Slacklining, or the parallelogram of force
As the consulting business is really exhausting, we had to go to Miami Beach to relax a bit:
If you now get the impression that we are lazy, then you’re totally wrong. We are doing university level math and solving the largest mystery of life. Let me introduce to you:
This is Christoph. He is from the “Law school” in Hamburg. And the blue thingy which he wears around his neck, that’s a slackline.Whenever Christoph finds two trees in a distance of about 6.561 yards (6m) with nothing in between, he will wrap one end of the slackline around one tree and the other end around the other tree, so that he will finally have a tight rope in between. Then he justs jumps on it and walks back and forth until he will finally fall down. This sounds easy, but as he told me, you need a really good sense of balance and a lot of training.
So, what has this to do with the “largest mystery of life”? Well, as we were having our Miami Beach sunbathes, we were drawing little parallelograms of force into the sand. We tried to figure out what force would act on the tree while Christoph was trying to keep upright. Well, while we were on the beach, we did not really figure it out. The main question was: How far is the slackline deflected while Christoph is standing on it?
While Christoph is slacklining, he has better things to do than measuring distances, so he couldn’t tell us. And we were, of course, to lazy to go to the next tree. But later that day (when it was already to dark to get sharp photos) we did the experiment and finally got our data. Here are my calculations:
From the distances you get the angles α1 = 7.59° and α2 = 18.43° which gives the following forces in the ropes:
F1 = 1513N
F2= 1581 N
where F1 is the force in the left rope and F2 is analogously the force in the right rope.
Well, that’s not that much. Sorry, Christoph, you will not chop any trees down this way, but your 7tons rope will stand it forever (it was like 7 tons, wasn’t it?) .
PS/1: I once again did not manage to include the SVG drawing into this blog post. Wp.com does not allow me to upload svg files and it seems like I cannot include files from foreign hosts. Does anyone know a way around this?
PS/2: It was easy to draw the palm trees following these instructions.
Wieso kriege ich 151,4 kP für das linke Seilstück raus und 158,28 kP für das rechte Seilstück?
Der linke Bum trägt übrigens 20 kg des netten Jungen und der rechte 50 kg, beide Bäume werden mit je 150 kp nach innen gezogen,denke ich. Aber wo ist mein Fehler?
Moin min Morn!
Tjo, mit 1kP = 10N haben wir doch die gleiche Lösung. Oder übersehe ich etwas?
Hier kurz mein Rechenweg: Ich habe eine Kräftebilanz an dem Schnittpunkt der beiden Seilabschnitte gemacht. In horizontaler Richtung müssen sich die beiden Seilkräfte gerade aufheben:
0N = cos(alpha1)*F1 – cos(alpha2)*F2
Und in vertikaler Richtung müssen beide Seile zusammen die 700N von Christoph tragen:
700N = sin(alpha1)*F1 + sin(alpha2)*F2
Die Lösung dieses Gleichungssystems ist F1=1513N und F2=1581N. Oder, wenn man mit diesen Kilopond rechnet (das tun ja nichtmal die Amis!), dann eben mit dem Komma eine Stelle weiter links ;-).
[...] yet introduced our group: We, that is my fellow Alexander (who I might call Pingu later on), the slackline walker Christoph and Clemens. Clemens seemed not to be very familiar with my way of driving a car: when he [...]