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Two Mass In Contact On Slope

  1. Mar 19, 2009 #1
    1. The problem statement, all variables and given/known data

    Two packages at UPS start sliding down the 20° ramp shown in Figure P8.25. Package A has a mass of 5.0 kg and a *coefficient of friction of 0.20. Package B has a mass of 10 kg and a coefficient of friction of 0.15. How long does it take package A to reach the bottom?

    I call the larger mass m2 and the smaller mass m1.

    http://img166.imageshack.us/img166/1899/p825.gif [Broken]


    2. Relevant equations

    F=ma
    fkkn


    3. The attempt at a solution

    I have drawn two free body diagrams.

    m1

    up: n1
    down:m1gCos(θ)
    left:F2 on 1 , m1gSin(θ)
    right: fk1

    m2

    up: n2
    down: m2gCos(θ)
    left: m2gSin(θ)
    right:F1 on 2 , fk2

    From this I wrote.

    Σ μ θ

    ΣFx1=fk1-F2 on 1-m1gSin(θ) = m1ax1

    ΣFy1=n1-m1gCos(θ)


    therefore

    n1=m1gCos(θ)

    ΣFx2=F1 on 2+fk2-m2gSin(θ)=m2ax2

    ΣFy2=n2-m2gCos(θ)


    therefore

    n2=m2gCos(θ)





    Now where to go from here? I am not sure but I believe I need to find the acceleration of the system since both are bound together by this law. By finding that I would then know ax1 and ax2.

    I'm just not sure how... two different μk confuse me, how can I find the total acceleration of the system if both are dragged by different coefficients?
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Mar 19, 2009 #2
    Σ μ θ

    Perhaps if I draw a third free body diagram representing the total mass being oppossed by two forces of friction. fk1 and fk2?

    fk1k1m1gCos(θ)

    fk2k2m2gCos(θ)

    Using total mass to find the force pressing against those two forces..?

    w12=m12gCos(θ)

    Maybe? Its my idea at the moment.

    It would allow me to find a=ΣF/m12
     
  4. Mar 19, 2009 #3
    Using the method in the quote I got.

    fk1k1m1gCos(θ)=9.21N

    fk2k2m2gCos(θ)=13.81N

    Oops below, it was Sin not Cos! (Had it wrong in the post above)

    w12=m12gSin(θ)=50.277

    So would that mean ΣF12=50.277N-13.81N-9.21N=27.575N

    a12=F/m12=1.817m/s2?
     
  5. Mar 19, 2009 #4
    And..

    F2 on 1=fk1-m1gSin(θ)-m1ax1

    F1 on 2=m2ax2-fk2+m2gSin(θ)


    Is that completely off base?
     
  6. Mar 19, 2009 #5
    If the acceleration I found for the total system was correct...

    a12=a1=a2=1.817m/s2

    then

    vf=(2aΔx)(1/2) <-- sqrt

    vf=4.263m/s

    Δt=vf/a12

    Δt=2.346s?
     
  7. Mar 19, 2009 #6
    The system didn't accept 2.346s.

    I just checked all my work and worked through it all again and got the same answer.

    So is something wrong with my original free-body diagram, logic or both?

    I think I need to factor in F1 on 2 and F2 on 1 to find the acceleration but I can't think of how I would do that..
     
    Last edited: Mar 19, 2009
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