Conceptual Question about Static Friction

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The discussion centers on understanding why the coefficient of static friction is independent of the weight of a block on an incline. The original poster suggests that the coefficient, represented as height over length, does not change with varying weight. However, a response clarifies that this representation is not the correct definition of the coefficient of static friction. The coefficient is defined based on the ratio of the maximum static friction force to the normal force, which is indeed influenced by weight. The conversation highlights the need for a proper understanding of the definitions and relationships involved in static friction.
TyroneTheDino
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Homework Statement


Today we were studying a block on an incline and determining the angle needed for the block to start sliding. The question asked is: Why is the equation for the coefficient of static friction independent of the weight of the block?

Homework Equations


ΣFx=Fapplied-fs=0
ΣFy=Fn-mgcosθ=0
μs=height/length

The Attempt at a Solution


My reasoning is that because the coefficient of static friction is reliant by height over length, weight has no involvement in this matter. If you double the weight of an object on an incline, the coefficient of static friction would still be proportional to coefficient if you were to leave the weight the same. I feel like I am on the right track, I just feel like my reasoning is missing proof.
 
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TyroneTheDino said:
Why is the equation for the coefficient of static friction independent of the weight of the block?
That's strange wording. I would have expected a question like ""why is the coefficient of static friction independent of the weight of the block?"
Is it referring to the equation you quote, μs=height/length? That is not the definition of the coefficient of friction. (It's none of the equations you listed.) To answer the question properly, you need to reference the definition.
 

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