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Statics Problem

  1. Aug 28, 2007 #1
    1. The problem statement, all variables and given/known data
    See the attachment


    2. Relevant equations



    3. The attempt at a solution

    I am pretty lost where to start... no relationship pops out at me to help in finding the components of the PR and QR to find the magnitude of QR and PR. I know lots of angles, but don't see how I can quite relate them to the 250 N force. If anyone could help me pointed in the right direction, it would be appreciated.
     

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  2. jcsd
  3. Aug 28, 2007 #2

    mgb_phys

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    Find the torques (Moments) from the points where the rods attach to the wall.
    Draw in the forces and directions acting on the rods where they meet the wall.
     
  4. Aug 28, 2007 #3
    Is there another way? This problem is in the chapter before moments and such are discussed. This chapter just deals with vectors and their components.
     
  5. Aug 28, 2007 #4

    learningphysics

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    Dealing with PR first... draw the component of the 250N force along PR, and perpendicular to PR... you have a right triangle formed by these 2 componenets and the 250N force... analyze that right triangle... do you see how to use the angles given?
     
  6. Aug 28, 2007 #5
    Alright, so for PR, I took 250/sqrt(2) to get approx 177 since I formed a 45-45-90 triangle.

    Then, to solve for QR, I took 177/sqrt(2) = 125 to get the bottom of a new triangle I made, with QR as the hypotenuse. Then took 125/cos(70) to get the value of 365.5 N for QR.

    Is what I did correct?
     
    Last edited: Aug 28, 2007
  7. Aug 28, 2007 #6

    learningphysics

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    The part of PR looks correct. But the QR part doesn't. Do exactly the same thing you did with the PR part... the only thing that is different is the angle.

    draw the component of the 250N force along QR, and perpendicular to QR... you have a right triangle formed by these 2 componenets and the 250N force... analyze that right triangle...
     
  8. Aug 29, 2007 #7
    I started off with the knowledge that the vertical components of PR and QR had to equal 250.

    So from that, I planned on drawing a vector diagram where this was reflected. I flipped PR vertically, and then joined it onto the bottom of QR to make a triangle, with a vertical side on the right of magnitude 250.

    Then I worked out interior angles, and used the sine rule.

    My answers were:
    QR = 195N
    PR = 94N
     
  9. Aug 29, 2007 #8

    learningphysics

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    Ah, after reading BlackWyvern's post I see I was completely wrong. CoasterGT, disregard my previous comments.

    So a force acting along PR added to a force acting along QR must add to 250N acting downwards... I'm getting different numbers though.
     
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