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Please check my work (friction on an incline plane)

  1. Sep 2, 2013 #1
    two blocks of unequal mass are connected by a string over a smooth pulley. one mass of mass 2m rests on the incline while the second mass of mass m is hung over a pulley over the incline if the coefficient of friction is μk, what angle of the incline allows the masses to move at a constant speed?


    |F1| = |F2|

    F1= weight of mass hanging over the pulley

    F2=Ʃforce acting on mass (2m) ie.. the x component of force along the direction of the incline + friction

    mg=(2m)gsinθ+(μk)(2m)gcosθ

    mg=(2mg)(sinθ+(μk)cosθ)

    mg's cancel..

    (1/2)=(sinθ+(μk)cosθ)

    but since i know that.. μk=tanθ,

    1/2=sinθ+tanθcosθ = sinθ+sinθ= 2sinθ

    sinθ=(1/4)

    θ=14.4775°

    is the way i substituted tanθ for the coefficient valid? if not, why is it not valid?
     
    Last edited: Sep 2, 2013
  2. jcsd
  3. Sep 2, 2013 #2

    mfb

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    Why should the force from the hanging mass be equal to the horizontal component of the force on the other mass? I would assume the string goes along the incline.

    This looks like the mass m is hanging now, instead of the mass 2m. How did you get that formula?

    The result will certainly depend on μk, a specific number cannot be right as result.

    Please add explanations to your formulas, otherwise it is hard to understand what you did and what was wrong.
     
  4. Sep 2, 2013 #3
    i fixed the m and 2m error.

    i set mg equal to the rest of that because the problem wanted me to find the theta when the net force is zero.

    so starting here,

    mg=(2m)gsinθ+(μk)(2m)gcosθ

    factoring mg..

    mg=(2mg)(sinθ+(μk)cosθ)

    mg's cancel..

    (1/2)=(sinθ+(μk)cosθ)

    now i'm supposed to solve for theta but i don't know how.. :(
     
  5. Sep 2, 2013 #4

    mfb

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    You have to start earlier. Did you draw a sketch? Did you identify all forces acting on the system?
    Which forces have to be in equilibrium?

    (sinθ)^2+(cosθ)^2=1 might be handy somewhere in the solving process, but first you need correct equations for the forces.
     
  6. Sep 2, 2013 #5
    here is the free body diagram with the arbitrary direction-sign specification (although that part wasn't needed)

    http://i.imgur.com/BCmluRZ.png
     
  7. Sep 2, 2013 #6

    Doc Al

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    You left out the tension force and the friction.

    Hint: If the speed is constant, what must be the tension?
     
    Last edited: Sep 2, 2013
  8. Sep 2, 2013 #7
    i essentially set the two tensions equal to one another. where..

    T1=mg
    T2=(2m)gsinθ+(μk)(2m)gcosθ

    does this depend on the direction of the blocks? since if the blocks move to the right..

    mg=(2m)gsinθ+(μk)(2m)gcosθ

    and if the blocks move to the left..

    mg=(2m)gsinθ-(μk)(2m)gcosθ

    so this is actually [itex]\hat{v}[/itex] dependent?
     
  9. Sep 2, 2013 #8

    Doc Al

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    Staff: Mentor

    Solve it both ways and see if you get reasonable answers.
     
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