How do I calculate masses predicted by the force law equation

[jonathanM111]

I have attached the problem and my attempt at solving it, there is a strange equation under the data analysis section that I do not understand. I'm and suppose to calculate a theoretical value for a slope and compare it to the one I have.

 

[haruspex]

https://www.physicsforums.com/conversations/locked-brainly-thread.131039/goto/post?id=5833956#post-5833956
I posted the question here on brainly https://brainly.com/question/4998386
first 3 attachments are the laboratory manual and the 4th is my attempt.

I wrote this response to your original thread, but it got locked before I could post...

I don't understand the directions you are given. Seems to me you should be plotting the log of the mass difference against the log of the distance.
Looks like you needed more precision in the mass measurements. The first half of your datapoints are useless because the mass difference is beyond the precision limit.
To answer your question, you need to compute the force predicted by the quoted law and infer the mass difference from that. However, the quoted law only specifies "is proportional to" so there will be an unknown constant factor. Since you are taking logs, that will show up on the graph as an unknown vertical displacement. This means you should slide the predicted line vertically to get the best fit with the plotted line from the measurements

 

[jonathanM111]

https://www.physicsforums.com/conversations/locked-brainly-thread.131039/goto/post?id=5833956#post-5833956
I posted the question here on brainly https://brainly.com/question/4998386
first 3 attachments are the laboratory manual and the 4th is my attempt.

I wrote this response to your original thread, but it got locked before I could post...

I don't understand the directions you are given. Seems to me you should be plotting the log of the mass difference against the log of the distance.
Looks like you needed more precision in the mass measurements. The first half of your datapoints are useless because the mass difference is beyond the precision limit.
To answer your question, you need to compute the force predicted by the quoted law and infer the mass difference from that. However, the quoted law only specifies "is proportional to" so there will be an unknown constant factor. Since you are taking logs, that will show up on the graph as an unknown vertical displacement. This means you should slide the predicted line vertically to get the best fit with the plotted line from the measurements

hmm, you see, I was thinking the same but I can't seem to find a way to use the formula provided, I might be misunderstanding it but I've also looked up the equation Fαd^-3 but I just can't find it anywhere. Finding the new data for the plot that's proportional to the log of mass vs distance is where I am stuck.

 

[haruspex]

I've also looked up the equation Fαd-3

You mean F∝d^-3, right? This looks like a dipole equation. This is because a bar magnet is effectively made of little dipoles all pointing the same way. See https://en.m.wikipedia.org/wiki/Force_between_magnets.
This means mg=F=kd^-3 for some unknown constant k.

Trust me, reinterpret the problem as plotting log(m) against log(d).
If you take logs of both sides of this equation, what do you get?

 

[jonathanM111]

https://www.physicsforums.com/conversations/locked-brainly-thread.131039/goto/post?id=5833956#post-5833956
I posted the question here on brainly https://brainly.com/question/4998386
first 3 attachments are the laboratory manual and the 4th is my attempt.

I wrote this response to your original thread, but it got locked before I could post...

I don't understand the directions you are given. Seems to me you should be plotting the log of the mass difference against the log of the distance.
Looks like you needed more precision in the mass measurements. The first half of your datapoints are useless because the mass difference is beyond the precision limit.
To answer your question, you need to compute the force predicted by the quoted law and infer the mass difference from that. However, the quoted law only specifies "is proportional to" so there will be an unknown constant factor. Since you are taking logs, that will show up on the graph as an unknown vertical displacement. This means you should slide the predicted line vertically to get the best fit with the plotted line from the measurements

okay so I think I figured it out F is proportional to the inverse distance cubed, that's what the expression says. Therefore I can calculate the the inverse of the distance cubed and take the log from it, if I plot this new line on the graph that I have they're pretty similar in terms of their slope, which is the dipole force.

 

[haruspex]

okay so I think I figured it out F is proportional to the inverse distance cubed, that's what the expression says. Therefore I can calculate the the inverse of the distance cubed and take the log from it, if I plot this new line on the graph that I have they're pretty similar in terms of their slope, which is the dipole force.View attachment 210325

Ok, but please try plotting each against log(distance) instead of against distance.

 

[jonathanM111]

Ok, but please try plotting each against log(distance) instead of against distance.

this is how the two graphs look(theres more tables in this experiment) it does look correct how you say, why would the manual say otherwise?

 

[jonathanM111]

Ok, but please try plotting each against log(distance) instead of against distance.

BTW I apologize for my sloppiness.

 

[haruspex]

this is how the two graphs look(theres more tables in this experiment) it does look correct how you say, why would the manual say otherwise?View attachment 210332

As you can see, the d^-3 line is now dead straight. This is why I thought log(d) is the right way. But, interestingly, the curves from the experiment look less convincing now. Either this is down to experimental error or the exact equation to match the experiment is not F∝d^-3.
I think I see reasons it won't be quite that. For one thing, the far end of the horizontal magnet will have some effect. I'll get back to you.

 

[jonathanM111]

As you can see, the d^-3 line is now dead straight. This is why I thought log(d) is the right way. But, interestingly, the curves from the experiment look less convincing now. Either this is down to experimental error or the exact equation to match the experiment is not F∝d^-3.
I think I see reasons it won't be quite that. For one thing, the far end of the horizontal magnet will have some effect. I'll get back to you.

hmm, for your reference this is how they look they they're against plain distance.

 

[haruspex]

Having looked into this some more, I suspect the whole question is flawed.
The dipole formula is an approximation which can be used when the lengths of the magnets are small compared with the distance between them, and even then assumes the magnets are end-to-end.
In the present case, it would be more accurate to treat the magnets as having a magnetic monopole at each end. This is still an approximation (the "Gilbert model"), but should work better here.
Unfortunately the vertical force equation gets quite messy, but to a first approximation it would be proportional to d^-2, not d^-3. Note that if you ignore your measurements with the largest d values the slope of the log-log graph is -2, or even a bit less.

 

[haruspex]

Treating each magnet as consisting of a point magnetic "charge" (i.e. monopole) at each end, I obtained an expression for the force.
The image below shows the expression [as the Excel formula, 1/A2/A2-1/(1+A2)^2-A2/(1+A2*A2)^1.5+(1+A2)/(1+(1+A2)^2)^1.5], and a graph of ln(F*L^2) against ln(d/L), where L is the length of each magnet and d is the separation between their closest poles.
It exhibits a curvature similar to the one that you plotted from measurements. The straight line fit has a gradient closer to -2 than to -3.
For larger values of d/L the gradient steepens, with no obvious limit. I believe this is because the far pole of the horizontal magnet becomes significant, rapidly weakening the force as it counteracts the force on the near end.
For very small values of d/L it should converge to -2.

To get the d^-3 relationship, the vertical magnet should be short compared with d, making it a simple dipole, but the horizontal magnet long compared with d, making it more like a monopole.