you are pulling a 10 pound kayak over ice. you are exerting 2 pounds of force on the rope that you are pulling connected to the front of the kayak, the angle of the rope to the horizontal is 25 degrees... how would you determine the coefficient of friction here and what is it?
Best Answer to first correct response
2007-11-12 15:07:09 · 2 answers · asked by Anonymous in Science & Mathematics ➔ Physics
the kinetic friction that is
2007-11-12 15:13:17 · update #1
Consider the horizontal component of motion. You're providing a pulling force of 2 lbs * cos 25. The resisting force is friction, which is equal to (coefficient of friction) (weight).
So (cof) (weight) = 2 cos 25
cof = 2 cos 25 / (weight)
cof = (2) (.91) / (10)
cof = .18
which is about right for rough ice
2007-11-12 15:15:46 · answer #1 · answered by jgoulden 7 · 0⤊ 0⤋
F = nR
F is the force applied to slide the kayak over ice,n is the coefficient of sliding friction and R is the normal reaction.
R = mg
F= Force applied at angle x cos 25(horizontal component)
n = F/R
= 2 pounds x cos 25/1o pounds x 9.8
= approximately .015
2007-11-12 15:27:43 · answer #2 · answered by Anonymous · 0⤊ 0⤋