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Determining Angle with Vector Components

  1. Sep 19, 2014 #1


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    Greetings women and men,

    I have a problem in which I have to find an angle [itex]\phi[/itex].

    http://srg.sdf.org/images/PF/StaticsHW.png [Broken]

    A horizontal force of [itex]\vec{F}=400 lbs[/itex] is placed on the structure at point A. Find angle [itex]\phi[/itex] to give the AB component of [itex]\vec{F}[/itex] a magnitude of 600 lbs.

    To solve this, I drew a diagram
    http://srg.sdf.org/images/PF/StaticsHW2.png [Broken]

    I used the Pythagorean theorem to find x: [tex]600^2=400^2+x^2 \rightarrow x=\sqrt{600^2-400^2} \therefore x=447.2[/tex]

    Then I used the law of sines to find [itex]\phi[/itex]: [tex]\frac{447.2}{\sin{30}}=\frac{400}{\sin\phi} \rightarrow \frac{400\sin{30}}{447.2}=\sin\phi \rightarrow \arcsin{\frac{400\sin{30}}{447.2}}=\phi=26.6\deg[/tex]

    However, [itex]\phi=26.6\deg[/itex] is not the same as the answer in the back of my textbook. I'm not sure where I went wrong. The correct answer in the textbook is [itex]\phi=38.3\deg[/itex].

    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Sep 19, 2014 #2


    Staff: Mentor

    Where's the right angle? The pythagorean theorem is predicated on using a right triangle.
  4. Sep 19, 2014 #3


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    Ah! Duh! That's why. Law of cosines it is. Can't believe I overlooked that.

  5. Sep 19, 2014 #4


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    Gold Member

    Uh ... do you understand the Pythagorean Theorem? What kind of triangles does it apply to?

    EDIT: I see that jedishrfu beat me to it.
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