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flutieflakes

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- Thread starter flutieflakes
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flutieflakes

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Integral

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If you have a question, post it. If not get to work on your lab.

- #3

flutieflakes

M1 M2

| |

22cm| 78cm|

----------------------------------

6.7cm| 68cm|

| |

70g 260g

The horizontal line is a 100cm yard stick weighing 120g

The verticle lines are strings attached to weights (the ones going up are attached to pulleys)

The Lengths next to the strings are the distance from the left end of the stick to the strings.

What would I do to determine the weight of M1 and M2?

The apparatus is hanging in equilibrium.

Thanks

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flutieflakes

- #5

HallsofIvy

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What you need to do here is multiply each weight time the distance of that weight from the left end (just since that is the info you are given- since this object ISN'T "twisting", it is "balanced" about any point). Be sure to take those weights hanging down to be negative and those weight over the pulleys to be positive (their pull is upward). Use "M1" and "M2" for the unknown weights. Add all of those to get the total "torque". Since the object is in equilibrium, it is not twisting and that torque must be equal to 0. That gives you one equation. Because the object is also not moving directly up or down, the net force must also be 0: ignoring the distances, add all of the weights (again, positive for those going over the pulley, negative for those hanging down) and set that equal to 0.

You now have two equations to solve for M1 and M2.

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