# Choosing different axes in the same system

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1. Dec 8, 2016

### Carpetfizz

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. Dec 8, 2016

### BvU

Forget the math and look at it as a physicist: would the tension go to infinity if the two masses go to the same $M$ ?

3. Dec 8, 2016

### Carpetfizz

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?

4. Dec 8, 2016

### Buffu

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

5. Dec 8, 2016

### Carpetfizz

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?

6. Dec 8, 2016

### BvU

Do the exercise for the case M1 = M2 first...

7. Dec 8, 2016

### Buffu

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.