My Math Forum data gathering and dependent/independent

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 September 9th, 2019, 12:17 PM #1 Senior Member   Joined: Mar 2019 From: TTF area Posts: 144 Thanks: 1 data gathering and dependent/independent Carry out an investigation into the relationship between the mass of an object on a plank, m. and the distance from the fulcrum, d. at which it will cause the plank to tip. You will be using a 30 cm rule to model the plank, and a pile of quarters (coins) to represent the mass of the cyclist. Can someone tell me what the dependent variable and the independent variable is in this investigation? also how should I set up a data gathering table?
 September 9th, 2019, 12:41 PM #2 Senior Member   Joined: Jun 2019 From: USA Posts: 213 Thanks: 90 It depends on how you decide to set up the experiment -- there's more than one way to skin a cat in this example. The independent variable is the one you control directly. For example, if you change the number of coins you put on the rule, then the number of coins (or mass) is the independent variable. The dependent variable (or variables) is/are what you will measure that depend on the independent variable. For example, you could place one coin on the rule, and move it until it just starts to tip the rule. The independent variable (the input, or what you control) is the mass, m. The dependent variable (the output, or what you measure) is the tipping distance, d. In this case, I would set up a table with one column for the masses, and one column for the distance measurements. (Or rows, if you prefer.) Incidentally, in real life we also have to deal with so-called extraneous variables, or things we can't necessarily control. Things like vibrations from people walking or breezes blowing on the rule would fall into this category, and if we're not careful about identifying and minimizing them, they can throw off our results. Go forth and practice, future experimental investigator. Thanks from helpmeddddd
September 9th, 2019, 12:47 PM   #3
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That's exactly how I wanna do the experiment! So usually you record at least 3 trials and get the average In this case, am I recording say 1 coin at the point it makes the ruler tip and recorded this 3 times?

attached is what I mean. @darnItJimImAnEngineer

My coins weigh 5.0g each.

My next question is..... If my data table is correct I need to create a non-linear graph from my raw data my data looks as tho the relationship is y=k/x is this correct? if so how do I transform my data to get a linear graph? Thanks
Attached Images
 Screen Shot 2019-09-09 at 12.33.38 PM.jpg (20.0 KB, 8 views)

Last edited by helpmeddddd; September 9th, 2019 at 12:59 PM.

 September 9th, 2019, 01:13 PM #4 Senior Member   Joined: Jun 2019 From: USA Posts: 213 Thanks: 90 What happens if you plot (1/d) versus m? Thanks from helpmeddddd
September 9th, 2019, 01:26 PM   #5
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I'll graph that but first, might sound stupid but how do I transform to 1/d is that just (from my table) 1/10 ect ect?

I transformed my data doing that 1/10 1/5.5 1/4 ect ect vs 5, 10, 15 ect ect

My linear graph is attached
Attached Images
 Screen Shot 2019-09-10 at 9.41.05 AM.jpg (15.6 KB, 7 views)

Last edited by helpmeddddd; September 9th, 2019 at 01:44 PM.

 September 9th, 2019, 02:51 PM #6 Senior Member   Joined: Mar 2019 From: TTF area Posts: 144 Thanks: 1 how do I write it in the equation form? my gradient being 5 ? y=mx+b y=Mass, M= gradient,5. whats x? and b?
 September 9th, 2019, 03:23 PM #7 Senior Member   Joined: Jun 2019 From: USA Posts: 213 Thanks: 90 Look at your graph. If y is mass, then x is (1/d). b is the y-intercept (theoretically zero). By the way, it's traditional to put the independent variable as the abscissa (horizontal axis) and the dependent variable as the ordinate (vertical axis). Thanks from helpmeddddd
September 9th, 2019, 03:36 PM   #8
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Umm... I think I've done something wrong. Can you show me how you transform the data into 1/d? because Now plotting mass on the x-axis and y on the vertical axis doesn't give me a linear line. I'll attach my non-linear graph.
Attached Images
 Screen Shot 2019-09-10 at 11.33.35 AM.png (17.7 KB, 4 views)

 September 10th, 2019, 10:42 AM #9 Senior Member   Joined: Jun 2019 From: USA Posts: 213 Thanks: 90 You already had it. If the distance d is a dependent variable, then the inverse distance (1/d) is also a dependent variable. Thanks from helpmeddddd
 September 10th, 2019, 11:19 AM #10 Senior Member   Joined: Mar 2019 From: TTF area Posts: 144 Thanks: 1 Ok, my graph is now linear it was my error. My gradient is now 0.0162 and my y-intercept is 0.016 the equation is ? m=0.0162 1/d+0.016? @DarnItJimImAnEngineer Last edited by helpmeddddd; September 10th, 2019 at 11:47 AM.

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