# Thread: Moment vector of a force at some point

1. ## 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)

i)

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

$\displaystyle M_0= r \times F$

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.

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

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

Is this correct?

ii)

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

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

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?

2. ## Re: Moment vector of a force at some point

(1) unit vector in the direction of 3i - j + 4k is

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

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

so, recalculate $\displaystyle r \times F$

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

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

(3) use a cartesian grid ... sketch in the vectors

3. ## 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 $\displaystyle \left(\frac{1}{|F|}\right)(3i - j + 4k)$

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

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

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

$\displaystyle = 0i + 0j + 0k$

For $\displaystyle |r|$ I've got $\displaystyle \sqrt{1^2 + 2^2 + 3^2} = \sqrt{14}$ and $\displaystyle |F| = 3$ which gives me $\displaystyle 3\sqrt{14}\sin(\phi)$ ?

4. ## Re: Moment vector of a force at some point

Go to the link ... scroll down to Let's Practice (iii)

Unit Vectors

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# find the movment of force F=3i-2j 2k acting at point (1,2,1) about origin

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