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Inclined plane revisited

  1. Nov 2, 2004 #1
    Movers want to set the ramp of their truck so that the work they do against the combination of gravity and friction is a minimum for crates moving up the ramp with constant velocity. µ is the coefficient of kinetic friction and θ is the angle between the ramp and the ground. For the work to be a minimum, they must choose:

    a. tan θ = µ

    b. tan θ = -µ

    c. tan θ = -1/µ

    d. tan θ = 1/µ

    e. tan θ = 1 - µ

    The obvious solution of setting the derivative of work equal to zero finds the maximum work. The problem asks for the minimum?
     
  2. jcsd
  3. Nov 3, 2004 #2

    HallsofIvy

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    The word done is "Force*distance".

    The force necessary to move an object (I'm assuming they are sliding it) is the component of weight along the ramp: mg sin(θ) (draw a picture and look at the right triangles!) plus the friction force: μ times the component of weight perpendicular to the ramp, mg cos(θ).
    Force= mg sin(θ)+ mg μ cos(θ).

    The distance is the length of the hypotenuse of the right triangle formed by the ramp: h/cos(θ) where h is the height of the ramp (which I am assuming is fixed: the height of the truck bed).

    The work done is hmg(sin(θ)+ μcos(θ)/cos(θ).

    What value of θ minimizes that?
     
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