# How do I find the magnitude of the sum of these force vectors in 3 dimensions?

Inertigratus

## Homework Statement

[PLAIN]http://img406.imageshack.us/img406/3756/fmp1.jpg [Broken]
How do I calculate the magnitude of the sum of these 3 vectors?
Also, how do I find the angles the sum of the vectors make with each axis?
F1 = 260N
F2 = 75N
F3 = 60N

F = F1 + F2 + F3

## The Attempt at a Solution

The thing is that I'm still new to vectors and not sure how to proceed since the vectors aren't beginning at origo. Plus, the magnitude is in Newton, how do I find out the components of each vector using the magnitude?

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cupid.callin
find their i, j, k component and add them (take care of - signs of forces in -X,-Y,-Z directions)

Inertigratus
I know that, but how?
The x component of F2 = |F2|*cos(0) = 75N ?
How do I find the z component?
Also, the F3 only has a y component, which is 60N, right?

cupid.callin
The x component of F2 = |F2|*cos(0) = 75N ?

Its -75N

Z component is |F2|*sinθ

yes F3 has only Y component

Edit: its -60N

take care of +/- signs or you'll end up with wrong answer

Inertigratus
What angles do I use?
For F2 it's simply 90/2 degrees, or pi/4.
For F3 it's 0 degrees.
How do I get the angles for F1?

Inertigratus
Anyone?

I can't use the dot product since that requires me to know the end point of the force vector right?

cos(a) = AB/|A||B|

AlexChandler
Does this figure have all right angles?

Inertigratus
What do you mean? the figure got rotated a little, the z-axis is supposed to point straight upwards. I guess all angles are right angles. It doesn't say, but it looks like it.

Mentor
If you can write a vector for the diagonal of the "box" that F1 lies on, then F1 is just a magnitude scaled version of it.

AlexChandler
Does this figure have all right angles?

Staff Emeritus
Homework Helper
Gold Member

## Homework Statement

[PLAIN]http://img406.imageshack.us/img406/3756/fmp1.jpg [Broken]
How do I calculate the magnitude of the sum of these 3 vectors?
Also, how do I find the angles the sum of the vectors make with each axis?
F1 = 260N
F2 = 75N
F3 = 60N

F = F1 + F2 + F3

## The Attempt at a Solution

The thing is that I'm still new to vectors and not sure how to proceed since the vectors aren't beginning at origo. Plus, the magnitude is in Newton, how do I find out the components of each vector using the magnitude?
Hi Inertigratus.

$$\vec{F}_3$$ is in the negative y direction, so:

$$\vec{F}_3=60\hat{j}}\ \text{N}\,.$$

The displacement vector to corner B from corner A is:

$$\vec{d}_{BA}=(-3\hat{i}-12\hat{j}+4\hat{k})\ \text{m}\,.$$

The unit vector (magnitude = 1) in the direction of $$\vec{d}_{BA}\ \text{ is }\ \hat{d}_{BA}=\frac{1}{\sqrt{(-3)^2+(-12)^2+(4)^2}}\,(-3\hat{i}-12\hat{j}+4\hat{k})=\frac{-3}{13}\,\hat{i}+\frac{-12}{13}\,\hat{j}+\frac{4}{13}\,\hat{k}\,.$$

The unit vector, $$\hat{d}_{BA}$$ is in the same direction as the force vector, $$\vec{F}_1\,.$$

Therefore, $$\vec{F}_1=(260 \text{ N })\hat{d}_{BA}=(260 \text{ N }) \left(\frac{-3}{13}\,\hat{i}+\frac{-12}{13}\,\hat{j}+\frac{4}{13}\,\hat{k}\right)=(-60\hat{i}-240\hat{j}+80\hat{k})\ \text{N}\,.$$

Do similarly for F2. Then add F1 + F2 + F3 as vector quantities.

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Inertigratus
Thanks a great deal! I thought the force didn't take up the whole line from A to B. Guess I have to read up on displacement vectors a little.
But if the force was from A to B, shouldn't the arrow extend all the way up to B? or is it implicit?
Also, what about the angle between the resultant force and the axis?
Do you use the dot product between the resultant force vector and the unit vector of one axis?
Doesn't that just give you the resultant force vector divided by the magnitude of that vector?
Since the unit vector is just 1.

Mentor
Thanks a great deal! I thought the force didn't take up the whole line from A to B. Guess I have to read up on displacement vectors a little.
But if the force was from A to B, shouldn't the arrow extend all the way up to B? or is it implicit?
The force simply has a magnitude and direction; there's nothing else implied. All you know is that at point A, there is a force of magnitude F1 in a given direction.

In general, forces drawn on a diagram are sketched to show direction and approximate relative magnitude (i.e. they are not to scale) unless otherwise indicated.

Also, what about the angle between the resultant force and the axis?
Do you use the dot product between the resultant force vector and the unit vector of one axis?
Doesn't that just give you the resultant force vector divided by the magnitude of that vector?
Since the unit vector is just 1.

$$\vec{F} \cdot \hat{i} = |\vec{F}||\hat{i}|cos(\theta)$$

but $$|\hat{i}| = 1$$, so that

$$cos(\theta) = \frac{\vec{F} \cdot \hat{i}}{|\vec{F}|}$$

Note that if you have the rectangular components of F, <Fx, Fy, Fz>, then

$$\vec{F} \cdot \hat{i} = F_x$$

AlexChandler
We have
Fnet=<F1x+F2x,F1y+F3y,F3z>
It will help to find all of the components of F3
We have three simple triangles... two 3 4 5 triangles forms each of the two small faces, and the triangle ADX is a 5 12 13 triangle. (X is the corner closest to the x)
F1 has the same direction as <-3,-12,4>
and magnitude 260 N
Thus F1=260 <-3,-12,4>/√(9+144+16)=260/13<-3,-12,4>=<-60N,-240N, 80N>
we know F2 has direction of <-3,0,4>/√(9+16) and magnitude 75 N
So F2= <-9N,0,12N>
and F3 of course is <0,-60N,0>
Thus Fnet=<-69N, -300N, 92N>
with magnitude √(69N2+300N2+92N2>
Fnet=321.286N

Inertigratus
Thanks, I think I get it. I got the same answer as you Alex. The net force, 321.3N.
However the book states that the answer is supposed to be 347.3N ?

Mentor

Inertigratus
Ohh, you're right. I forgot to take the square root. Thanks!

Inertigratus
Got the angles too, thanks all. Just wondering though, how come the angles are in radians?

Mentor
Got the angles too, thanks all. Just wondering though, how come the angles are in radians?

That's like asking why the distances are in meters! Feel free to convert radians to degrees or grads or mils or turns or points or binary radians or even hour angles!

Have a browse http://en.wikipedia.org/wiki/Angle" [Broken].

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AlexChandler
Ahh yes F2 = <-45,0,60>N

AlexChandler
Use clock angles... 90 degrees = 15 minutes = 15 seconds = 3 hours
everything is so much more simple this way.
yet nobody listens to me