September 10th, 2019, 11:28 AM  #11 
Senior Member Joined: Jun 2019 From: USA Posts: 213 Thanks: 90 
It should be linear. It's your first graph with the axes flipped. And yes, m = 5*(1/d) + 0, or (1/d) = 0.2*m + 0. (Don't forget the units, though!) 
September 10th, 2019, 11:56 AM  #12 
Senior Member Joined: Mar 2019 From: TTF area Posts: 144 Thanks: 1 
Yes, thanks I do know that! What units would it be? and what would be my mathematical relationship between M(coin) and d for the linear graph? My nonlinear graphs relationship was M is proportional 1/d @DarnItJimImAnEnigineer Last edited by helpmeddddd; September 10th, 2019 at 12:17 PM. 
September 10th, 2019, 12:20 PM  #13 
Senior Member Joined: Jun 2019 From: USA Posts: 213 Thanks: 90  
September 10th, 2019, 12:28 PM  #14 
Senior Member Joined: Mar 2019 From: TTF area Posts: 144 Thanks: 1 
Mass is in grams, d is in cm so I guess 1/d is in cm^1 are you saying that 0.0161 in grams? @DarnItJimImAnEngineer Last edited by helpmeddddd; September 10th, 2019 at 12:51 PM. 
September 10th, 2019, 04:33 PM  #15 
Senior Member Joined: Jun 2019 From: USA Posts: 213 Thanks: 90 
What do you have to multiply by $cm^{1}$ to get $g$?

September 10th, 2019, 04:51 PM  #16 
Senior Member Joined: Mar 2019 From: TTF area Posts: 144 Thanks: 1 
No. Is it cm^1 than? I'm confused

September 10th, 2019, 05:35 PM  #17 
Senior Member Joined: Jun 2019 From: USA Posts: 213 Thanks: 90 
$\displaystyle m = \left( 0.0162~ g\cdot cm \right) \frac{1}{d} + \left( 0.016~ g \right) $ Now, for example, let's say $\displaystyle d = 10~ cm$. Your equation gives $\displaystyle m = \left( \frac{0.0162~g\cdot cm}{10~cm} \right) + \left( 0.016~g \right) = (0.00162~g)+(0.016~g) \approx 0.0176~g$ Believe it or not, I have thirdyear engineering students who struggle with getting units correct on things like this, so don't feel too bad. 
September 11th, 2019, 11:29 AM  #18 
Senior Member Joined: Mar 2019 From: TTF area Posts: 144 Thanks: 1 
how Do I relate this to torque? At the tipping point: At the tipping point:refer to attachment Last edited by helpmeddddd; September 11th, 2019 at 11:32 AM. 
September 11th, 2019, 01:22 PM  #19 
Senior Member Joined: Jun 2019 From: USA Posts: 213 Thanks: 90 
The last equation in your attachment looks an awful lot like the equation of your line, right? If you rearrange them, they should be the same. Incidentally, look at your scenario and define what is D. Is it a constant, or does it depend on d? If so (for example, maybe D=Ld, where L is a constant?), then rewrite it that way. 
September 11th, 2019, 01:44 PM  #20 
Senior Member Joined: Mar 2019 From: TTF area Posts: 144 Thanks: 1 
Maybe this attachment will help? I'm not sure what D is. last thing I want to know is how do I can compare my results to the theory using that formula? Last edited by helpmeddddd; September 11th, 2019 at 02:36 PM. 

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