# Moment vector of a force at some point

• Jan 8th 2013, 08:53 AM
astartleddeer
Moment vector of a force at some point
Hi,

I need help with the following questions from which I'm asked to solve. My book gives a very brief illustration of moments associated with vectors but with no aided worked problems.

The question (s) :

A force of magnitude 3 units acts at the point with coordinates (1,2,3). The force is applied in the direction of the vector 3i - j + 4k.

i) Find the moment of the force about O (the origin)

ii) What is the moment of the force about the point with coordinates (1,2,3)

iii) Use a diagram to illustrate your answer

i)

For the moment around the origin (0), I've got :-

$M_0= r \times F$

Unfortunately I don't know how to construct matrices using Latex, so for this reason the following result below is after the determinant (s) have been computed.

$M_0 = (i + 2k + 3k) \times (3i - j + 4k)$

$M_0 = 10i + 5j -7k\ Nm$

Is this correct?

ii)

Now for (ii), the formula I've got is :-

$M_A = (r - r_A) \times F$

Where r is the components of the position vector for the intial point the force is acting.

Now the point where the force is acting is equal to the point where I'm asked to take the moment. I'm confused. If I go buy the aforementioned equation for (ii), then this tells me the moment equates to zero? Could someone please confirm this.

iii)

How do I illustrate these results by means of a diagram?

Sidenote - Why have I been given the magnitude of the force as 3 units? As far as I'm concerned I don't even seem to need this, except to illustrate the magnitude in the diagram?
• Jan 8th 2013, 10:41 AM
skeeter
Re: Moment vector of a force at some point
(1) unit vector in the direction of 3i - j + 4k is

$\frac{3}{\sqrt{26}} i - \frac{1}{\sqrt{26}} j + \frac{4}{\sqrt{26}} k$

$F = \frac{9}{\sqrt{26}} i - \frac{3}{\sqrt{26}} j + \frac{12}{\sqrt{26}} k$

so, recalculate $r \times F$

(2) moment is zero since $|r| = 0$ ... remember the other method of calculating a cross product?

$r \times F = |r| \cdot |F| \sin{\phi}$

(3) use a cartesian grid ... sketch in the vectors
• Jan 8th 2013, 11:37 AM
astartleddeer
Re: Moment vector of a force at some point
I don't understand how you've computed the unit vector.

I can derive that unit vector by $\left(\frac{1}{|F|}\right)(3i - j + 4k)$

Then you've multiplied it by the magnitude 3 to end up with F?

For $r \times F$ I've got :-

$\left[-\frac{12}{\sqrt{26}} - (4)(-\frac{3}{\sqrt{26}})\right](1) - \left[(3)(\frac{12}{\sqrt{26}}) -(4)(\frac{9}{\sqrt{26}})\right](2) + \left[(3)(-\frac{3}{\sqrt{26}}) - (-1)(\frac{9}{\sqrt{26}})\right](3)$

$= 0i + 0j + 0k$

For $|r|$ I've got $\sqrt{1^2 + 2^2 + 3^2} = \sqrt{14}$ and $|F| = 3$ which gives me $3\sqrt{14}\sin(\phi)$ ?
• Jan 8th 2013, 12:44 PM
skeeter
Re: Moment vector of a force at some point
Go to the link ... scroll down to Let's Practice (iii)

Unit Vectors