Determining mass of balance bar

  • Thread starter izmeh
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izmeh

Here's the situation proposed in a problem.
I have a balance bar that is a meter stick.
It has a fulcrum at 50cms. Mass A(100gms[.98nt]) is at 12.3cms and mass B(1.02nt) is located at 86.1cms.
Mass B has been removed and the fulcrum C shifts fo the left by 14cms.
I need to determine the mass of the meter stick. How do i do this?
 

HallsofIvy

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The first part "It has a fulcrum at 50cms. Mass A(100gms[.98nt]) is at 12.3cms and mass B(1.02nt) is located at 86.1cms." just tells you that the bar itself has center of mass located at its center (since A is 50-12.3= 37.7 cm from the center and B is 86.1-50= 36.1 cm from the center- 37.7*.98= 36.1*1.02 (well, close enough for experimental error)).

After B is moved, the fulcrum shifts "14 cm to the left" so it is now at 36 cm. We still have mass A at 12.3 cm, 23.7 cm from the fulcrum. Its torque is .98*23.7= 23.226 (actually 0.23226 N-m). The center of mass of the bar is still at 50, 14 cm from the fulcrum so, taking the weight of the bar to be W, the torque around the fulcrum is
14W. To be balanced we must have 14W= 23.226 or W= 1.659 N. That means that the mass of the bar is 1.659/9.81= 0.169 kg or 169 grams.
 

izmeh

thank you
 

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