Applications of Newton's laws - coefficient of friction

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Homework Help Overview

The discussion revolves around the application of Newton's laws in the context of friction, particularly focusing on the coefficient of friction for a skier moving up a hill. Participants explore the relationship between forces acting on the skier and the resulting acceleration, while also examining the nature of friction coefficients.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the calculation of acceleration using kinematic equations and the impact of gravity on the skier's motion. There is an emphasis on drawing free body diagrams to visualize forces. Questions arise regarding the expected range of friction coefficients and the implications of coefficients greater than 1, particularly in unusual cases like aluminum on aluminum.

Discussion Status

The discussion is active, with participants sharing calculations and diagrams. Some guidance has been provided regarding the nature of friction coefficients, and multiple interpretations of the results are being explored. There is no explicit consensus on the anomalies related to friction coefficients, but curiosity and inquiry are evident.

Contextual Notes

Participants mention constraints related to the expected values of coefficients of friction and the specific conditions of the problem, such as the skier's movement on an incline. There is also mention of varying sources for friction coefficients, raising questions about reliability and consistency in reported values.

Jujubee37
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Homework Statement
A skier traveling 12.9-m/s reaches the slope of a steady upward 14.9-degree incline and glides up 13.4-m up this slope before coming to rest. What was the average coefficient of friction between the skies and the slope?
Relevant Equations
fr = Fr/N is an equation to use but im not sure how that equation would work in this case. The answer should be a value between 0-1 but I keep getting something over that.
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Please show us the answer you got and the math that produced it. Then we will be able to diagnose if and where you went wrong.
 
0=12.9^2+2(a)(13.4) which got me -6.2 m/s/s
 
Jujubee37 said:
0=12.9^2+2(a)(13.4) which got me -6.2 m/s/s
That would be the acceleration if the skier were on a horizontal surface. Here the skier is going up a hill and gravity also has something to say about the acceleration. Draw a free body diagram of the skier.
 
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okay so I did the diagram and worked out the equation. 9.8sin(14.9)+u(9.8)cos(14.9)=6.2 which got me 0.388 which is the correct answer. thank you
 
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Jujubee37 said:
okay so I did the diagram and worked out the equation. 9.8sin(14.9)+u(9.8)cos(14.9)=6.2 which got me 0.388 which is the correct answer. thank you
You are welcome and also welcome to PF.
:welcome:
 
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Jujubee37 said:
The answer should be a value between 0-1
There's no upper limit to coefficients of friction, certainly nothing special about a value of 1.
But for a skier on snow, you would expect it to be rather lower.
 
haruspex said:
There's no upper limit to coefficients of friction, certainly nothing special about a value of 1.
But for a skier on snow, you would expect it to be rather lower.
That is my understanding too. Nevertheless, out of curiosity I visited several sites looking for coefficients of kinetic friction greater than 1. The only one I could find was aluminum on aluminum with μs = 1.05-1.35 and μk = 1.4. This is not a typographical error on my part. See e.g. here. I encountered the same numbers elsewhere which makes me wonder whether they copy from each other trusting that the initial report of these numbers is correct.

The question of why the coefficient of static friction is (generally) greater than the coefficient of kinetic friction has appeared on many PF threads of which I list only the first four:
https://www.physicsforums.com/threa...ways-smaller-than-the-static-friction.140426/
https://www.physicsforums.com/threads/kinetic-vs-static-friction-question.93180/
https://www.physicsforums.com/threads/static-friction-vs-dynamic-kinetic-friction.39630/
https://www.physicsforums.com/threads/kinetic-static-friction.102818/

Is aluminum a friction anomaly? I do not know. However, I do know that although aluminum skis exist, aluminum ski slopes do not.
 
kuruman said:
That is my understanding too. Nevertheless, out of curiosity I visited several sites looking for coefficients of kinetic friction greater than 1. The only one I could find was aluminum on aluminum with μs = 1.05-1.35 and μk = 1.4. This is not a typographical error on my part. See e.g. here. I encountered the same numbers elsewhere which makes me wonder whether they copy from each other trusting that the initial report of these numbers is correct.

The question of why the coefficient of static friction is (generally) greater than the coefficient of kinetic friction has appeared on many PF threads of which I list only the first four:
https://www.physicsforums.com/threa...ways-smaller-than-the-static-friction.140426/
https://www.physicsforums.com/threads/kinetic-vs-static-friction-question.93180/
https://www.physicsforums.com/threads/static-friction-vs-dynamic-kinetic-friction.39630/
https://www.physicsforums.com/threads/kinetic-static-friction.102818/

Is aluminum a friction anomaly? I do not know. However, I do know that although aluminum skis exist, aluminum ski slopes do not.
I see tables giving slightly different values, but still with that anomaly, e.g. https://engineeringlibrary.org/reference/coefficient-of-friction.
I cannot find any discussion of this, which in itself is strange.
I struggle to understand how such a result could be obtained. If ##N\mu_s## is just exceeded then it starts to slide, whereupon the frictional force becomes ##N\mu_k## and instantly stops it again. So did it slide or didn't it?
Maybe it creeps at a constant speed, with the frictional force rising from ##N\mu_s## to ##N\mu_k##, just matching the applied force?
 
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