Use of floor and ceiling functions in physics problems

[Agustin]

Homework Statement

explained on document attached

Homework Equations

Energy on a spring and work done by friction

The Attempt at a Solution

Included on document

https://docs.google.com/document/d/1FNrmIkkWzyZJNsbGbq_DYMyMZbpyMcAYKYk9iTbdR-4/edit?usp=sharing

 

[haruspex]

Homework Statement

explained on document attached

Homework Equations

Energy on a spring and work done by friction

The Attempt at a Solution

Included on document

https://docs.google.com/document/d/1FNrmIkkWzyZJNsbGbq_DYMyMZbpyMcAYKYk9iTbdR-4/edit?usp=sharing

What is your question to this forum?

 

[Agustin]

I want to know if my reasoning is right. Wether I made mistakes or not.

 

[haruspex]

I want to know if my reasoning is right. Wether I made mistakes or not.

Unfortunately it is very hard to read on my iPad because some of the special symbols are coming out as strings of question marks.
There appear to be sign errors in the early equations, but these might just be your transcription into the post.
Overall, I would say it is unnecessarily elaborate. You can write down straight away that the total distance traversed over the frictional area is $d=\frac {k\Delta x^2}{2F}$. Next, we can discard from d whole multiples of 2L, giving $d-2L\lfloor \frac d{2L}\rfloor$.

I'll take a look on a real computer later to see if the symbols become readable.

 

[Agustin]

Homework Statement

explained on document attached

Homework Equations

Energy on a spring and work done by friction

The Attempt at a Solution

Included on document

https://docs.google.com/document/d/1FNrmIkkWzyZJNsbGbq_DYMyMZbpyMcAYKYk9iTbdR-4/edit?usp=sharing

Unfortunately it is very hard to read on my iPad because some of the special symbols are coming out as strings of question marks.
There appear to be sign errors in the early equations, but these might just be your transcription into the post.
Overall, I would say it is unnecessarily elaborate. You can write down straight away that the total distance traversed over the frictional area is $d=\frac {k\Delta x^2}{2F}$. Next, we can discard from d whole multiples of 2L, giving $d-2L\lfloor \frac d{2L}\rfloor$.

I'll take a look on a real computer later to see if the symbols become readable.

That equation does not work for odd numbers. The problem asks the distance it stops relative to the point B. For example if d=7,25l the block will end up at 0.75l away from B, if we plug in those values in the equation you`ve given:
d-2l⌊d/2l⌋=7.25l-2l⌊7.25l/2l⌋=7.25l-2l⌊3.625⌋=7.25l-6l=1.25l

 

[jbriggs444]

That equation does not work for odd numbers. The problem asks the distance it stops relative to the point B. For example if d=7,25l the block will end up at 0.75l away from B, if we plug in those values in the equation you`ve given:
d-2l⌊d/2l⌋=7.25l-2l⌊7.25/2l⌋=7.25l-2l⌊3.625⌋=7.25l-6l=1.25l

Haruspex is using a different convention for the meaning of "d" than you are. He defines it as...

the total distance traversed over the frictional area is [d]

and then takes the remainder modulo 2L.

Translating that result to give the asked-for answer is a rather trivial exercise.

 

[Agustin]

Haruspex is using a different convention for the meaning of "d" than you are. He defines it as...
and then takes the remainder modulo 2L.

Translating that result to give the asked-for answer is a rather trivial exercise.

Maybe it is trivial, but I`m not looking for a trivial solution, but the general solution. That equation does not immediately give you the distance asked for, therefore it is a solution yet it isn`t the complete solution. The formula I present on the document immediately gives the distance asked for which is the point of the problem. And not just a simple formula.

 

[jbriggs444]

You do understand that the rules of this forum preclude the handing out of complete solutions?

 

[Agustin]

If I could post on any other forum I would, however I can´t so I have no choice but to post here. Am I to blame for that?

 

[haruspex]

Haruspex is using a different convention for the meaning of "d" than you are. He defines it as...
and then takes the remainder modulo 2L.

Translating that result to give the asked-for answer is a rather trivial exercise.

Thanks for jumping in, j. Yes, I was just illustrating how the first steps could be done much more simply and was leaving it to Agustin to take it from there.
Meanwhile, I could not verify Agustin's original treatment without finding a way to view it properly, and that has not happened yet.

I`m not looking for a trivial solution,

If I understand jbriggs' response, he was saying that obtaining the general solution from the point I got to was a trivial exercise. (But I would not have said it was that trivial )

 

[haruspex]

OK, I have now viewed the OP in a legible format and find that the final equation is incorrect.
Consider e.g. n=1. This produces L/2 instead of L. Similarly any odd integer value of n.
Likewise, an even integer gives L/2 instead of 0.

 

[Agustin]

OK, I have now viewed the OP in a legible format and find that the final equation is incorrect.
Consider e.g. n=1. This produces L/2 instead of L. Similarly any odd integer value of n.
Likewise, an even integer gives L/2 instead of 0.

You`re right. I forgot to check that part, that means my proof of the particular case on page 6 is wrong. I`ll check it and try working out a general formula that works. I greatly appreciate your time.

 

[Agustin]

I`ve already checked it, I think the formula is valid.

 

[haruspex]

I`ve already checked it, I think the formula is valid.

Then how do you explain that it gives the wrong answers for integer values of n?

 

[Agustin]

It does? It seems to work just fine for me, when n=1
dA=(l/2)(1+(-1) ^⌊n⌋(2(n-⌊n⌋)-1)=(l/2)(1-(2(1-1)-1))=(l/2)(1-(-1))=(l/2)(2)=l
If n=2
dA=(l/2)(1+(-1) ^⌊n⌋(2(n-⌊n⌋)-1)=(l/2)(1+(-1) ^2(2(2-2)-1)=(l/2)(1+(-1))=0

 

[haruspex]

It does? It seems to work just fine for me, when n=1
dA=(l/2)(1+(-1) ^⌊n⌋(2(n-⌊n⌋)-1)=(l/2)(1-(2(1-1)-1))=(l/2)(1-(-1))=(l/2)(2)=l
If n=2
dA=(l/2)(1+(-1) ^⌊n⌋(2(n-⌊n⌋)-1)=(l/2)(1+(-1) ^2(2(2-2)-1)=(l/2)(1+(-1))=0

You did not say that you had changed the equation. It looks right now.

 

[Agustin]

Sorry. Yes I`m very excited that it works. Doesn`t it seem interesting how much planning can some simple math do? It thrills me