Solve Coordinate System Problem: Find Acceleration & Tension

[frostchaos123]

Hi everyone, wondering if anyone can help me with this little problem...

An object with mass m1 , resting on a frictionless horizontal table, is connected to a cable that passes over a pulley and then is fastened to a hanging object with mass m2. Find the acceleration of each object and the tension in the cable.For this coordinate system, where the x-axis is pointing to the right, which is the positive, and the y-axis is pointing downwards, which is positive: http://www.imagehosting.com/show.php/1011793_Coordinate.bmp.html

Given that T is the tension, the equations should be:
m1*(a) = T;
m2*(a) = -T + m2*(g);

However, if i switch the x-axis from pointing rightwards to pointing leftwards, i will get this: http://www.imagehosting.com/show.php/1011808_Coordinate1.bmp.html

The equations become:
m1*(a) = -T;
m2*(a) = -T + m2*(g);

I am confused as to how to solve this using the 2nd coordinate system, as i feel that coordinates should not affect the answer.

My question is that what is it that is wrong in my equations that i cannot get the answer? Or is it possible that this assumption is flawed? I feel that maybe the tension have to opposite each other (having opposite signs) regardless of coordinates, but i am not very sure.

Thanks to all for your time.

 

[Doc Al]

However, if i switch the x-axis from pointing rightwards to pointing leftwards, i will get this: http://www.imagehosting.com/show.php/1011808_Coordinate1.bmp.html

The equations become:
m1*(a) = -T;
m2*(a) = -T + m2*(g);

What's the problem? You'll get a negative answer for "a", which means it points to the right.

Even better would be to let "a" be the magnitude of the acceleration. Since the tension points to the right of m1 and is called -T, the acceleration must also point to the right and should be called -a. (The same logic applies to the m2 acceleration.) That gives you the same equations you had before.

 

[frostchaos123]

So in other words, I should reformulate them to become:

m1*(-a) = -T;
m2*(a) = -T + m2*(g) ?

But if it is true in this case, wouldn't the expression m1*(a) and m2*(a) require one to know before hand which direction would the acceleration points to?

 

[Doc Al]

So in other words, I should reformulate them to become:

m1*(-a) = -T;
m2*(a) = -T + m2*(g) ?

You forgot the minus sign before the "a" in the second equation.

But if it is true in this case, wouldn't the expression m1*(a) and m2*(a) require one to know before hand which direction would the acceleration points to?

You don't have to know which way the acceleration points. If you solved your second set of equations in your first post, you'd get a negative answer. Either way you get the correct answer.

Of course you should know the direction of the acceleration. If the tension pulls m1 to the right, then it accelerates to the right.

The really important thing is to be consistent in your two equations, so that "a" means the same thing in both.

 

[learningphysics]

You made a mistake with your second set of equations in the new coordinate system... If you presume that "a" is in the positive x direction (to the left) for m1, then you must label "a" as being in the -y direction for m2 (upward). Because the two accelerations are dependent on each other...

So those equations would be:

The equations become:
m1*(a) = -T;
m2*(-a) = -T + m2*(g);

 

[learningphysics]

Ah, Doc Al answered before me.

 

[frostchaos123]

I see, thanks for all the replies. So basically, I have to consider that the acceleration for the m1 part and the m2 part as a whole and formulate them accordingly, while remembering what the "a" physically means.