Must d be Perpendicular Distance for Moment of Force?

[-Aladdin-]

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

 

[ZapperZ]

Moment of force= F*d*sin(alpha).
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.

 

[-Aladdin-]

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

 

[ZapperZ]

Then isn't it just the cross product of the two vectors? Once you have included the angle, then it doesn't matter if d and F are perpendicular to each other, does it?

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

Zz.

 

[-Aladdin-]

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

 

[ZapperZ]

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 one! So then why even bother writing "sin(alpha)"?

Zz.

 

[-Aladdin-]

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

 

[ZapperZ]

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

Zz.

 

[-Aladdin-]

So , it must be perpendicular.

 

[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.

 

[ZapperZ]

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.

 

[ZapperZ]

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.

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.

 

[rock.freak667]

Moment of force= F*d*sin(alpha).
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

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


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

 

[-Aladdin-]

Nope, we didn' took it this way.
M=F*d

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