# Obtain y=mx+b equation for Atwood's Pulley lab.

1. Sep 28, 2008

### Vasili

1. The problem statement, all variables and given/known data
How can the validity of Equation (1) for Atwood's pulley be investigated by means of a linear graph? What quantities would the slope m and intercept b represent here?

2. Relevant equations
Equation (1):

3. The attempt at a solution
Since the value for (m1+m2) never varies, It is a constant in this equation. The term g is also a constant. a and (m1-m2) are variables. So obviously I have to manipulate this equation into the y=mx+b form, with m and b as constants and x and y as variables, but for some reason I don't know how to approach it. Does the answer lie in substituting values for one of the variables from another formula?
Since a=2h$$^{2}$$:
2h$$^{2}$$ = (m1 - m2)g/(m1 + m2) . . . (I can't seem to figure out subscripts in the latex thing. They always seem to come out as superscripts).
But then I have just said that a=g, which doesn't help me. I think part of the problem is that having the same terms (the masses) on either side is what is confusing me here. I'm not used to having to manipulate an equation like this. I know it's probably simple but I'm stumped. I'd be very grateful for even a nudge in the right direction.

Last edited: Sep 28, 2008
2. Sep 28, 2008

### Vasili

Okay, so I've been thinking about this some more... Since I know that (m1+m2) is constant and (m1-m2) is variable, with the acceleration being the responding variable, the equation for acceleration:

Is already in the y=mx+b form, with y=a, m=g, x=(m1-m2)/(m1 + m2), and b=0. I don't know why that was so complicated last night... If x(the mass difference) is zero, the acceleration must also be, so there is no intercept, and I verified that the slope was 9.8 by making a table and graphing it. So is that all there is to it? Someone could at least post telling me whether I have actually answered the question or not...

3. Sep 28, 2008

### PhanthomJay

I don't know why you say m1 - m2 is variable, but m1 + m2 is constant? They are both variable, which you seem to have corrected in your mx +b form. Only g is constant (that's the 'm' in front of the 'x'), and the 'a' (called y) and (m1-m2/(m1+m2) (called 'x') are both variable. So youare correct that the slope is 'g'. (However , note that the graph stops abruptly when x>1 or x<-1).

4. Sep 28, 2008

### Vasili

(m1 + m2) is constant because the total mass is always the same number (mass removed from one mass is added to the other mass in the experiment). The ratio (m1-m2) of masses is what is manipulated in the experiment, and is not constant. Unless I misunderstand something. Thank you for the post, though.

5. Sep 29, 2008

### youngblood21

Ok I agree with you Vasili that (m1 + m2) is constant because as you said the total mass of the system does not change. I'm looking at slope and the y-int... I agree also that m=9.81 but why does b=0? I keep thinking that b is initial velocity but if it is initial velocity then the graph is no longer linear... any help here on what b is and why it is zero?

6. Sep 29, 2008

### Vasili

B is zero because if the x variable (m1-m2) is zero (both masses are the same), then the system will not accelerate in either direction, so the acceleration (y variable) equals zero. Which also means that the intercept is zero.