#### 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 − mω^{2}r

True weight = mg

Therefore, we have :

\[\text { Fraction less than the true weight } = \frac{\text{mg - (mg - m }\omega^2 r)}{\text{mg}}\]

\[ = \frac{\omega^2 r}{g} = \left( \frac{2\pi}{24 \times 3600} \right)^2 \left( \frac{6 . 4 \times {10}^6}{10} \right)\]

\[= 3 . 5 \times {10}^{- 3}\]

Is there an error in this question or solution?

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.