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Old Oct 15th 2012, 12:03 AM   #1
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atwood's machine

How to find out the upward force acting on the axle of a pulley in an atwood's machine with the masses in an acceleration?
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Old Oct 15th 2012, 02:35 AM   #2
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It should be the combined masses of the masses. Otherwise, the axle of the pulley would be falling down or going up, right? And the only forces that have the downward directions are caused by the masses.

[Assuming that the weight of the pulley itself is negligible]
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Old Oct 15th 2012, 10:03 AM   #3
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Atwood's Machine: a Solution

Hi,

Solution of "g" by use of Atwood's machine is not trivial. If there is an easier solution than mine - I would like to see it.

http://www.thermospokenhere.com/wp/0...od/atwood.html

Good Luck, Jim
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Old Oct 15th 2012, 10:57 AM   #4
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Hello Thermo. Your solution is a pretty complicated way of doing the math. Much simpler is to use Sum of F=ma:

(m+dm)g - mg = (2m + dm)a

which yields:

g = a(2m+dm)/dm

To the OP: the force exerted on the pulley is equal to twice the tension in the rope. If the two masses are m1 and m2, then the acceleration of the system is a=g(m2-m1)/(m2+m1). The tension in the rope is equal to the force required to accelerate the masses, which is the mass's weight plus or minus a force to accelerate them - be careful of signs!

Last edited by ChipB; Oct 15th 2012 at 01:38 PM.
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