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A Person Stands on a Spring Balance at the Equator. by What Fraction is the Balance Reading Less than His True Weight? - Physics

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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?

Solution

Balance reading = Normal force on the balance by the Earth.
At equator, the normal force (N) on the spring balance : 
N = mg − mω2

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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Solution A Person Stands on a Spring Balance at the Equator. by What Fraction is the Balance Reading Less than His True Weight? Concept: Circular Motion.
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