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Homework Help: Write F as a sum of an orthogonal and parallel vector

  1. Jun 30, 2014 #1
    an object is moving in the direction i + j is being acted upon by the force vector 2i + j, express this force as the sum of a force in the direction of motion and a force perpendicular to the direction of motion.

    the parallel would be [tex] \hat{i}+\hat{j}[/tex] and the orthogonal would be [tex]\hat{i} - \hat{j}[/tex]

    using projection of F onto the parallel and orthogonal

    [tex] \frac{<1,1>\cdot<2,1>}{||<1,1>||^2}<1,1> = <\frac{3}{2}, \frac{3}{2} >[/tex]

    [tex] \frac{<1,-1>\cdot<2,1>}{||<1,-1>||^2}<1,-1> = <\frac{1}{2} , \frac{-1}{2}>[/tex]

    [tex] \vec{F} = <\frac{3}{2}, \frac{3}{2} > + <\frac{1}{2}, \frac{-1}{2} > [/tex]

    [tex] = 2\hat{i} + 1\hat{j} [/tex]
  2. jcsd
  3. Jul 1, 2014 #2


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    Out of curiosity, what is the i+j direction? Do you mean the positive x and y directions? Where is the object exactly?

    I'm assuming a few things here, but:

    Place the force vector ##\vec F = 2 \hat i + 1 \hat j## at the position of the object; Then find a unit vector ##\vec u## in the direction of the object (using a position vector ##\vec r##).

    Use ##\vec F## and ##\vec u## to find the parallel projection ##F_{||}## (by using the dot product). Take a moment to express the parallel component in vector form, ##\vec F_{||}##, by using ##\vec u##.

    The perpendicular component is easily obtained by geometry afterwards, i.e:

    ##\vec F_{\perp} = \vec F - \vec F_{||}##
  4. Jul 1, 2014 #3


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    I know what you mean, but the sentence is not correct. It means as if the parallel and perpendicular components of the force vector would be i+j and i-j. You have to find the force component parallel with the direction of motion , say v and perpendicular to it. The direction of the motion is parallel to i+j and i-j is perpendicular to it.

    The result is correct.

  5. Jul 1, 2014 #4
    thats the only information given. this is in a calc class not a physics class so like most math classes they give the bare minimum with physics problems
  6. Jul 1, 2014 #5

    yeah I typed the question word for word, not the best wording
  7. Jul 1, 2014 #6


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    The question was correct. It was the first sentence of your solution which was not.

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