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Acceleration of two packages on an incline

  1. Oct 29, 2011 #1
    1. The problem statement, all variables and given/known data

    Two packages at UPS start sliding down a 20 degree incline. Package A has a mass of 5kg and a coefficient of friction of 0.20. Package B has a mass of 10kg and a coefficient of 0.15. Package A is in front of Package B according to a diagram given. The distance between Package A and the bottom of the ramp is 2m. How long does it take for Package A to reach the bottom?

    2. Relevant equations

    Kinematics Equations and Free-Body Diagrams yielded the following breakdown of all the forces, given in this form:

    (Force), (x hat) +/- (y hat)
    uk_(box) is the kinetic friction coefficient.
    Ff is the friction force
    Fg/Fn are obvious
    X_Y is the force X on Y

    3. The attempt at a solution

    The very first thing I did was make a table of forces symbolically:

    For package A, assume a tilted axis of 20 degrees, with +x in the direction of the packages' motion.

    Fn, 0 + Fn
    Fg, mg sin 20 - mg cos 20
    Ff, -uk_A(Fn) + 0
    B_A, B_A + 0
    Fnet, (M_a)(A_a) + 0

    For package B, assume an identical axis.

    Fn, 0 + Fn
    Fg, mg sin 20 - mg cos 20
    Ff, -uk_B(Fn) + 0
    A_B, -A_B + 0
    Fnet, (M_b)(A_b) + 0

    The acceleation is constrained by A_a = A_b, allowing us to use one acceleration "a".

    I then get these final equations for:
    Package A X: mg sin 20 - uk_a(Fn) + B_A = M_a(a)
    Package A Y: Fn = mg cos 20
    Package B X: mg sin 20 - uk_b(Fn) - A_B = M_b(a)
    Package B Y: Fn = mg cos 20

    Now, I haven't the foggiest idea what to do. All I see are endless streams of unsolvable equations with two unknowns.
     
    Last edited: Oct 29, 2011
  2. jcsd
  3. Oct 29, 2011 #2

    Delphi51

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    Homework Helper

    Each package has a force down the ramp and an opposing friction force.
    Write that the sum of the four forces is ma, where m = 15. You should be able to find the acceleration in a jiffy.
     
  4. Oct 29, 2011 #3
    That's exactly what I did:

    Package A X: mg sin 20 - uk_a(Fn) + B_A = M_a(a)
    Package A Y: Fn = mg cos 20
    Package B X: mg sin 20 - uk_b(Fn) - A_B = M_b(a)
    Package B Y: Fn = mg cos 20

    What I didn't understand was how they tied together. The solutions manual says to add the two equations, treating both forces as one big force; however, I don't understand how exactly they did that.
     
  5. Oct 29, 2011 #4

    Delphi51

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    Homework Helper

    F = ma
    5g*sin(20) - .2*5g*cos(20) + 10g*sin(20) - .15*10g*cos(20) = 15a
     
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