# How much will it weigh at the equator?

1. Feb 26, 2010

### huybinhs

1. The problem statement, all variables and given/known data

Take the earth to be a perfect sphere of 1.274×107 m. If an object has a weight of 96.9 N while on a scale at the south pole, how much will it weigh at the equator? take the equatorial spin speed to be 465 m/s.

2. Relevant equations

No ideas.

3. The attempt at a solution

2. Feb 26, 2010

### ideasrule

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.

3. Feb 26, 2010

### huybinhs

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?

4. Feb 26, 2010

### huybinhs

5. Feb 26, 2010

### ideasrule

No. You haven't drawn a free-body diagram. Do that, label all forces, and write out Newton's second law before doing anything else.

6. Feb 26, 2010

### huybinhs

If the person is standing at the equator, then force of gravity will be equal 0. Is this correct?

7. Feb 27, 2010

### huybinhs

any one?

8. Feb 27, 2010

### ideasrule

Do you really think there's no gravity for people at the equator? The people there don't just float around.

9. Feb 27, 2010

### huybinhs

You r right. It still has gravity even at the equator. What should I do next plz?

10. Feb 27, 2010

### ideasrule

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.

11. Feb 27, 2010

### huybinhs

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?

12. Feb 27, 2010

### ideasrule

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.

13. Feb 27, 2010

### huybinhs

Ok. So, 96.9 = mg => m = 96.9 / 9.81 = 9.88 kg.

Fnet = ma <=> 96.9 - mg = ma <=> 96.9 -(9.88) * g = 9.88 * 0.03394

=> g = 9.77 m/s^2 => F at equator = 96.5 N which less than F at the south poles. Is this correct?

14. Feb 27, 2010

### huybinhs

15. Mar 1, 2010

### mparsons06

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?

16. Mar 1, 2010

### huybinhs

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!

17. Mar 1, 2010

### mparsons06

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?!?!

18. Mar 1, 2010

### huybinhs

96.9 -(9.88)g = 9.88 * 0.03394 => -9.88g = 0.33353 - 96.9 => g = 96.56 / 9.88 = 9.77 m/s^2. Hope this helps! ;)

19. Mar 1, 2010

### mparsons06

I get it now. Thank you.

Do you know how to start out problem #2?

20. Mar 1, 2010