# Homework Help: Sign of acceleration in the pulley-type question

1. Aug 17, 2010

### tsw99

1. The problem statement, all variables and given/known data

suppose an object A is placed on a horizontal frictionless table, and connected with a light inextensible string with a light, fixed pulley at the edge, and object B is only pulled by the vertical string that connected to a pulley.

I tried to set the acceleration up and right to be positive,
then two equations:
$$T-m_{A}a=0$$
$$T-m_{B}g=m_{B}a$$
so the acceleration of the system is
$$a=\frac{m_{B}g}{m_{A}-m{B}}$$

however if I set the acceleration down (and right) to be positive, then
$$T-m_{A}a=0$$
$$m_{B}g-T=m_{B}a$$
the acceleration becomes
$$a=\frac{m_{B}g}{m_{A}+m{B}}$$

they are not off by a minus sign. I can't think of a satisfactory explanation for this, and which one is correct? thanks for any help!

2. Aug 17, 2010

### Mindscrape

I'll tell you which one is wrong, the first one, and now I want you to tell me why it is wrong.

3. Aug 18, 2010

### tsw99

I think I see the reason.
since the string is inextensible, so the direction of displacement must be set in the second way, in order to preserve the length of the string

please tell me if I am right or wrong. thanks

4. Aug 18, 2010

### Mindscrape

Right, I think you have, or at least are on to, the right idea. In your first set up you made the acceleration go to the right for the top block and up for the second block. That would mean the two blocks are going to meet each other, which intuitively isn't at all what happens. From physics laws, we know from Newton's 3rd law the the direction of force on the first object is opposite to the direction of force on the second object, so your tensions cannot have the same sign. Mathematically, we know that you'd create a discontinuity if m_A=m_B, and we should know that is definitely wrong.