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Box on an inclined surface with Force of Friction and angled applied Force.

  1. Mar 10, 2012 #1
    Incline Surface force

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    Last edited: Mar 10, 2012
  2. jcsd
  3. Mar 10, 2012 #2
    In that case:

    [itex]\vec{F}[/itex]f=-μ.N[itex]\hat{x'}[/itex]

    Being μ the coefficient of kinetic friction and N the normal reaction of the ramp.
    As the applied force makes 30º with the ramp, it's y component is going to contribute to a decrease in friction, and so being Fy=N=sin(30)F, you have:

    [itex]\vec{F}[/itex]f=-μ.sin(30)F[itex]\hat{x'}[/itex]
     
    Last edited: Mar 10, 2012
  4. Mar 10, 2012 #3
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    Last edited: Mar 10, 2012
  5. Mar 10, 2012 #4

    HallsofIvy

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    You are given the acceleration up the incline so you can calculate [itex]F_x[/itex]. Set up your force equations for the x and y components of all forces. Knowing the horizontal force will allow you to solve for the magnitude of [itex]F_a[/itex] and you can get [itex]F_a[/itex] from that.
     
  6. Mar 10, 2012 #5
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    Last edited: Mar 10, 2012
  7. Mar 10, 2012 #6
    Fnet=Fcos(30)-mgsen(20)-μN , with N=mgcos(20)-Fsen(30)
    ma=Fcos(30)-mgsen(20)-μmgcos(20)+μFsen(30)
    F=mg(sen(20)+μcos(20)+(a/g))/(cos(30)+μsen(30))
     
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