How Do You Calculate Static Friction on an Inclined Plane?

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Homework Help Overview

The discussion revolves around calculating static friction for a block on an inclined plane, specifically focusing on a scenario where a force is applied perpendicular to the surface. The problem involves determining the static force of friction and the coefficient of static friction for a block with a given mass and incline angle.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between normal force and static friction, with attempts to derive equations involving forces acting on the block. Questions arise regarding the correct interpretation of static friction and its coefficient.

Discussion Status

The discussion is active, with participants clarifying their understanding of the normal force and static friction. Some guidance has been offered regarding the relationship between static friction and the coefficient, though there is no explicit consensus on the final calculations or interpretations.

Contextual Notes

Participants note the importance of distinguishing between the normal force and the force unit of Newtons, and there is an ongoing exploration of the conditions under which static friction is calculated.

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Homework Statement



Suppose a block of mass = 10kg is on an incline plane that makes an angle = 45° from the horizontal. If a force of 10 N is being applied to the block, downward, and perpendicular to the surface, and the block is not moving, what is the static force of friction? What is the coefficient of static friction?



Homework Equations



\mu_{}s = Fn

The Attempt at a Solution




\mu_{}s = FN

Am I on the right track?
 
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laladude said:

Homework Statement



Suppose a block of mass = 10kg is on an incline plane that makes an angle = 45° from the horizontal. If a force of 10 N is being applied to the block, downward, and perpendicular to the surface, and the block is not moving, what is the static force of friction? What is the coefficient of static friction?

3.jpg


Homework Equations



\mu_{}s = Fn

The Attempt at a Solution



ƩFy = ma
ƩFy = N + -mg + - 10N = 0
N = (10N)mg * cos45°
should be N -mgcos45 -10 = 0, N = mgcos45 +10. Don't confuse the normal force, N, with the force unit of Newtons (N)!
friction = mg * sinθ
f = 69.3 N
This is the correct answer for the first part.
\mu_{}s = FN

Am I on the right track?
you mean F_f is less than or equal to \mu_sN, don't you?
 
PhanthomJay said:
should be N -mgcos45 -10 = 0, N = mgcos45 +10. Don't confuse the normal force, N, with the force unit of Newtons (N)!This is the correct answer for the first part.you mean F_f is less than or equal to \mu_sN, don't you?

LOL Yes, I did mean that. So it will be \mu_{}s = Ff / N?

Which will be 69.3N/79.3N..correct?
 
laladude said:
LOL Yes, I did mean that. So it will be \mu_{}s = Ff / N?

Which will be 69.3N/79.3N..correct?
No-o. Since static friction force is always less than or equal to \mu N , then \mu_{}s is greater than or equal to[/color] Ff / N.
 
PhanthomJay said:
No-o. Since static friction force is always less than or equal to \mu N , then \mu_{}s is greater than or equal to[/color] Ff / N.

Oh, okay! Thank you so much :biggrin:
 
You're welcome. Your calculation for the Normal force is also correct. Note that only if the block was at rest but just on the verge of moving would the static friction coefficient be equal to Ff/N. Otherwise, it would be greater than Ff/N.
 

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