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

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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".

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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.

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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.

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Andresito
2006-03-30 13:01:46

@astro_allison, thank you for the alternative.

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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.

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Post A Comment!

You are replying to:

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 $distance to fulcrum from CM = \frac{10}{7}$

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