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Third Law of Newton

  1. May 18, 2014 #1
    Please look at picture.

    Here are my equations:
    For 10kg:
    -uk m2 g cosθ -T -m2 g sinθ=m2 a

    For 20kg:
    T-m1 g =m1 a

    Add them:

    -uk m2 g cosθ -m2 g sinθ -m1 g =(m2+m1) a

    a= 5 m/s^2 (kinematics equations)

    Isolating uk gives 3.66!

    The real answer is 0.16.

    Where am I going wrong?

    Attached Files:

  2. jcsd
  3. May 18, 2014 #2


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    The gravitational force on m2 (along the incline) points in the wrong direction in your first equation.
  4. May 18, 2014 #3
    Sorry I wrongly typed it.

    My original calculations also gave me this formula : -uk m2 g cosθ +m2 g sinθ -m1 g =(m2+m1) a

    But, I got 3.66, which makes no sense.
  5. May 18, 2014 #4
  6. May 18, 2014 #5


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    You're not getting 3.66, only -3.66.

    You're getting that because you have the wrong sign for the acceleration. If it's positive, it increases velocity, not decreases it. It pops up from the kinematic equations anyway. Go ahead and check it again.
  7. May 18, 2014 #6


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    For mass 2, m2 if you choose to use a negative sign to go with the tension, T, then the sign convention for m2 is that motion to the left & down the ramp is positive motion. That's perfectly fine & it will work out very nicely for this problem. Friction will oppose the motion, so you used the correct sign with it. In what direction will gravity tend to make m2 move?

    Your chosen sign convention also makes the sign of a2 be consistent with the sign of a1.

  8. May 18, 2014 #7

    Andrew Mason

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    It is pretty hard to figure out where you went wrong when you don't show us how you got your answer.

    Take the suggestion from the question and do a free body diagram for each mass. Then write the equation of motion for each:

    (1) T-m1g = m1a

    (2) m2gsinθ -μk m2 g cosθ - T = m2a

    Then all you have to do is solve the system of two equations for μk and a, which appears to be what you have done. Show us how you got your answer and we might be able to help you.

    Last edited: May 18, 2014
  9. May 18, 2014 #8


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    He got his answer by substituting 5m/s^2 instead of -5m/s^2 for a.
  10. May 19, 2014 #9

    Andrew Mason

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    Yes. Another approach to solving this is by using the change in energy:

    change in energy = ΔKE + ΔPE = work done against friction.

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