Calculating Weight at Earth's Equator: 600.0 N to 597.9 N

[poohead]

Homework Statement

Suppose the Earth is a perfect sphere with R=6370 km. if a person weighs exactly 600.0 N at the north pole how much will the person weigh at the equator? (hint: the upward push of the scale on the person is what the scale will read and is what we are calling the weight in this case) ans: 597.9 N

Homework Equations

Fg=Gm1m2/r^2
Fg=ma
Fc=Mv^2/r?

The Attempt at a Solution

I have no real clue on how to solve such a question, i plugged in for Fg but the answer i seem to get is higher than that of 597.9 N. Need help for a solution please!

 

[Dadface]

I will give you a hint...A person standing on the equator experiences a centripetal force towards the centre of the Earth .Work out what this is and draw a free body force diagram for the person.

 

[Doc Al]

Apply Newton's 2nd law. What's different about being on the equator compared to being on the north pole?

 

[chrisk]

Think about the angular velocity at the north pole and the equator. Then calculate the centripetal acceleration at the equator and use this value to find the effective acceleration of gravity at the equator.

 

[poohead]

how could i determine the angular velocity though? i am not given a time

 

[chrisk]

The time is 24 hours for one revolution! Check out this website for an explanation

http://www.ac.wwu.edu/~vawter/PhysicsNet/Topics/Gravity/AccOfGravity.html

 

[poohead]

so i should use Fc=4pie^2Mr/ T^2 to solve for Force centripetal

 

[chrisk]

Yes, use this equation but it's a little easier to find the effective acceleration of gravity at the equator, g - a_centripetal, then multiply by the mass.

 

[poohead]

okay so it would be easier to just get the centripetal acceleration, then take gravity- acceleration centripetal then multiply by mass for F=ma right? but how would i get the acceleration centripetal when Ac=V^2/r and i don't have V?

 

[chrisk]

V= Earth's angular velocity times the Earth's radius.

 

[poohead]

i am not familiar with Earth's angular velocity?

 

[gabbagabbahey]

i am not familiar with Earth's angular velocity?

Really? You don't know how long a day is? Or you don't know how to calculate angular velocity from knowledge of the period of rotation?

 

[poohead]

don't know how to calculate angular velocity from knowledge of the period of rotation? I've never heard of the term for one so maybe that's why I am lost with what your saying

 

[gabbagabbahey]

Angular frequency is what you really need for this problem; and it is defined by:

[tex]\omega\equiv2\pi f=\frac{2\pi}{T}=\frac{|v|}{r}[/tex]

Angular velocity is a vector quantity with magnitude [itex]\omega[/itex] and direct determined by a cross product.

 

[poohead]

okay thank you i have gotten the correct answer, much appreciated

 

[Dadface]

It takes 24 hours(one day)for one rotation of 360 degrees or 2pi radians.The centripetal force is m vsquared/r or mr omega squared,use the equation you are happiest with.At the poles the upward push of the scale (R) will equal the persons weight(600N).At the equator there must be a resultant force(the centripetal force)because the Earth is spinning and the person is accelerating towards its centre.It follows that R cannot be equal to the weight because the resultant force will then be zero and it further follows that weight minus R equals the centripetal force.From this you can calculate the recorded weight(R)