Jun 19, 2013
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A 0.5 kg uniform meter stick is suspended by a single string at the 30 cm mark. A 0.2 kg mass hangs at the 80 cm mark. What mass hung at the 10 cm mark will produce equilibrium?

I don't understand how to do it. If someone could please explain this to me in a simple yet comprehensive way I would really appreciate it. Thanks a bunch to all you Physics experts!
 
May 10, 2013
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Ok... I have not done this in a really, really long time... but isn't this a rotational equilibrium problem? Aka balance all the torques?

-You have to pick a pivot point, I guess use the 30 cm mark because they don't tell you what the tension on that string is.
-The meter stick's mass is uniformly distributed, so you can draw the vector for force of gravity at the halfway point (50 cm). So you can make that torque 1 = (20cm from pivot)*(0.5 kg) * (9.8).
-you can have the 0.2 kg mass at 80 cm be torque 2, so torque 2 = (50 cm from pivot)(0.2)(9.8)
-you can then have the unknown mass (let's call it 'x') be torque 3, so torque 3= (20 cm from pivot) (x) (9.8)

-torques 1 and 2 are on one side of the pivot, and torque 3 is on the opposite side.
So you can set up: torque 1 + torque 2 = torque 3 , and then solve for x.

Again, it's been a while since I've done this kind of stuff but I hope that helps a bit.
 
May 10, 2013
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I made 30 cm the pivot (where the string holding everything up is attached).

EDIT: I just learned that there's an MCAT Q&A section where students post questions/problems. You'll probably have more success posting any future questions there.
 
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gettheleadout

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Moving to MCAT Q&A.
 

NervousNerd

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I briefly skimmed the long answer posted by aquivoy (skipped his math) but he answered the concept right.

I would approach as a static equilibrium problem. For an object or system to be in static equilibrium, the net force must be 0 and the net torques must be 0.

F=0
T=0

To make a problem like such easier, set your pivot to a point which could eliminate one of the torques (where a mass is hanging)

Sorry for not solving it out but if this reiterates aquivoys post above, I'm sure his math is sound.