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Evaluating the z component

  1. Aug 21, 2007 #1
    1. The problem statement, all variables and given/known data
    Hey,

    I had a question about resolving the z-component of a vector.

    Lets say that there is a problem that asks us to find the resultant force vector in three-dimensions.

    I know that the the resultant force vector, [itex]\vec{F}_{R}[/itex] is given by the sum of its vector components as follows,

    [tex]
    \vec{F}_R = \vec{F}_{x}\hat{i} + \vec{F}_{y}\hat{j} + \vec{F}_{z}\hat{k}
    [/tex]

    I know how to find [itex]{F}_{x}[/itex] and [itex]{F}_{y}[/itex],

    [tex]
    {F}_{x} = \left|\vec{F}\right|{cos}{\theta}
    [/tex]

    [tex]
    {F}_{x} = \left|\vec{F}\right|{sin}{\theta}
    [/tex]

    but how do I find [itex]{F}_{z}[/itex]?

    [tex]
    {F}_{z} = ???
    [/tex]

    I tried looking for this result for a while over the internet with no luck.

    Is there a general way. in the same sense that the [itex]x[/itex] and [itex]y[/itex] components use [itex]cos\theta[/itex] and [itex]sin\theta[/itex], to always evaluate the [itex]z[/itex] component in terms of [itex]\theta[/itex]?

    Thanks,

    -PFStudent
     
    Last edited: Aug 21, 2007
  2. jcsd
  3. Aug 21, 2007 #2

    Dick

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    Homework Helper

    If you are in three dimensions you need two angles to specify direction, the theta you already know about and an angle phi which measures the angle from the x-y plane. Usually taken as pi/2 on the +z axis and -pi/2 on the -z axis. In terms of this F_z is |F|*sin(phi) and F_x and F_y need to be multiplied by cos(phi). Look up 'spherical coordinates'.
     
  4. Aug 21, 2007 #3
    Hey,

    Thanks for the reply.

    Does that mean the only way I can resolve the z component is by introducing an angle [itex]\phi[/itex] that measures an angle from the x-y plane?

    So, is there no way to resolve the z component just in terms of [itex]\theta[/itex] and the magnitude of the force?

    Thanks,

    -PFStudent
     
    Last edited: Aug 22, 2007
  5. Aug 21, 2007 #4

    Dick

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    Well, no. How could you? (BTW, you mean angle FROM the x-y plane, right?). theta is the angle IN the x-y plane. There are lots of vectors in three dimensions that have the same x-y plane angle, but very different z components.
     
    Last edited: Aug 21, 2007
  6. Aug 21, 2007 #5

    learningphysics

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    You can also have cylindrical coordinates, which uses the 2D polar coordinates of R and [tex]\theta[/tex] and then z as the third component.
     
  7. Aug 21, 2007 #6
    A three dimensional length would be

    [tex] 3D^2 = x^2 + y^2 + z^2[/tex]

    This can also be expressed in cylindrical coordinates, which might be somewhat familiar from trig, where

    [tex] 3D^2 = r^2 + z^2[/tex]

    [tex] r^2 = x^2 + y^2[/tex]

    [tex] x = rcos(\theta)[/tex]

    [tex] y = rsin(\theta)[/tex]

    Then there are spherical coordinates, which take a while to get used to, where we have to define another angle which is measured from the z-axis to the 3D vector and called the azimuthal angle

    [tex] 3D^2 = \rho^2 = x^2 + y^2 + z^2 = r^2 + z^2[/tex]

    [tex] x = \rho cos(\theta) sin(\phi)[/tex]

    [tex] y = \rho sin(\theta) sin(\phi)[/tex]

    [tex] z = \rho cos(\phi)[/tex]

    Physics gets really confusing with spherical coordinates because everyone uses a different notation. Sometimes rho and r are switched, and theta and phi are switched, and sometimes one is but not the other. When we fix the electron sign convention, we should also fix spherical coordinates.
     
  8. Aug 22, 2007 #7
    Hey,

    What I meant to ask was, Is the only way to determine the z-component for the scenario I described by introducing a second angle, [itex]\phi[/itex]?

    In addition, am I right to say I cannot solve for [itex]{F}_{z}[/itex] in cylindrical coordinates as I would need to know the z-component, but since the z-component is itself in cylindrical coordinates, then that would lead me know no where.

    So is the spherical coordinate system the only way to solve for the z-component (that is, by introducing [itex]\phi[/itex])?

    Would you say that the most "standard" notation for,

    cylindrical coordinate system is:
    [tex]
    \left(r, \theta, z\right)
    [/tex]

    spherical coordinate system is:
    [tex]
    \left(\rho, \theta, \phi\right)
    [/tex]

    Thanks,

    -PFStudent
     
    Last edited: Aug 23, 2007
  9. Aug 24, 2007 #8
    [tex]{F}_{r}^2 = {F}_{x}^2 + {F}_{y}^2 + {F}_{z}^2[/tex]
    You know all the values except [tex]F_{z}[/tex], can you solve for it?
    Here, we are talking about magnitudes and components of the vectors.
     
    Last edited: Aug 24, 2007
  10. Aug 24, 2007 #9
    Hey,

    Thanks for the reply. I meant to say in the original post, that [itex]{F}_{r}[/itex] is unknown. So, then using the equation you mentioned to solve for [itex]{F}_{z}[/itex], would not work because you would have two unknowns: [itex]{F}_{z}[/itex] and [itex]{F}_{r}[/itex].

    Thanks,

    -PFStudent
     
  11. Aug 25, 2007 #10
    If you want to find the resultant, you must know FX, FY and FZ.
    If you want to resolve Fr and find the components, you must at least know Fr and theta.
    If you do not know Fr but only theta, I don't think you can find the components even though you know that
    [tex]
    {F}_{x} = \left|\vec{Fr}\right|{cos}{\theta}[/tex]
    and
    [tex]{F}_{x} = \left|\vec{Fr}\right|{sin}{\theta} [/tex]

    since you say that you do not know Fr.
     
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