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Minimum Force (roller over step)

  1. Nov 7, 2013 #1
    qUgpWXX.jpg

    On an inclined plane (φ = 12°), calculate the minimum force P required to pull the roller over the step. The radius of the roller, r = 0.685m

    My attempt

    The sum of the Moment force about the system = 0

    0 = mg(sin(φ)*(r-h) + mg(cos(φ)*√[r2-(r-h)2] - P(cos(θ)*r-h)

    P(cos(θ)*r-h) = mg(sin(φ)*(r-h) + mg(cos(φ)*√[r2-(r-h)2]

    P(cos(32.9)*0.575m) = (26kg)(9.8)(sin(12)*(0.575) + (26)(9.8)(cos(12)*√[0.6852 - (0.575)2]

    0.483*P = 30.46 + 92.79

    P = 255.17



    This answer appears to be wrong. Any help on this? I can't seem to figure out where I went wrong.
     
  2. jcsd
  3. Nov 7, 2013 #2

    haruspex

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    I guess you meant ##0 = mg\sin(φ)(r-h) + mg\cos(φ)\sqrt{r^2-(r-h)^2} - P\cos(θ)(r-h)##.
    What about P sin(θ)?
     
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