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Homework Help: Hemisphere on a ramp

  1. Jan 26, 2010 #1
    [/tex]1. The problem statement, all variables and given/known data
    http://emweb.unl.edu/negahban/em223/sexam3/sexam3.htm" [Broken]

    number 3...the one about the semicylinder on a ramp

    2. Relevant equations
    [tex]\Sigma[/tex] F_x = 0
    [tex]\Sigma[/tex] F_y = 0
    [tex]\Sigma[/tex] M=0
    f = [tex]\mu[/tex] * N

    3. The attempt at a solution
    I got the first part to be 16.7 degrees. However, I can't find \phi I just need to be pointed in the right direction...
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jan 26, 2010 #2
    If you were to draw a line from G to the point of contact between the cylinder and ramp, phi occurs when that line coincides with the direction of mg.
     
  4. Jan 26, 2010 #3
    So then I know that the distance between G and the ramp is r*(1-4/(3*pi)), but I don't know how that helps me solve the problem...
     
  5. Jan 26, 2010 #4
    I wouldn't say that would be the distance...that would be the case when the line from G to contact falls on a radial line, which wouldn't be the same case as when the line from G to contact falls on the vector mg.
     
  6. Jan 26, 2010 #5
    right...that makes sense. I'm still having a hard time picturing how this fact will help me solve the problem...
    isn't the distance between G and the contact surface going to be in the same direction as mg, otherwise the semi-cylinder would try and move back to the place where they are in the same direction (because g will cause a torque???) sorry if that doesn't make sense...
     
  7. Jan 26, 2010 #6
    Yup...that is what is going on...and the line covering that distance for this particular case does not necessarily coincide with a radial line (by radial line, I'm referring to the line from the center of what would have been the complete cylinder to the point of contact).
     
    Last edited: Jan 26, 2010
  8. Jan 26, 2010 #7
    So right before it is going to tip, torque = zero...maybe?
     
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