How does kinetic friction change around a curve

[kingkongsr.]

If I am running in a straight line, and I am experiencing kinetic friction that is proportional to my weight, i.e. my normal force, then when I run around a curve, does my normal force increase by my velocity squared divided by the radius of the turn as if it were a perfect circle ( {(v^2/r)+([weight of runner]*gravity)}*(kinetic friction coefficient) = kinetic friction)?

 

[Cyrus]

You experience static friction when you run. There is no relative motion between your shoes and the ground.

As for your main question, yes. The normal force, and thus the static or dynamic friction, increases due to the turn. The radius is the local radius of curvature at that instant along the curve. You can think of it as a perfect circle of radius, r, at that point along the curve at that instant. (Each point along the curve has a different radius for any general curve, if the curve is a circle the the radius is the same at all points along the circle)

 

[kingkongsr.]

thank you Cyrus.

thank you very much Cyrus. I just needed a second opinion other than my own