# Constant string length question with pulleys

1. Jan 18, 2017

### ual8658

1. The problem statement, all variables and given/known data
Given Ma = Mb =Mc and muk between c and the surface is 0.30

2. Relevant equations
(rope length) = constant
Maaa = Mag - T
Mbab = Mbg - 2T
Mcac = T - (mu)Mcg

3. The attempt at a solution
The equations I came up on my own and the answer key did as well. The problem is I also need an equation relating the length of the rope as constant. My convention is to measure the length of the rope in the direction of positive motion as defined by my equations.

In this case I make the assumption mass A and mass B accelerate down (the answer later gives B as negative indicating it goes up) and made that the positive direction. I also made going to the left positive for mass C.

The problem is when I go to write the rope length equation I get
Xa + 2 Xb + (X - Xc) = constant

and obviously differentiate to get the acceleration equation. I measured X as the distance from the right side to the pulley nearest C and thus X - Xc is the rope length between that pulley and C. This is so that I measure positive distances in the direction of positive acceleration.

The answer key simply lists the equation as
Xa + 2 Xb + Xc) = constant

with all other equations I gave as true.

Where am I picturing this wrong?

2. Jan 18, 2017

### Orodruin

Staff Emeritus
The picture defines $x_C$ as increasing when the mass moves to the right. Of course, if you define things different from the problem you will get different results.

3. Jan 18, 2017

### haruspex

How do you mean? Some arbitrary point fixed at the right side? But the string doesn't go there. It goes from the pulley to C, a distance XC.
A length does not have a direction. A displacement has direction. X-XC can be a displacement of C, but that does not make it a component of the string length.
Take a very simple model, two blocks connected by one straight string, moving to the left at the same speed. Their displacements are in the same direction, so the sum will not be constant.

4. Jan 19, 2017

### ual8658

Indeed it turns out the professor's answer was wrong. But thank you!

I think I was trying to get at displacement.

5. Jan 19, 2017

### Orodruin

Staff Emeritus
Would you mind elaborating. What you posted from the solution seemed correct.

6. Jan 20, 2017

### ual8658

Of course. So if you do define the displacement to the right as positive, then tension will pull in the negative direction and friction works in the positive direction. Acceleration then must be assumed positive to the right. In my relevant equations, tension is assumed positive to the left despite direction being define positive to the right. The direction of positive axes for both the displacement (rope length) and the kinematic equations must match. The way I did it in my explanation is correct so long as the rope length equation is NOT

Xa + 2Xb + Xc = constant

but rather the one with the negative Xc.