# Gravity with Galileo's inclined plane?

1. Feb 1, 2004

### flamedfordbronco

Hey everyone,
I'm new here, so I'll let you know a little about me. I'm going to uconn right now, and my major is mechanical engineering. I have a question about a lab we did for my engineering class. We replicated Galileo's experiment with inclined planes, and our teacher told us to find gravity, but my physics class was a long time ago so I'm a little fuzzy on the equations I have to use (our teacher didn't give us any, we have to find it all out on our own). Here's the data we have:

-Mass of two metal balls (used to roll down the plane)
-length of plane
-height of plane
-angle the plane is at compared to horizontal.
-time it took for the balls to roll the full length, 3/4, 1/2, and 1/4.

Now, we have to calculate the force of gravity from this data, which I know is 9.8 m/s^2. but, how do I get there? we did two trials of the experiment, with two different angles, so I'm sure that you have to use the angles in there, I just have no idea how. I plotted a graph of distance over time squared, and found the slope of the best fit line, so that's the acceleration of the incline right? How should I get to gravity from this? thank you all very much, I hope I was clear on this.

~Frank~

2. Feb 1, 2004

### Staff: Mentor

What you are trying to calculate is the acceleration due to gravity (called g) based on your data. Since the ball is constrained to move along the plane, what you will measure is the acceleration along the plane. This acceleration is related to g like so: a = g sin&theta;, where &theta; is the angle the plane makes with the horizontal.

3. Feb 1, 2004

### flamedfordbronco

Doc Al,
Thanks a lot for your post! I thought that acceleration due to gravity and acceleration of gravity were different things, but I wasn't sure because my professor is from poland and can't speak english too well lol.
How do I get the acceleration along the plane? Is that by taking the slope of the line of best fit of that graph that I made? I read on another website that 2*slope=acceleration along plane. do I just divide that by the sin of the angle to find the acceleration due to gravity? should I get huge numbers for this?
I'll give you an example of what I've done so far. I plotted the graph for a ball traveling down a 1.938 meter plane, at a 9.626 degree incline. The slope of the best fit line of that graph is 4.64 m/s^2. So, the acceleration along the plane is 9.28 m/s^2 right? then, if I divide that by the sin of 9.626, I get 55.5 m/s^2... am I doing something wrong? That seems like a giant number. Thanks a bunch Doc Al, I've been trying and trying to figure this out all weekend, but I kept getting those huge numbers, and thought I was wrong. Thanks again!

~Frank~

4. Feb 1, 2004

### flamedfordbronco

Whoa Doc Al I have a correction to make,
I think I made my graphs wrong, I had distance on the x axis, time on the y axis. When I switch them around and get the slope then, it turns out to be .214, so the acceleration along the incline would be .428, and then the acceleration due to gravity is 2.56, which seems more correct to me! is that what I was screwing up with? thanks!

~Frank~

5. Feb 1, 2004

### Staff: Mentor

What you should be measuring is the acceleration along the plane. What you literally measure is the time (t) at various distances (d). The equation linking these together is d = 1/2 a t2. So, if your "slope" is the slope of a "d" versus "t2" graph, then you must double the slope to get "a".
Yes.
No. You must be making an error somewhere.

6. Feb 1, 2004

### Staff: Mentor

That's part of the problem, but your numbers still seem way off to me.

7. Feb 1, 2004

### flamedfordbronco

yeah still seems way off to me too... would you mind if I ran my results by you to see if you can pick up where I'm messing up?
Our plane was 75.5 inches long, I converted this to 1.938 meters. The plane was also 13.125 inches high, which I converted to .3328 meters. Then, our heavy ball went down the full length of the track in an average of 2.956 seconds, or 8.738 square seconds. We also have the times for 3/4 of the track, 1/2, and 1/4 (when you square them and plot them on the graph, it's a straight line, so I'm assuming that those are ok). When I figure out the acceleration due to gravity, all the numbers come out to around 2.5, no matter what the angle or weight, so it seems like I have one place where I'm making a big error in all of them. shouldn't the acceleration due to gravity still be around 9.8? Thanks again doc, I really appreciate this.

~Frank~

8. Feb 1, 2004

### Staff: Mentor

I would hope so. Those times seem long to me. There must be something you (or I) am missing.

9. Feb 1, 2004

### flamedfordbronco

if I don't square the time, then it comes out to around 11, which is closer, but I thought the times had to be squared... I'll keep looking for things we might be missing, thanks!

~Frank~

10. Feb 2, 2004

### flamedfordbronco

Doc,
I've been talking with my group members, and I found out what my problem was. Since there was so much noise in the class, I misheard the length of the ramp! it was 107.5 inches, instead of 77.5 *d'oh!*. So now that I changed the length, that changes the angle, and so the equations work out and I get around 10 for every trial. Thanks a bunch for all your help!

~Frank~

11. Feb 2, 2004

### Staff: Mentor

You are most welcome. I'm glad you figured out the problem.