(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

I am running to school at 6m/s when I hit a patch of icy sidealk. The coefficient of friction between the ice and my shoes is 0.16. I wish to turn my 55kg frame to the left. What is the smallest turn I can make?

v = 6 m/s

μ = 0.16

m = 55 kg

2. Relevant equations

Fnet = mv^2/r

Ff = μFn

Fnet = ma

3. The attempt at a solution

Assume that up and left are positive.

Fnet = ma

Fg + Fn = ma

-539 + Fn = 0

Fn = 539

Ff = μFn

Ff = 0.16 (539)

Ff = 86.24

Fnet = mv^2/r

Ff = mv^2/r

86.24 = 55(6)^2/r

86.24 = 1980/r

r = 23 m

I'm not sure if that's the answer or what to do next, or even if what I did is the right method. The question is kind of confusing me because I don't fully understand how the person is moving.

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Centripetal Force/Changing direction on ice - Turn length

**Physics Forums | Science Articles, Homework Help, Discussion**