An  equal arm balance is made by placing a meter stick at its center so it looks like a seesaw. The first mass m=3 kg is hung at r= 0.25 m To the left of the center. A second mass equaling 1.4 kg is hung at 0.4 m to the right of the center. At what location with respect to the center of the meter stick must a third mass of 0.8 kg be hung for the meter stick to be in equilibrium? to the left of the center. A second mass equaling 1.4 kg is hung at 0.4 m to the right of the center. At what location with respect to the center of the meter stick must a third mass of 0.8 kg be hung for the meter stick to be in equilibrium? (All angles are 90 degrees)

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Asked Nov 12, 2019
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An  equal arm balance is made by placing a meter stick at its center so it looks like a seesaw. The first mass m=3 kg is hung at r= 0.25 m To the left of the center. A second mass equaling 1.4 kg is hung at 0.4 m to the right of the center. At what location with respect to the center of the meter stick must a third mass of 0.8 kg be hung for the meter stick to be in equilibrium? to the left of the center. A second mass equaling 1.4 kg is hung at 0.4 m to the right of the center. At what location with respect to the center of the meter stick must a third mass of 0.8 kg be hung for the meter stick to be in equilibrium? (All angles are 90 degrees)

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Expert Answer

Step 1

To maintain the state of equilibirum,

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ти-т,у, - т,, -0

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Step 2

Re-arranging and substit...

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mz- mrz m2 (3kg) (0.25m)-(1.4kg) ( 0.4 m) 0.8kg =0.2375m

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