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Newton's 3rd law

  1. Nov 1, 2004 #1
    Newton's 3rd law....

    I'm having trouble with this problem...

    Two packages at UPS start sliding down the 20 degree ramp shown in the figure. Package A has a mass of 3.50 kg and a coefficient of kinetic friction of 0.200. Package B has a mass of 8.50 kg and a coefficient of kinetic friction of 0.130.

    How long does it take package A to reach the bottom?

    Picture is attached...

    Now since both have different masses and friction coefficients I'm thinking they each take a different amount of time to reach the bottom. However because at the beginning they're an action/reaction pair I ADDED the x-component of weight for B to the forces acting on A when calculating the acceleration... then used kinematics :

    [tex]x_f = x_i + v_i t + 1/2 a (t)^2[/tex]

    where x final is 2 meters and initial x and v are zero and acceleration is the one I got from force analysis...

    I've been at this problem for a few hours-- going back to it after every while and still have no idea on what to do. btw does it have calculus? It could be that. My professor puts in one or two questions with calculus in HW/midterm preperation questions but we never actually study physics problems with calculus. :(

    thanks in advance for any help. :)

    Attached Files:

  2. jcsd
  3. Nov 1, 2004 #2
    ok now I think I did a big mistake...

    I rewrote the free-body diagrams for each block and labeled the action/reaction force as F_a on b which is positve and F_b on a which is negative... now I have two equations with two unknows: the acceleration which is the same at the start and the action/reaction force. I can't subtract second equation from the first because that won't cancel out the action/reaction force since they have opposite signs...

    stuck again. :(
  4. Nov 2, 2004 #3


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    First assume that the blocks do not interfere. After drawing the freebody diagrams and writing the equations you will arrive at
    [tex] a_1 = g\sin \alpha - \mu_{1}g\cos \alpha[/tex]
    [tex] a_2 = g\sin \alpha - \mu_{2}g\cos \alpha[/tex]

    Because [tex] \mu_{1} > \mu_{2} [/tex]
    we have [tex] a_1 < a_2 [/tex]

    The assumption that the bodies behave seperately is not correct.
    Therefore both the bodies have the same acceleration "a"

    Then once again drawing the corresponding free body diagrams which also involve the force of contact between the blocks, we arrive at (Correct me if am wrong)

    [tex] a= \frac {(m_1 + m_{2})g\sin \alpha - g\cos \alpha(m_{1} \mu_{1} + m_{2} \mu{2})}{(m_{1} + m_{2})}[/tex]

    After this it is easy to calculate the time.
  5. Nov 2, 2004 #4


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    More importantly the concept behind this problem is that both the blocks move as one unit. Now take your co-ordinate system as perpendicular and parallel to the plane. (This is beacuse the distance to be covered is given parallel to the plane). Also while drawing the free body diagrams include the force exerted by block A on block B and vice versa. Notice how this force cancels when you add the two equations?. If you cant subtract.... just add :)
  6. Nov 2, 2004 #5
    Thanks for the help!

    I learned a lot from this problem thanks to you. :biggrin:
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