Question

Explain how a triple-beam balance works. Would such a balance that functions properly on the earth...

Explain how a triple-beam balance works. Would such a balance that functions properly on the earth yield the correct mass of an object on the moon? Why or why not?

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Answer #1

It works by finding balance.

It has a plate on which objects are placed to determine the mass.

This object together with the plate exert a torque

\tau _{ob}=M\cdot g\cdot d_{obj}

where:

mass of the object

M
acceleration of gravity

g
distance from the fulcrum.

d_{obj}

each of the three arms corresponds to a scale
The idea is that the sum of the torques generated by each arm equals the torque generated by the unknown mass. In this way, knowing the distances and the masses, the value of M is obtained

each beam corresponds to a scale.
First scale of 0 to 10 grams   

\tau _{1}=m_{1}\cdot g\cdot d_{1}

where:
reference mass.

m_{1}
It is the position of the graduated scale, which indicates the equivalent mass (object) to be in balance

d_{1}
the acceleration of gravity

g


Second scale of 10 grams a100

\tau _{2}=m_{2}\cdot g\cdot d_{2}

This scale is equal position represents only the unknown object mass between 10 and 100 g
third scale of 100 to 500 grams

\tau _{3}=m_{3}\cdot g\cdot d_{3}

This scale is equal position represents only the unknown object mass between 100 and 10 g

when positioning the three masses so that this balanced with the object, measures the scales are added and the unknown mass is obtained

example
with the system in balance

Assuming each mass is located in a position that indicates the following

d_{1}\rightarrow 2g

d_{2}\rightarrow 34g

d_{3}\rightarrow 200g

Finally

M=d_{1}+d_{2}+d_{3}=2g+34g+200g=236g

physically it holds that

M\cdot g\cdot d_{obj}=m_{1}\cdot g\cdot d_{1}+m_{2}\cdot g\cdot d_{2}+m_{3}\cdot g\cdot d_{3}

simplifying g

M\cdot d_{obj}=m_{1}\cdot d_{1}+m_{2}\cdot d_{2}+m_{3}\cdot d_{3}

the equation that governs the machine does not depend on gravity but the ratio of torque produced by the masses, thus the measurement on the moon should be equal to the measured on earth

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