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Past paper moments question

  1. May 14, 2005 #1
    I'm stuck! This is the question:
    A uniform rod AB has length 3m and weight 120N. The rod rests in equilibrium in a horizontal position, smoothly supported at points C and D, where AC=0.5m and AD=2m. A particle of weight W Newtons is attached tto the rod at a point E where AE=x metres. The rod remains in equilibrium and the magnitude of the reaction at C is now twice the magnitude of the reaction at D. Show that W=60/(1-x).

    I've worked out the reactions at C and D before W is applied, (40N and 80N I think) and by substituting 2Rc=Rd into 180+Wx=0.5Rc+2Rd (where Rc and Rd are the reactions at C and D) I got 180+Wx=4.5Rc, but now I'm stuck. I can't figure out how to bring the 'pre-W' results and the 'post-W' results together, and I'm not even really sure if that's what I'm supposed to be doing! Could someone point me in the right direction please?
     
    Last edited: May 14, 2005
  2. jcsd
  3. May 14, 2005 #2

    OlderDan

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    I think you are missing a 2 in this equation

    180+Wx=0.5Rc+Rd

    and it looks like you used the 2 to get the equation that followed

    To bring your results together you need one more piece of information. What is the final Rc + Rd in terms of the weight of the rod and the added weight, W?
     
    Last edited: May 15, 2005
  4. May 14, 2005 #3
    That missing 2 was a typo, sorry. I've corrected it in the original question. I can't try to do the question now cos I'm piiiiiissed but I'll try again in the morning. Or maybe the afternoon. Thanks for the advice though.
     
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