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Normal force on an inclined plane

  1. Sep 25, 2006 #1
    A crate of mass m = 100 kg is pushed at constant speed up the frictionless ramp (theta= 27.0°) by a horizontal force F.
    What is the magnitude of the force exerted by the ramp on the crate?

    My x-y coordinate system has x+ pointing up the ramp.

    I assume that the force in question is the Normal force, which points directly along my positive y-axis.

    No acceleration in the y direction (or x direction either, but that's irrelevent for this part of the problem), so
    [itex]\Sigma\\F_{y}=N-mg_{y}=N-mg\cos\theta=0[/itex]
    which gives [itex]N=mg\cos\theta=(100kg)(9.81m/s^2)(cos27^o)=874N[/itex]

    But that's not the right answer... (and I don't know what the right answer is, either)

    Help?
     
  2. jcsd
  3. Sep 25, 2006 #2

    radou

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    Homework Helper

    Constant velocity implies that the resulting force exerted on the crate has to equal zero. First write the equation of equilibrium for the assumed x-direction, to obtain the magnitude of the force F. Then write the equilibrium equation for the y-direction, to obtain the magnitude of the normal force N exerted on the crate from the ramp. Your mistake was that you did not include the pushing force F in your equation.
     
  4. Sep 25, 2006 #3
    Got it; even after I tilted my coordinate system, I assumed that a horizontal force has no y-component. Bad thing.

    Thanks!
     
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