Two Ice Skaters Force On Each Other

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SUMMARY

The discussion centers on calculating the force exerted between two ice skaters of equal mass (56.0 kg) who are spinning in a circle with a period of 2.4 seconds. The initial approach using the gravitational force equation F = G((M1)(M2))/radius² was incorrect, as it pertains to gravitational attraction rather than centripetal force. The correct method involves using Newton's second law, F = ma, where 'a' is the centripetal acceleration derived from the skaters' circular motion. The final force calculation requires determining the radius of their circular path, which is 0.75 m.

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thulling
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1. On an ice rink, two skaters of equal mass grab hands and spin in a mutual circle once every 2.4 s.
If we assume their arms are each 0.75 m long and their individual masses are 56.0 kg, how hard are they pulling on one another? (answer is F= __N)



2. I thought I should use the equation F= G((M one)(M two))/radius squared
where G=6.67*10^-11 N*(m/kg)squared

3. I attempted to plug in the numbers to this equation and got F=3.7*10^-7
This answer was incorrect.

I know I am given time=2.4s, but I am not sure where to use this. ?

Please help
 
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That equation you used was for the force of gravity between the skaters. It's not significant here. Try to use F=ma
 
I got it! Thanks so much!
 
Last edited:

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