# Finding Projection of Force onto line

1. Aug 28, 2014

### Bluestribute

1. The problem statement, all variables and given/known data
Determine the magnitude of the projection of force F = 700N along the u axis.

2. Relevant equations
Cosθ = (A • B)/(||A|| * ||B||)

3. The attempt at a solution
I'm guessing I have to use the above equation, but my problem is finding the B vector. A is easy enough (<-2, 4, 4> I believe), but what about that B? Or am I thinking of this in the wrong way?

2. Aug 28, 2014

### LCKurtz

You have a generic formula for $\cos\theta$. What two vectors are A and B in your problem?

3. Aug 28, 2014

### Bluestribute

Well, I'd say A would be from the origin to A. I have no idea what B would be as there's only two lines, one point, and no other coordinates besides for the beam.

4. Aug 28, 2014

### HallsofIvy

Staff Emeritus
It looks to me like the "B" vector would be a vector from the origin along the direction u.

5. Aug 28, 2014

### LCKurtz

It looks like A is the point at the end of the hardware. But you are given a force vector $\vec F$. That would be your A. And it asks for the projection along the $u$ axis. Isn't that a hint for B? And you are going to need the formula for the magnitude of the projection of one vector on another.

6. Aug 28, 2014

### Bluestribute

Well there's no coordinates anywhere along the u axis so I don't know how to find the vector for it . . . And are you talking about the formula F$_{}parallel$ = Fcos$\theta$??? How would I use that in this problem since it doesn't give the required answer . . .

7. Aug 28, 2014

### Fredrik

Staff Emeritus
There's an angle. That's all you need.

8. Aug 28, 2014

### Bluestribute

You're gonna have to explain how I use it in the problem . . .

Do I use Fcosθ to find the magnitude of the B vector? If I'm using the equation:

F$^{}p$ = (A • B)/(A*B) (vectors on top, magnitudes on bottom)

How does the Fcosθ equation help at all? It's not the answer they're looking for, I know that much . . .

9. Aug 28, 2014

### Bluestribute

My problem is, how the hell do I use a dot product with only one vector known? And what do I do with Fcos(theta)?

10. Aug 28, 2014

### LCKurtz

Your vector u is in the xy plane. You have an angle given. Can you figure out the components of a unit vector in the direction of u?

11. Aug 28, 2014

### Bluestribute

Alright, so my unit vector is <(1/2),(rt(3)/2,0>? My A vector is <-2,4,4>? Where do I use the 700N? Is it really:

<(1/2), (rt(3)/2), 0> • <-2, 4, 4> = 700cos(30)B?

12. Aug 28, 2014

### LCKurtz

Ok, you now have a unit vector in the direction of u. As I told you before, A is not the vector you want for the force. You are given $\vec F$ which is in the same direction as A but it isn't A because $\vec F$ is 700 units long. You need to to two things:

1. Figure out the vector $\vec F$ (its components).
2. Find the formula in your text that tells how to find the component of one vector on another.

13. Aug 28, 2014

### Bluestribute

You're assuming I have resources . . . which I don't except for this forum since Google returns nothing.

U = < 1/2, $\sqrt{}3$/2, 0 > (unit vector on the "u" line)
A = < 233.33, 466.67, 466.67 > (unit vector of the force multiplied by 700)

U • A = ||A||*||B||*cosθ, where ||A|| = 700 and θ = 30°

Is this right?

14. Aug 28, 2014

### Fredrik

Staff Emeritus
OK, let's talk about projections in general then. You need to know how to find the projection of an arbitrary vector $\vec x$ onto a 1-dimensional subspace U. Let $\vec y$ and $\vec z$ be the unique vectors such that $\vec y$ is in U, $\vec z$ is orthogonal (perpendicular) to every vector in U, and $\vec x=\vec y+\vec z$.

We're looking for a formula for $\vec y$. If $\vec u$ is a unit vector in U, there's a unique real number r such that $\vec y=r\vec u$. So we can write $\vec x=r\vec u+\vec z$. Now what do you get if you use this formula to compute $\vec u\cdot\vec x$? The result is simply r. (You should verify that). So we have $\vec y=r\vec u=(\vec u\cdot\vec x)\vec u$.

15. Aug 28, 2014

### Bluestribute

Ok, let me write out my interpretation and thinking:

$\stackrel{\rightarrow}{z}$ is essentially a normal vector but for a line. The addition, I'm assuming, is basic trig (it's 10:00 here and my work is away).

I don't understand why we're searching for $\stackrel{\rightarrow}{y}$, unless it has to do with the specific problem here.

So the reason it equals simply r is because $\stackrel{\rightarrow}{y}$ and $\stackrel{\rightarrow}{z}$ are perpendicular? The ending is a bit fuzzy, but what I'm getting is that r = $\stackrel{\rightarrow}{u}$ • $\stackrel{\rightarrow}{x}$

So if I was relating this to my problem, I would have to get the unit vector of u, which is simple enough because it's a 30-60-90 triangle, dot it with the unit(?) vector for the Force (aka point A), and multiply by the force? It's knowing when to use a unit vector and when to multiply by the force that I know for sure I'm struggling with.

16. Aug 29, 2014

### LCKurtz

That's the idea, but it's a bit easier to calculate it this way. If $\hat u$ is a unit vector in the direction of $\vec u$ then the magnitude of the component of $\vec F$ along $\vec u$ is just $|\vec F\cdot \hat u|$. Remember, this is the scalar magnitude of the vector projection, and that's what the problem asked for.

17. Aug 29, 2014

### Fredrik

Staff Emeritus
Because $\vec y$ is by definition the projection of $\vec x$ onto U.

Right, two vectors are perpendicular if and only if their dot product is zero. So we get
$$\vec u\cdot\vec x=\vec u\cdot (r\vec u+\vec z)=r(\vec u\cdot\vec u)+\vec u\cdot\vec z=r|\vec u|^2+0=r.$$ And since $\vec y=r\vec u$, this means that $\vec y=(\vec u\cdot\vec x)\vec u$.

Do you understand this so far? Do you see how to use this result to find $|\vec y|$?

By the way, the only reason I'm using the $\vec x$ notation is that I found \vec easier to type than something that makes the text bold, like \mathbf. (Not sure what is the best way to write vectors in bold). I see (when I'm quoting you) that you use a more complicated way to create that arrow. You can just type \vec x, or if you prefer, \mathbf x to get $\mathbf x$.

You have found $\vec F$. You have found $\vec u$. (I think. I didn't check those results). And now you know the formula for the projection of an arbitrary vector onto an arbitrary 1-dimensional subspace, such as the straight line marked u in the picture.

Last edited: Aug 29, 2014
18. Aug 29, 2014

### Bluestribute

Ok, so here's what I ended up doing.. I have two more problems for this section, but I want to know if I'm doing it right (so expect more topics . . . or hopefully not if I'm understanding).

I took $\vec F$ (the coordinates of point A multiplied by the magnitude of F) and dotted it with 30-60-90 unit triangle (which had no z components). Essentially $\vec u$ • $\vec x$, though my x in this case was $\vec F$. Then I multiplied by my unit vector and got 290, the answer.

And I'm gonna start using that notation because I'm gonna be here for a while. With all the money I get from my scholarship, I should be paying you guys. This forum saved all of us last year . . .

19. Aug 29, 2014

### LCKurtz

I hope that isn't what you mean. You have to multiply a unit vector in the direction of A by the magnitude of $\vec F$ to get $\vec F$.

20. Aug 29, 2014

### Bluestribute

Yes yes, I meant unit vector. I've just been writing it so much I started writing sloppy shortcuts but yes, unit vector A.