GREPhysics.NET: GR9277 #100

GR9277 #100

Problem

GREPhysics.NET Official Solution

\prob{100}
9277_100

A uniform rod of length 10 meters and mass 20 kilograms is balanced on a fulcrum with a 40-kg mass on one end of the rod and a 20-kg mass on the other end, as shown above. How far is the fulcrum located from the center of the rod?

0m
1m
1.25 m
1.5 m
2m

Mechanics $\Rightarrow$ }Sum of Moments

Take the sum of the moments (or torque) about the triangular pivot fulcrum and set it to 0.

$\sum M = 20gd +20gq - 40gx =0$ , where $d$ is the distance from the fulcrum to the 20kg mass, $x$ is the distance from the fulcrum to the 40kg mass and $q$ is the distance from the fulcrum to the center of mass of the rod.

From conservation of length, one also has $d+x=10$ and $q+x=5$ .

Plug everything into the moment equation. Shake and bake at 300 K. Solve for q to get choice (C).

Alternatively, one can solve this problem in one fell swoop. Taking $q$ as the distance from the fulcrum to the center of mass of the rod, one sums the moment about the fulcrum to get $\sum M = 20gq + 20g(5+q) - 40(5-q)g =0$ . Solve for $q$ to get choice (C). (This is due to the user astro_allison.)

Alternate Solutions

There are no Alternate Solutions for this problem. Be the first to post one!

Comments

Albert
2009-11-05 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 $20Kg$ and other side has $40kg$ . The rod weighs $20kg$ , so I divide $20kg$ in half and give both weights $10kg$ each. And now they weigh $30kg$ and $50kg$ .

Once reached so far, just put 'em up in ratio and get it,

$50/30 = x/(10-x)$

$x$ comes out $6.25$ , which is of course $1.25 meters$ further from the center.

The god-damned technique works every freakin' time!

Just make sure you set the ratio to "more mass: lesser length".

apr2010
2010-04-08 15:26:08

That is interesting that you do not know which has less length...Keeping the indexes right should work.

SurpriseAttachyon
2015-09-10 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 bar-mass portion changes but the weight-mass doesn\'t.\r\n\r\nYet when you plug in different answers, it\'s still correct. That\'s odd.

nikog
2015-10-20 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= $\\frac{F}{2}$ and F2= $\\frac{F}{2}$ symmetrical to the point p and you measure the torgue at x+p then\r\nτ= $\\frac{F}{2}$ ( l+x) - $\\frac{F}{2}$ (l-x) = Fx where l the distance from the point p

Reply to this comment

matonski
2009-03-24 01:57:48

I just put the origin at the center and found the center of mass. COM = $\frac{20 \text{kg} (-5\text{m}) + 40 \text{kg}(5\text{m})}{80\text{kg}$

kroner
2009-09-29 14:09:00

I think it's safe to say that matonski wins.

adjwilley
2009-11-04 10:40:16

Beautiful.

torturedbabycow
2010-04-09 10:14:17

Seconding the "beautiful." In fact, I think I might need a cigarette after reading it. ^__^

alemsalem
2010-09-26 09:32:41

fuck yeah :) im also gonna smoke now

neon37
2010-11-12 09:11:16

Damn!......Elegant!...

elenaPh
2013-04-15 11:40:16

amazing solution ! thanks!

aaa2
2014-09-25 12:06:00

wow just wow

justacomment
2014-10-07 23:19:21

Salute.

blabla0001
2017-03-25 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

Reply to this comment

vsravani
2008-11-01 10:51:15

why is minus sign taken in the equation for 40kg mass
20gd+20gq-40gx=0?

elzoido238
2008-11-06 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.

Reply to this comment

Andresito
2006-03-30 13:01:46

@astro_allison, thank you for the alternative.

Reply to this comment

astro_allison
2005-11-26 23:32:09

the term "10 $gg$ " doesn't make sense to me...

I solved it this way:

if $d$ is the distance from the center of mass to the fulcrum,

$20d + 20(5+d) = 40(5-d)$

solving, you get $d = \frac{5}{4} = 1.25$

for some reason when I solved it with the $10gg$ term in your equation, I got $distance to fulcrum from CM = \frac{10}{7}$

yosun
2005-11-27 02:28:28

astro_allison: thanks for the typo-alert; it has been fixed. ( $10gq$ should be $20gq$ ... note that the first g is a q.) also, thanks for the alternate solution.

gotfork
2008-10-17 11:31:32

I'm a bit confused, where does the 20d come from?

gotfork
2008-10-17 12:23:09

Whoops, just realized the rod has a mass.

his dudeness
2010-09-05 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)

Reply to this comment

Post A Comment!

You are replying to:

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 $20Kg$ and other side has $40kg$ . The rod weighs $20kg$ , so I divide $20kg$ in half and give both weights $10kg$ each. And now they weigh $30kg$ and $50kg$ . Once reached so far, just put 'em up in ratio and get it, $50/30 = x/(10-x)$ $x$ comes out $6.25$ , which is of course $1.25 meters$ further from the center. The god-damned technique works every freakin' time! Just make sure you set the ratio to "more mass: lesser length".

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