Gravitational Attraction at Equator

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At the equator, a person's weight measured by a spring balance is 725N, but this value differs from the true gravitational force due to the Earth's rotation. The centrifugal force caused by the Earth's rotation at the equator reduces the effective weight experienced. To calculate the difference, one must consider the additional centrifugal force acting on an object at the equator, which is absent at the poles. Understanding this force is crucial for determining the true gravitational attraction in that location. The discussion emphasizes the importance of recognizing the effects of Earth's rotation on weight measurements.
Felin
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The weight of a person at the equator, as determined by a spring balance, is 725N.

By how much does this differ from the true force of gravitational attraction at the same point? Assume that the Earth is spherically symmetric.

This question is on a section that we havn't even done yet so I am completely lost as to what to do to figure this one out.
 
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There is an additonal force at the equator that isn't present at the poles. Once you realize what that is, you can calculate its magnitude.

Dorothy
 
Thread 'Chain falling out of a horizontal tube onto a table'

Thread 'Chain falling out of a horizontal tube onto a table'

My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...
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