Finding Projection of Force onto line

[Bluestribute]

Homework Statement

Determine the magnitude of the projection of force F = 700N along the u axis.

Homework Equations

Cosθ = (A • B)/(||A|| * ||B||)

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?

 

[LCKurtz]

Homework Statement

Determine the magnitude of the projection of force F = 700N along the u axis.

Homework Equations

Cosθ = (A • B)/(||A|| * ||B||)

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?

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

 

[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.

 

[HallsofIvy]

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

 

[LCKurtz]

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.

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.

 

[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 . . .

 

[Fredrik]

Well there's no coordinates anywhere along the u axis so I don't know how to find the vector for it . . .

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

 

[Bluestribute]

You're going to 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 . . .

 

[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)?

 

[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?

 

[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?

 

[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.

 

[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?

 

[Fredrik]

You're assuming I have resources . . . which I don't

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 $\vec y=r\vec u=(\vec u\cdot\vec x)\vec u$.

 

[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.

 

[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.

 

[Fredrik]

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

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

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 = $\stackrel{\rightarrow}{u}$ • $\stackrel{\rightarrow}{x}$

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 $\mathbf x$.

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.

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.

 

[Bluestribute]

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.

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 going to start using that notation because I'm going to 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 . . .

 

[LCKurtz]

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)

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 $\vec F$.

 

[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.

 

[Fredrik]

The explanation of what you did got weird in the next paragraph too:

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.

It sounds like you took $\vec u\cdot\vec F$ (a number) and multiplied it by $\vec u$ (a vector), and got the final answer $(\vec u\cdot\vec F)\vec u$ (a vector), and then you said that this is equal to 290 (a number).

I still think you did the right thing, which is to calculate $|\vec u\cdot\vec F|$. I got a result close to 290, but not quite that. It's of course possible that I've made a mistake, but I have to ask if you're rounding off to the nearest multiple of 10 or something like that.

 

[Bluestribute]

I most definitely rounded at the end (I got like 291.652319 . . . I don't know the decimal).

Actually, reading my work, I ended up finding the unit vector from the force. Then I multiplied the magnitude of the force through its unit vector. I took this new $\vec F$ and dotted it with the unit vector (along the path of projection).

That worked for the rest of the problems in the set too.

 

[Fredrik]

Hm, rounded off to the nearest integer, I got 287. I guess you also rounded off intermediate results. If you're going to do that, you should include a few extra decimals in the intermediate results.
\begin{align}
&\vec F=\frac{700}{\sqrt{36}}(-2,4,4)\\
&\vec u=\bigg(\frac{1}{2},\frac{\sqrt 3}{2},0\bigg)\\
&\|\vec F\cdot\vec u\| =\frac{700}{\sqrt{36}}\big((-2)\frac{1}{2}+4\frac{\sqrt{3}}{2}+4\times 0\big) =\frac{700}{\sqrt{36}}(-1+2\sqrt{3})\approx 287.
\end{align}