# Homework Help: Vector Calc Cross-product

1. Dec 19, 2008

### ZachN

1. The problem statement, all variables and given/known data

If a force f acts at a point x show that its moments about the three coordinate axes are the components of a vector, x x f.

2. Relevant equations

Uh... Euclidean space R3 in cartesian system.

x = <x1, x2, x3>
f = <f1, f2, f3>

3. The attempt at a solution

Okay, so far I have just taken the vector product.

(x2f3-x3f2)e(1) + (x1f3-x3f1)e(2) + (x1f2-x2f1)e(3).
This is where I have gotten and it shows the moments as components of the vector x x f. There has to be more though, this answer just seems to fall a bit short of a proof.

Last edited: Dec 19, 2008
2. Dec 19, 2008

### rock.freak667

Right so you calculated xxf.

You need to find the moment of that force about the three coordinate axes.
So do you know a formula for find the moment of a force about a specified axis? (Think triple scalar product)

3. Dec 20, 2008

### ZachN

tau = r x f. But i am having trouble defining it in terms of velocity - angular and linear velocity. v = (omega x r). linear velocity is proportional to the distance from the axes of rotation and the angular velocity.

I'm wondering if I can say that the linear velocity is directly proportional to the angular velocity cross the distance of the point x. Can proportionality exist for cross products? Lets say if (though it isn't) P (pressure)=(RxT)/V that the pressure was proportional to the cross product of R and T. I'm rambling...

4. Dec 20, 2008

### HallsofIvy

Why would you want to? This problem has nothing to do with velocity.

5. Dec 20, 2008

### ZachN

In order for there to be a moment there is velocity and acceleration - so it must be able to be expressed in terms of angular or linear velocity.

Are you guys saying I have the right answer.

6. Dec 20, 2008

### HallsofIvy

Then I strongly recommend that you review the definition of "moment of a force about an axis". It has nothing whatsoever to do with velocity and acceleration!

No! Where would you get that idea?

The problem was to show that the moment of a given force around a given axis is equal to rxf. rock.freak667 agreed that you had correctly calculated rxf. You have said nothing about the moment of the force around the axis!

7. Dec 21, 2008

### ZachN

I don't want to argue about it cause I came here for help but a moment is directly related to acceleration. There can be no moment without an acceleration. A moment requires a force and a force is mass times acceleration.

The moment about an axis M = r (length of moment arm) cross F (or times the force in the same plane as the particular axis perpendicular to the moment arm). I have already given that:

(x2f3-x3f2)e(1) - moment about x axis (1)
(x1f3-x3f1)e(2) - moment about y axis (2)
(x1f2-x2f1)e(3) - moment about z axis (3)

Other than that I don't know...

8. Dec 21, 2008

### HallsofIvy

After you have calculated the vector product and shown that its components are the moments about the axes, then you are done. I said before that you hadn't shown that because you had just stated "and it shows the moments as components". Once you have shown that x2f3- x3f2 is indeed the moment about the x- axis, etc. you have done that.

9. Dec 21, 2008

### rock.freak667

Last edited by a moderator: Apr 24, 2017