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Finding Projection of Force onto line

  1. Aug 28, 2014 #1
    1. The problem statement, all variables and given/known data
    Determine the magnitude of the projection of force F = 700N along the u axis.

    Hibbler.ch2.p121.jpg


    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. jcsd
  3. Aug 28, 2014 #2

    LCKurtz

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    You have a generic formula for ##\cos\theta##. What two vectors are A and B in your problem?
     
  4. Aug 28, 2014 #3
    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.
     
  5. Aug 28, 2014 #4

    HallsofIvy

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    It looks to me like the "B" vector would be a vector from the origin along the direction u.
     
  6. Aug 28, 2014 #5

    LCKurtz

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    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.
     
  7. Aug 28, 2014 #6
    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[itex]_{}parallel[/itex] = Fcos[itex]\theta[/itex]??? How would I use that in this problem since it doesn't give the required answer . . .
     
  8. Aug 28, 2014 #7

    Fredrik

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    There's an angle. That's all you need.
     
  9. Aug 28, 2014 #8
    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[itex]^{}p[/itex] = (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 . . .
     
  10. Aug 28, 2014 #9
    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)?
     
  11. Aug 28, 2014 #10

    LCKurtz

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    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?
     
  12. Aug 28, 2014 #11
    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?
     
  13. Aug 28, 2014 #12

    LCKurtz

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    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.
     
  14. Aug 28, 2014 #13
    You're assuming I have resources . . . which I don't except for this forum since Google returns nothing.

    U = < 1/2, [itex]\sqrt{}3[/itex]/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?
     
  15. Aug 28, 2014 #14

    Fredrik

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    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##.
     
  16. Aug 28, 2014 #15
    Ok, let me write out my interpretation and thinking:

    [itex]\stackrel{\rightarrow}{z}[/itex] 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 [itex]\stackrel{\rightarrow}{y}[/itex], unless it has to do with the specific problem here.

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

    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.
     
  17. Aug 29, 2014 #16

    LCKurtz

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    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.
     
  18. Aug 29, 2014 #17

    Fredrik

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    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
  19. Aug 29, 2014 #18
    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 . . .
     
  20. Aug 29, 2014 #19

    LCKurtz

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    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##.
     
  21. Aug 29, 2014 #20
    Yes yes, I meant unit vector. I've just been writing it so much I started writing sloppy shortcuts but yes, unit vector A.
     
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