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Statics: Forces in Space

  1. May 29, 2015 #1
    2.61_Pic.jpg
    I'm having trouble number 2.61. I have the answers, but I don't understand why they are doing what they're doing.

    They have this for (Fx)1:(Fx)1=(80 lb)*cos(30)*cos(40)=53 lb
    I'm familiar with the formula Fx=Fcosθ , but I'm confused about using it in 3D. Is there a formula that I need to memorize specifically for 3D?
     
  2. jcsd
  3. May 29, 2015 #2

    SteamKing

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    Not really. You should study carefully the diagram which is included with the problem.

    Calculating (80 lb) * cos 30° gives the projection of the force vector in the x-y plane. Multiplying this projection of the force vector in the x-y plane by cos 40° gives the x-component of the force. Similar calculations can give the components along the y and z axes.

    It's just a matter of applying 2-D trigonometry and not being distracted because the force vector is 3-D.
     
  4. May 29, 2015 #3
    Ohh ok, that makes sense! I'm still a little confused about finding Fz. It says Fz=80*sin(30). I kind of have an idea of why, but I'm not super confident about it. Do you use sin(30) in this situation because the the line on the far left of the triangle is parallel to the z axis?
     
  5. May 29, 2015 #4

    Bandersnatch

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    This is just basic trigonometry. Look at this right triangle:
    Snapshot.jpg
    From trigonometry you know that ##sin\alpha =F_z/F## and ##cos\alpha = F_x/F##.
    You can rearrange these to get:
    $$F_z=F sin\alpha$$
    $$F_x=F cos\alpha$$
    These have the intuitive meaning of projections: a vector times cosα means a projection of that vector onto the plane from which the angle alpha is measured, and sinα means a projection onto a plane perpendicular to the previous one.
     
  6. May 29, 2015 #5
    Perfect! Thanks so much, its been a minute since I had trig! :)
     
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