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Choosing different axes in the same system

  1. Dec 8, 2016 #1
    1. The problem statement, all variables and given/known data

    Just doing some practice problems from past finals and I needed some help on this one. Sorry if my question doesn't exactly fit the template.


    2) Relevant Equations / Information

    For part a) and for M_1, I drew the axes such that the x-axis points to the top right, in the direction of motion of M_1, and the y-axes points up perpendicular to it. For M_2 I drew the axes such that the x-axis points to the bottom right and the y-axis points up perpendicular to it.

    This is the answer the solution guide provided


    The difference is that my axes for M_1 was flipped such that the x-axis pointed in the opposite direction to what is presented in the solution.

    I'm curious as to why they chose to make the direction opposite to M_1's motion positive.

    3) Attempt at Solution

    I tried doing it both ways and it yielded two different answers

    1. The way in the answer sheet:

    $$T = \frac{M_1M_2g(\mu_1cos(\phi)+sin(\phi)+sin(\theta)}{(M_1+M_2)}$$

    2. The way with the different axis:

    $$T = \frac{M_1M_2g(\mu_1cos(\phi)+sin(\phi)+sin(\theta)}{(M_1-M_2)}$$

    I'm not exactly sure how switching the axes gives the right answer or what the rationale behind doing that is. Any advice would be much appreciated.
  2. jcsd
  3. Dec 8, 2016 #2


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    Forget the math and look at it as a physicist: would the tension go to infinity if the two masses go to the same ##M## ?
  4. Dec 8, 2016 #3
    Sorry, I'm not sure what you mean. Where in my question does it imply the two masses go to the same M and what does that mean?
  5. Dec 8, 2016 #4
    $$T = \frac{M_1M_2g(\mu_1cos(\phi)+sin(\phi)+sin(\theta)}{(M_1-M_2)} = \infty$$ if $$M_1 = M_2$$
  6. Dec 8, 2016 #5
    Okay that makes sense. I feel like you wouldn't get the quantity (M1-M2) unless you solved for T all the way through. Is there any way you can account for this edge case before you solve the problem?
  7. Dec 8, 2016 #6


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    Do the exercise for the case M1 = M2 first...
  8. Dec 8, 2016 #7
    The FBD in book is correct.
    I can't tell your mistake because I have not seen your FBD and the description you provided is ambiguous.
    Can you please provide your FBD.
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