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A problem that has force, friction on an incline

  1. Sep 24, 2008 #1
    A block of mass m, acted on by a force of magnitude F directed horizontally to the right as shown above, slides up an inclined plane that makes an angle [tex]\theta[/tex] with the horizontal. The coefficient of sliding friction between the block and the plane is [tex]\mu[/tex].

    The picture is that of an incline, [tex]\theta[/tex] on the left of the incline, mass m on the incline, and a horizontal force is applied to m directed right.

    Develop an expression for the magnitude of the force F that will allow the block to slide up the plane with constant velocity. What relation must [tex]\theta[/tex] and [tex]\mu[/tex] satisfy in order for this solution to be physically meaningful.

    This is what I did. (wrong according to the answer)

    F = ma
    mgsin[tex]\theta[/tex] + [tex]\mu[/tex]mgcos[tex]\theta[/tex] - Fcos[tex]\theta[/tex] = 0
    Fcos[tex]\theta[/tex] = mg(sin[tex]\theta[/tex] + [tex]\mu[/tex]cos[tex]\theta[/tex])
    F = mg(sin[tex]\theta[/tex] + cos[tex]\theta[/tex]) / cos[tex]\theta[/tex]

    This is how I reasoned: If it's moving up at a constant velocity, Fcos[tex]\theta[/tex] should equal mgsin[tex]\theta[/tex] + friction ([tex]\mu[/tex]gcos[tex]\theta[/tex])

    This is what the answer is supposed to be:

    F = mg([tex]\mu[/tex]cos[tex]\theta[/tex] + sin[tex]\theta[/tex]) / (cos[tex]\theta[/tex] - [tex]\mu[/tex]sin[tex]\theta[/tex])

    F > 0 => cos[tex]\theta[/tex] > sin[tex]\theta[/tex]
    tan[tex]\theta[/tex] = 1/[tex]\mu[/tex]

    What mistake did I make in my reasoning and where did the answer come from?
  2. jcsd
  3. Sep 24, 2008 #2

    Doc Al

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    Staff: Mentor

    Since the applied force has a component perpendicular to the plane, the normal force is not simply [itex]mg \cos\theta[/itex].

    (Also: It's much easier if you write your entire equation using Latex, not just some pieces.)
  4. Sep 24, 2008 #3
    Ok, I see. Thanks for the tip. It worked.

    I have a general/stupid question. How do you know which component of F is || or perpendicular to the incline. I tried several problems like this one (with numbers) and had trouble figuring out the axis

    For the last part of the problem, how do I know that cos[tex]\theta[/tex] > sin[tex]\theta[/tex]?

    I'm still getting used to latex. It's coming to me.
  5. Sep 24, 2008 #4

    Doc Al

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    Staff: Mentor

    Always draw yourself a diagram and mark the angles. (That's the only way I know.) Since we know the incline makes an angle [itex]\theta[/itex] with the horizontal, it must make the same angle with the applied force F.
    Since [itex]\cos\theta - \mu\sin\theta[/itex] (the denominator) must be positive, you can deduce that [itex]\cos\theta > \mu\sin\theta[/itex]. (Not [itex]\cos\theta > \sin\theta[/itex].)
  6. Sep 24, 2008 #5
    I see now. Thanks for all the help.
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