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Thread: mechanics, vectors components help

  1. #1
    Super Member
    Sep 2008

    mechanics, vectors components help

    express the force vectors in terms of components parallel and perpendicular to the plane shown.

    Can I say the angle the particle make with the plane is also 43 degrees , because they are alternate angles?

    how do I make a triangle with the 10N force being the hypotenuse from this diagram ?

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  2. #2
    Super Member Aryth's Avatar
    Feb 2007
    You're going to want to keep the axes as is, but a quick translation shows that you are correct in assuming the 43^{\circ} angle.

    Now, you may or may not know that the components of the vectors are scalars, represented by F_x for the force in the \hat{i} direction and F_y for the force in the \hat{j}.

    We also know that the magnitude of the force (I'll call F_T), which is 10N, is the hypotenuse of a right triangle, meaning that we can define \cos{\theta} and \sin{\theta} in terms of the force:

    \cos{\theta} = \frac{F_x}{F_T}

    Because in the unit circle, we define it to be cos = x/hyp.

    This means that:

    F_x = F_T\cos{\theta}


    F_y = F_T\sin{\theta}

    To prove it, we will rederive the magnitude of the force by using the pythagorean theorem, that tells us that:

    F_T^2 = F_x^2 + F_y^2

    F_T^2 = (F_T\cos{\theta})^2 + (F_T\sin{\theta})^2

    F_T^2 = F_T^2\cos^2{\theta} + F_T^2\sin^2{\theta}

    We know that \cos^2{\theta} + \sin^2{\theta} = 1 so we get:

    F_T^2 = F_T^2

    |F_T| = |F_T|

    And there you go.

    The final statement is:

    \vec{F} = F_x\hat{i} + F_y\hat{j}
    Last edited by Aryth; Mar 26th 2009 at 11:35 AM.
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