My Math Forum Wall, floor and stick

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 June 26th, 2011, 11:59 AM #1 Newbie   Joined: Jun 2011 Posts: 6 Thanks: 0 Wall, floor and stick First of all, forgive me for my terrible English, im from Ukraine and studding in conservatorie. I hope you will understand something from my writing. I dont know mathematic at all, but i find old USSR book in my house about mathemetic and, at that book i find mathemetic not so borring as it seems to be. Apply mathematic in daylly life can be verry fun and interesting, this was exploration for me. So i have one question: Imagine wall and floor. Angle between them is 90 degree. Also, we have a stik 1 meters long. Stick have two ends- A and B (oposite sites). A- is touching the floor, B- touching the wall (looks like triangle, where stick is median). If stick touching the wall and AB is parallel to the wall- thats mean that B is 1 meter high from the floor. If stick is parallel to the floor- A is 1 meter far from the wall. If we shove A end of the stick in the direction of a wall, then the B-end will slide up the wall.Interesting- the movement that point A and B makes , no match.If we move A from 40 cm point to 0 cm point, B will slide from 90 cm to 1 meter on the wall. Its look like some axeleration. How to describe this in mathematic?
 June 26th, 2011, 01:27 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Wall, floor and stick If we let A represent the distance end A is from the wall and B represent the distance end B is from the floor, then by the Pythagorean theorem, we have: (1) $A^2+B^2=1$ Now, to describe the motion of one end relative to the other, we may use calculus (hence, I moved this topic to the calculus sub-forum). Suppose we move end A towards the wall at a uniform rate r. Then we have (2) $\frac{dA}{dt}=-r$ Implicitly differentiating (1) with respect to t, we find: $2A\frac{dA}{dt}+2B\frac{dB}{dt}=0$ Divide through by 2: $A\frac{dA}{dt}+B\frac{dB}{dt}=0$ Solve for dB/dt: $\frac{dB}{dt}=-\frac{A}{B}\frac{dA}{dt}=\frac{A}{B}r$ Separating variables, we have: (3) $B\,dB=Ar\,dt$ From (2) we have: $\int_1\,^{A(t)}\,dA=-r\int_0\,^t\,dt$ $A(t)=-rt+1$ Substituting into (3) we get: $\int_0\,^{B(t)}B\,dB=\int_0\,^t -r^2t+r\,dt$ $\frac{1}{2}B^2(t)=-\frac{1}{2}r^2t^2+rt$ $B^2(t)=2rt-(rt)^2=rt(2-rt)$ $B(t)=\sqrt{rt(2-rt)}$ $\frac{dB}{dt}=\frac{A}{B}r=\frac{1-rt}{\sqrt{rt(2-rt)}}r=\frac{\sqrt{r}(1-rt)}{\sqrt{t(2-rt)}}$ where $t\ne0$
 June 26th, 2011, 01:50 PM #3 Newbie   Joined: Jun 2011 Posts: 6 Thanks: 0 Re: Wall, floor and stick MarkFL OMG it will took a lot of time to understand this Is it realy so difficult? I dont know anythink about calculus (please recomend some book). I was tried to find the answer by myself, but all what i get- butiful picture. Different positions of mediana form some shape of circle. http://allmatematika.ru/e107_files/publ ... 17_796.jpg
 June 26th, 2011, 01:56 PM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Wall, floor and stick If we wish to discuss the velocities of the two ends then calculus is called for, however if we wish only to discuss the positions, then all you need is the Pythagorean theorem. If you know A, then B is given by: $B=\sqrt{1-A^2}$
 June 26th, 2011, 02:38 PM #5 Newbie   Joined: Jun 2011 Posts: 6 Thanks: 0 Re: Wall, floor and stick MarkFL, Its funny: what you imagin when you see a^2 ?? The square or the quantity? http://xmages.net/show.php/2901212_jpg.html
June 26th, 2011, 02:47 PM   #6
Senior Member

Joined: Jul 2010
From: St. Augustine, FL., U.S.A.'s oldest city

Posts: 12,211
Thanks: 521

Math Focus: Calculus/ODEs
Re: Wall, floor and stick

I see both...I see the quantity if I am after B, but for the Pythagorean theorem I see squares:

[attachment=0:3vv1mi25]pythagorean.jpg[/attachment:3vv1mi25]
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 pythagorean.jpg (2.1 KB, 552 views)

 June 26th, 2011, 05:03 PM #7 Senior Member   Joined: Apr 2007 Posts: 2,140 Thanks: 0 Interesting physics problem is let angle between the floor and the stick be greater than or equal to 0° and less than or equal to 90°. Let an object with mass stand at point B when the angle is 0° to start. Assume the gravity takes effect, at what angle greater than 0° will the object start to slide towards the point A? This is one of those friction problems.

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