- #1

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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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- Thread starter -Aladdin-
- Start date

- #1

- 45

- 0

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

Thanks in advance,

Aladdin

- #2

- 35,980

- 4,695

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

Thanks in advance,

Aladdin

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

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

If they both have to be perpendicular, then the angle will always be 90 degrees, and it is a constant equal to

Zz.

- #7

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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.

- #11

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So , it must be perpendicular.

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.

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

Zz.

- #12

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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.

We need to be careful here because that is the perpendicular

Zz.

- #13

rock.freak667

Homework Helper

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

Thanks in advance,

Aladdin

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

[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.

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