Alpha is the angle betwen the force and d.

- #4

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http://hyperphysics.phy-astr.gsu.edu/hbase/vctorq.html#vvc6

Zz.

- #5

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You mean that it doesn't matter if d is the perpendicular distance, because you included the angle?

- #6

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If they both have to be perpendicular, then the angle will always be 90 degrees, and it is a constant equal toYou mean that it doesn't matter if d is the perpendicular distance, because you included the angle?

Zz.

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- #8

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Did you look at the figure in the link that I showed? I thought that is self-explanatory?

Zz.

Zz.

- #9

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So , it must be perpendicular.

- #10

cjl

Science Advisor

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To be slightly more accurate, your equation already takes that into account. The moment is the force multiplied by the perpendicular distance. In your equation, d*sin(alpha) is the perpendicular distance.

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I'm going to correct this and say no, it doesn't, which is what I said already. But obviously, it is not getting through to you, but I'm going to make sure others reading this do NOT get the same wrong information.So , it must be perpendicular.

I have no idea why you are fixated with this "perpendicular".

Zz.

- #12

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We need to be careful here because that is the perpendicular

To be slightly more accurate, your equation already takes that into account. The moment is the force multiplied by the perpendicular distance. In your equation, d*sin(alpha) is the perpendicular distance.

Zz.

- #13

rock.freak667

Homework Helper

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isn't the moment of a force about an axis given by

Now, Must d be the perpedicular distance from the force to the axis , or any distance?

Thanks in advance,

Aladdin

[tex]M=\hat{\lambda} \cdot (\vec{r} \times \vec{F})[/tex]

where [itex]\hat{\lambda}[/itex] the unit vector in the direction of the axis?

- #14

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M=F*d

Ohhhhh, I got it , thanks ZappperZ, and for all :d.

Sorry for confusion.