# Moment of force

1. May 19, 2009

Moment of force= F*d*sin(alpha).
Now, Must d be the perpedicular distance from the force to the axis , or any distance?

2. May 19, 2009

### ZapperZ

Staff Emeritus
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. May 19, 2009

No, Moment of force with respect to an axis,
Alpha is the angle betwen the force and d.

4. May 19, 2009

### ZapperZ

Staff Emeritus
5. May 19, 2009

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

6. May 19, 2009

### ZapperZ

Staff Emeritus
If they both have to be perpendicular, then the angle will always be 90 degrees, and it is a constant equal to one! So then why even bother writing "sin(alpha)"?

Zz.

7. May 19, 2009

Yeah I know, but thats not my point, my point is if d is any distance or it must me the perpendicular distance? You got me Mr.

8. May 19, 2009

### ZapperZ

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

Zz.

9. May 19, 2009

So , it must be perpendicular.

10. May 19, 2009

### cjl

Actually, it doesn't need to be 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.

11. May 19, 2009

### ZapperZ

Staff Emeritus
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. May 19, 2009

### ZapperZ

Staff Emeritus
We need to be careful here because that is the perpendicular component of the distance vector. If you read what the OP wrote, he/she is simply not considering that, and somehow, refuses to accept that it can be ANY direction.

Zz.

13. May 19, 2009

### rock.freak667

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

$$M=\hat{\lambda} \cdot (\vec{r} \times \vec{F})$$

where $\hat{\lambda}$ the unit vector in the direction of the axis?

14. May 19, 2009