Kinetic friction.

  • Thread starter Bob Loblaw
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  • #1
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Homework Statement



An 60-kg skier is sliding down a ski slope at a constant velocity. The slope makes an angle of 11° above the horizontal direction.
(a) Neglecting any air resistance, what is the force of kinetic friction acting on the skier?


(b) What is the coefficient of kinetic friction between the skis and the snow?

Homework Equations



Fr = μ*Nf

The Attempt at a Solution



I was able to calculate Nf from the information given as mg*cos(24) or 537.16N. Without knowing the force I was unable to answer the problem. Any ideas where to begin? Thanks!
 
Last edited:

Answers and Replies

  • #2
bel
157
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If he is sliding down with constant velocity, he is not accelerating as he would be if there were no friction, hence the force of friction is equal in magnitude and opposite in direction to the component of gravitational force parallel to the incline.
 
  • #3
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So I took mg*sin(11) which gave me 112.31 which is correct.

Now as I need to find the coefficient of kinetic friction between the skis and the snow. Using my equation, I tried Fr = μ*Nf where Fr is 112.31 and Nf was 537.16. I arrived at .021 which was not correct. Where did I go wrong?
 
  • #4
bel
157
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Calculate the magnitude of your normal force again, I got a different answer.
 
  • #5
69
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Nomal force should be mg*cos(11) in this case. which would be 577.8. I tried using that in my equation and came up with .19 as the kinetic coefficient for part B which is still not the correct answer. Am I using the right equation?
 
  • #6
bel
157
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I got 60*9.81*cos(11)= 577.79 N, keep a few more digits in calculation, by the way, and round at the end, especially since you are looking at small numerical answers.
 
  • #7
69
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0.1943 was accepted as an answer in web assign. Thanks for your help!
 

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