The problem involves calculating the weight of an object at the equator, given its weight at the south pole and the equatorial spin speed of the Earth. The context is centered around gravitational forces and centripetal acceleration due to Earth's rotation.
Several participants have provided guidance on necessary steps, such as drawing diagrams and applying relevant equations. There is ongoing exploration of the relationship between gravitational force and apparent weight at the equator, with some participants questioning their calculations and assumptions.
Participants are working under the constraints of a homework assignment, which may limit the information they can use or the methods they can apply. There is some confusion regarding the application of gravitational acceleration and its effect on weight at different locations.
ideasrule said:Start by drawing a free-body diagram of a person standing at the equator. Remember that part of gravity goes into providing the centripetal force.
huybinhs said:Ok. I got:
a = v^2 / r = 465^2 / (6378100) = 0.0321 m/s^2 (I took 6378.1 km as radious. Is this still correct?)
If yes, then:
m = F / a = 96.9 / 0.0321 = 3020 kg. Correct?
huybinhs said:If the person is standing at the equator, then force of gravity will be equal 0. Is this correct?
ideasrule said:Do you really think there's no gravity for people at the equator? The people there don't just float around.
ideasrule said:Again, draw a free-body diagram, label all the forces, and write down Newton's second law. If you've done any of these steps, you haven't shown them here.
huybinhs said:There will be 2 forces in the diagram: gravity and the normal force. The normal force represents the apparent weight of the of the object = 96.9 N.
the acceleration of the object = v2 / r = 4652 / (1.274×10^7 /2) = 0.03394 m/s^2.
Then I got equation forces:
96.9 - m (0.03394) => m = 2855 kg. Then my normal force = 2855 * 9.780 = 27922 N = final answer. (Note that 9.780 is gravity at equator). Is this correct?
ideasrule said:The normal force is the apparent weight of the object, but it isn't 96.9N. That's the normal force at the pole, which is just equal to mg since Earth's rotation has no effect there. From this you can find m, and from there the normal force at the equator (which is what the question asks for).
Seems right.
Where's your Newton's second law? Newton's second law says Fnet=ma. Fnet is composed of two forces: the normal force and gravity. a is just the acceleration that you calculated earlier.
mparsons06 said:How would the equation be set up to get g = 9.77 m/s^2?
I have different numbers, but I was just curious how the equation would look on the other side of g.
Then where did you plug g back into?
huybinhs said:After I found out my g is 9.77 m/s^2. I use the equation F at equator = m*a = 9.77 * 9.88 = 96.5 N . Hope this helps!
mparsons06 said:How did you get the 9.77?
Is it like:
96.9 - (9.88) * g = (9.88)(0.03394)
g = (9.88)(0.03394)/96.9 - 9.88
I'm so confused??
mparsons06 said:I get it now. Thank you.
Do you know how to start out problem #2?