GR9277 #100



Alternate Solutions 
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Comments 
Albert 20091105 04:38:54  Alternate solution:
No center of mass needed, no torque, no directions, nothing! I solve it by simply taking the ratio of the masses with the lengths of the rod on either side of the fulcrum. And here's how:
On one side you have and other side has . The rod weighs , so I divide in half and give both weights each. And now they weigh and .
Once reached so far, just put 'em up in ratio and get it,
comes out , which is of course further from the center.
The goddamned technique works every freakin' time!
Just make sure you set the ratio to "more mass: lesser length".
apr2010 20100408 15:26:08 
That is interesting that you do not know which has less length...Keeping the indexes right should work.

SurpriseAttachyon 20150910 15:00:03 
That\'s so strange that this works. The mass on the bar on each side and the mass from the weight on each side should be treated differently because as you move the bar, the barmass portion changes but the weightmass doesn\'t.\r\n\r\nYet when you plug in different answers, it\'s still correct. That\'s odd.

nikog 20151020 19:04:27 
it\'s not that odd. The torque produced from a force F at the a point p at a distance x (i.e at the point x+p) is τ=Fx. Instead if you consider a problem where you have two forces F1= and F2= symmetrical to the point p and you measure the torgue at x+p then\r\nτ=( l+x)  (lx) = Fx where l the distance from the point p

  matonski 20090324 01:57:48  I just put the origin at the center and found the center of mass. COM =
kroner 20090929 14:09:00 
I think it's safe to say that matonski wins.

adjwilley 20091104 10:40:16 
Beautiful.

torturedbabycow 20100409 10:14:17 
Seconding the "beautiful." In fact, I think I might need a cigarette after reading it. ^__^

alemsalem 20100926 09:32:41 
fuck yeah :) im also gonna smoke now

neon37 20101112 09:11:16 
Damn!......Elegant!...

elenaPh 20130415 11:40:16 
amazing solution ! thanks!

aaa2 20140925 12:06:00 
wow just wow

justacomment 20141007 23:19:21 
Salute.

blabla0001 20170325 09:26:30 
this solution is too good for me. it is also so good that it becomes dangerous. hope i ll never need it again

  vsravani 20081101 10:51:15  why is minus sign taken in the equation for 40kg mass
20gd+20gq40gx=0?
elzoido238 20081106 11:37:44 
There is a negative sign for the torque due to the 40 kg mass because its contribution to the torque is in the opposite direction of the torques arising from the 20 kg mass and the mass of the rod.

  Andresito 20060330 13:01:46  @astro_allison, thank you for the alternative.   astro_allison 20051126 23:32:09  the term "10" doesn't make sense to me...
I solved it this way:
if is the distance from the center of mass to the fulcrum,
solving, you get
for some reason when I solved it with the term in your equation, I got
yosun 20051127 02:28:28 
astro_allison: thanks for the typoalert; it has been fixed. ( should be ... note that the first g is a q.) also, thanks for the alternate solution.

gotfork 20081017 11:31:32 
I'm a bit confused, where does the 20d come from?

gotfork 20081017 12:23:09 
Whoops, just realized the rod has a mass.

his dudeness 20100905 16:27:20 
Yeah I made the same mistake as gotfork... it's tough to get used to things like rods and ropes having masses after so many years of physics :) ... luckily the answer came out close enough to be right anyway(1.667~1.5)

 

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