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Component of weight perpendicular to bar

  1. Mar 31, 2015 #1
    1. The problem statement, all variables and given/known data
    http://postimg.org/image/tlvhadljz/
    I am trying to find the component of weight perpendicular to the bar.

    2. Relevant equations


    3. The attempt at a solution
    I tried as shown and I got mgcos(theta) but I am supposed to be getting mgsin(theta)
     
  2. jcsd
  3. Mar 31, 2015 #2
    Could you please explain the question. Even the picture seems vague.
     
  4. Mar 31, 2015 #3

    Delta²

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  5. Mar 31, 2015 #4

    SammyS

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    It's fairly basic trigonometry. The angle, θ, is obtuse, so you might be better off considering the angle π - θ which is acute. You can then more directly see that you need to use sin(π - θ) by considering opposite/hypotenuse.

    Untitled.jpg
     
  6. Mar 31, 2015 #5
    The theta you considered is not the same with the theta that is shown in the figure on the left.
     
  7. Mar 31, 2015 #6

    Delta²

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    Oh dear they are right. theta is measured from the vertical to the ground plane (not from the ground plane itself) so the theta we considered is equal to the actual theta minus π/2. So the perpendicular component of weight is mgcos(theta-π/2)=mg(cos(theta)cos(pi/2)+sin(theta)sin(pi/2))=mgsin(theta).
     
  8. Mar 31, 2015 #7
    thanks everone
     
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