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Need help with 3D vector; force directed along a line

  1. Sep 22, 2007 #1
    1. The problem statement, all variables and given/known data


    2. Relevant equations 3. The attempt at a solution

    I really have no idea. I'ce been trying to do this thing for almost two hours, and can't even figure out how to start it. Any help that could be given would be greatly appreciated.
  2. jcsd
  3. Sep 22, 2007 #2
    If you know the total length is 4=r and that the height (z) is 2, could you figure out the azimuthal angle and the polar angle?

    Would it help if I told you that

    [tex] x = r sin \theta cos \phi[/tex]

    [tex] y = r sin \theta sin \phi[/tex]


    [tex] z = r cos \theta[/tex]

    where theta is the azimuthal angle, and phi is the polar angle? You have some distances that relate to angles, and some forces that relate to angles. If you can figure out those angles you can figure out everything. Also, just as a reminder, if this really gets you going, the x coordinate will be whatever you solve for plus the 1m offset from the wall.
  4. Sep 22, 2007 #3

    In this case you can see that the Force will have zero Z component. So the remaining are only the x and the y components.Hence you can use the folowing equation

    [tex]F[/tex] = [tex]\sqrt{F^2_x + F^2_y}[/tex]
  5. Sep 22, 2007 #4
    Nope, wouldn't help. I've never heard the terms azimuthal angle or polar angle before, and we've never had any thetas or phis in that class. We use alpha, beta, and gamma for the angle between the vector in question and the x, y, and z axis, if that indicates where our knowledge differs.
  6. Sep 22, 2007 #5
    This is because you are not familiar with polar co-ordinates and all.So try doing it with simple rectangular cartesian system.
  7. Sep 22, 2007 #6
    My problem is that I have no idea how to do that. The whole process has me completely confused.

    Where do I start?
  8. Sep 22, 2007 #7
    Zero z-component? I don't see it.

    Three angles? That's sort of strange. So the angle between the x and y resultant is the polar angle, and the angle coming down from the z axis to the vector r, the 4m in this case, is the azimuthal. This is the easiest way I see of doing the problem, but there may be a trick. I'll have to think about it a little, with your angles there would be a way too.
  9. Sep 22, 2007 #8
    Wow, I hadn't even seen that FedEx had originally posted.

    Anyway, yeah, the force has a z component, doesn't it? The point B doesn't have a z value, but A does so there has to be some displacement.
  10. Sep 22, 2007 #9

    Well i am talking about the foce. The force is directed towards the hook or we nay say towards the ground.Hence if we drop the perpediculars at that point z comp would be zero.

    And i think that your way is much simple.But the OP must know polar coordinates.(If i am not wrong)
  11. Sep 22, 2007 #10
    But the force is directed towards B and not towards A.
  12. Sep 22, 2007 #11
    Alright, I don't understand it, but we'll try it your way.
    So this gives me F_y = 16.58

    How do I use the F_x and F_y components to find point B?
  13. Sep 22, 2007 #12
    Oh, I see what you are saying. I guess it is a matter of interpretation because even though the z force will be zero at B because of the ground's normal force, the ground's normal force is not actually included in the force from the wall, which is what the problem is asking about. I don't think you can say it is zero.

    Also, I'm not using polar coordinates despite the familiar polar terms. Everything I have said applies to cartesian coordinates. This is the way you do cartesian coordinates in three dimensions.
    Last edited: Sep 22, 2007
  14. Sep 22, 2007 #13
    My guess is that F_x = A_x + B_x , so B_x = F_x - A_x = 25 - 1 = 24

    Is this right?

    Sorry for being stupid tonight. My dorm is partying, and my professor fails a teaching this stuff. Not a good combination when you just got you textbooks, and have an assignment due Monday.
    Last edited: Sep 22, 2007
  15. Sep 22, 2007 #14
    Don't worry, this is a relatively difficult question for an intro class. I'm going to tell you the deal with the formulas I've given you, and hopefully it will make sense.

    Let's look at this guy

    [tex]x = r sin \theta cos \phi[/tex]

    Picture the projection of the cord onto the x y axes. You are dropping a perpindicular down, so to say. If we had that projection then we would be back in our familiar 2 dimensional space, which everyone knows and loves. So, what we need to do is get that projection.

    This is where we define the azimuthal angle. By convention, it is defined as the angle from the z-axis to the vector, as shown here

    So, from trig, we know that the vector multiplied by the sine of the azimuthal angle would give the very projection we are looking for (the red line in the picture).
    [tex]proj_r = r sin \theta[/tex]

    *As a quick sidenote, do you now see that z, the other leg, is given by
    [tex] z = r cos \theta[/tex]

    Look at the equation like this
    [tex]x = proj_r cos \phi[/tex]

    where the proj_r is simply the perpendicular that has been dropped down so that we are back in 2D.

    So now that we have the projection, we might as well call this vector p, then all the regular 2D stuff applies to it.
    [tex]x = p cos \phi[/tex]
    [tex]y = p sin \phi[/tex]

    Does this all make sense? The sooner you learn it, the better off you will be for later math or physics courses.
  16. Sep 23, 2007 #15
    That...does make some sense...

    so theta = acos(2/4) and phi = acos(1/4)?

    meaning that:
    p = r*sin(theta) = 4*sin(60) = 3.464
    x = 3.464*cos(75.52) = 0.8661
    y = 3.464*sin(75.52) = 3.354

    Am I right?
    Last edited: Sep 23, 2007
  17. Sep 23, 2007 #16
    Yeah, you pretty much have it. So, knowing z and r immediately gives you the azimuthal angle, the 3D one.

    [tex]\theta = Arccos(\frac{z}{r}) = Arccos(\frac{2}{4})[/tex]

    You can figure out the polar angle, the 2D one, from the forces.

    [tex]F_x = F_r sin\theta cos\phi[/tex]

    So, do the algebra (with the phi substitution) to find that

    [tex]\phi = Arccos[\frac{F_x}{F_r sin (Arccos(\frac{2}{4}))}][/tex]

    You have the two angles, and now you have everything. Cool?

    *I haven't actually solved the numerics, so I don't know if your answer is good or not (just my estimates give me the impression you have a made mistake somewhere though). But go through my algebra and make sure it gives the same.
    Last edited: Sep 23, 2007
  18. Sep 23, 2007 #17
    Well solved,Mindscrape. The answer is correct. I understood my fault.
  19. Sep 23, 2007 #18
    Okay, hold up. Somethings wrong in that last post by Mindscrape.

    one of those sines or cosines is wrong.

    Shouldn't [tex]F_x = F_r sin \phi cos \theta[/tex] be [tex]F_x = F_r sin \theta cos \phi[/tex]
    Last edited: Sep 23, 2007
  20. Sep 23, 2007 #19
    Well I didnt see mindscrapes last post i just saw the answers you got and they are right.

    And yes now as i have seen the post, mindscrape has made a mistake in a hurry,he has taken sine instead of cosine and vice versa in the second equation but in the third equation everything is right.
  21. Sep 23, 2007 #20
    Argh, that can't be. My answer for phi was quite a bit off from his, so we can't both be right.

    Oh well, I'll trust that he is correct and my results were in error, as he knows more about this stuff than I do.

    Thanks you two for your help. This wasn't the way we were taught, but I understand this better than what was lectured, so I'll stick with this.
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