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Static equilibrium of a wheel

  1. May 17, 2015 #1
    1. Problem statement
    In figure, what magnitude of (constant) force F applied horizontally at the axle of the wheel is necessary to raise the wheel over an obstacle of height h=3.00cm ? The wheel's radius is r=6.00cm , and it's mass is m=0.800 kg.


    2. Relevant equations
    Balance of forces: Fnet=0 (xy plane)
    Balance of torques: τnet=0 (z axis)

    3. The attempt at a solution
    By balance of torques about point of contact with obstacle,
    F(r-h)=mgr
    I chose that point so as to not account for forces of contact there!
    The answer i got
    F=(mgr)/(r-h)
    F= 15.7 N
    The answer at the back of the text is 13.6N. :cry:
    What did i do wrong?

    Also i have this lingering doubt, will the normal forces at the point of contact with the horizontal floor , be included?:confused:
    I worked another problem where the solution i found ignored it and when i tried putting it in, it simply nullifies the weight , which lead to unreasonable answers... But still why is it not to be included? I mean it's also a force no.. Of course it disappears as soon as the wheel lifts off due to that horizontal F .. Thanks in advance..:smile:
    Transtutors001_adfe0810-1fc3-46da-b30c-b263490666fb.PNG
     
  2. jcsd
  3. May 17, 2015 #2

    Orodruin

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    What is the moment arm of the gravitational force wrt the point of contact?
     
  4. May 17, 2015 #3
    The moment arm was (r^2 - (r-h)^2)^(1/2)
    I mistook it to be r as i was doing the eye approximations :p
    Thanks a lot @Orodruin
    I'll be more careful!
     
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