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Question
A person stands on a spring balance at the equator. By what fraction is the balance reading less than his true weight?
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Solution
Balance reading = Normal force on the balance by the Earth.
At equator, the normal force (N) on the spring balance :
N = mg − mω2r
True weight = mg
Therefore, we have :
\[\text { Fraction less than the true weight } = \frac{\text{mg - (mg - m }\omega^2 r)}{\text{mg}}\]
\[ = \frac{\omega^2 r}{g} = \left( \frac{2\pi}{24 \times 3600} \right)^2 \left( \frac{6 . 4 \times {10}^6}{10} \right)\]
\[= 3 . 5 \times {10}^{- 3}\]
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