Determining what the y-intercept means in a force lab

[alyssad]

Homework Statement

In lab, we put a glider on an airtrack and used a pulley to move the glider. The mass on the pulley was changed with each iteration. I have created a graph to find the mass of the glider, but I don't know what the y-intercept is supposed to mean. The graph I have is linear, and the slope is in units of kg.

Homework Equations

(g/a) = (m1)(1/m2) + 1
(g/a) --> 9.8/acceleration that was found experimentally
m1 --> mass of the glider and the slope of the line
m2 --> mass that was attached to the pulley

The Attempt at a Solution

Part of the lab is to predict the y-intercept. I predicted that the y-intercept would be 1 based on the equation that we were supposed to use, but the y-intercept ended up being -2.57. I know that leads to a large percent error. What I need to know is what the y-intercept is supposed to represent in this context.

Thanks for any help that you can give!

 

[TSny]

So, you are plotting g/a on the vertical axis and 1/m₂ on the horizontal axis. The y-intercept corresponds to making 1/m₂ equal to zero. This would mean making m₂ infinite. In other words, the y-intercept is essentially the value of g/a for an extremely large value of m₂. Intuitively, what would you expect g/a to be in this case? Thinking about this might help you understand the predicted value of 1 for the y-intercept.

It could be that your experimental value for the y intercept is sensitive to error in your data.

Can you upload a figure showing your graph? If not, can you list your experimental data points (1/m₂, g/a)?

 

[alyssad]

I've added a picture of my data as well as the graph.

 

[TSny]

OK. As you can see from your graph, the difference between +1 and -2.57 for the y-intercept is actually very small considering that your y-data extends over the range of about 8 - 54. So, I think a y-intercept of -2.57 can be considered "close" to a y-intercept of +1 even though the % difference of these two numbers is large.

I noticed that you apparently used the values for $\frac{1}{m_2}+1$ rather than the values of $\frac{1}{m_2}$ for your "x" values for the graph. Did you mean to do that?

 

[alyssad]

The equation in the lab manual is g/a = m_1*(1/m_2) +1. Given this, figured that I should be adding 1 to my x values, but now I'm thinking that shouldn't be doing that.
Also, I apologize for the awful formatting. I'm still trying to figure out how to format equations.
I just changed my graph so that the data used for the x-values didn't include the +1, and my y-intercept is now -1.2

 

[TSny]

The equation in the lab manual is g/a = m_1*(1/m_2) +1. Given this, figured that I should be adding 1 to my x values, but now I'm thinking that shouldn't be doing that.
Also, I apologize for the awful formatting. I'm still trying to figure out how to format equations.
I just changed my graph so that the data used for the x-values didn't include the +1, and my y-intercept is now -1.2

The equation for a straight line on x-y axes with y-intercept b is y = mx + b. A graph of this line would plot y on the vertical axis and x on the horizontal axis. It would not plot x + b on the horizontal axis.

Likewise, if you have (g/a) = m₁(1/m₂) + 1, this would give a straight line with y-intercept +1 if you plot (g/a) on the vertical axis and 1/m₂ on the horizontal axis. You do not want to plot 1/m₂ + 1 as the "x" values.

So, making this change does help a little in moving the y-intercept up. I think a value of -1.2 is actually very good compared to the expected value of +1.

 

[alyssad]

Thank you so much for all of your help! I really appreciate you taking the time to explain it to me.

 

[TSny]

You are welcome. Enjoy your course.