Maximal static friction problem

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Homework Help Overview

The discussion revolves around the concept of static friction in mechanics, particularly in the context of an object on a sloped surface. The original poster is exploring the relationship between the minimum force required to move a stationary object and the maximum static friction force.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to understand whether the force needed to keep an object stationary on a slope should be greater than the maximum static friction force. They question the interpretation of static friction as a breaking point versus a threshold for movement.

Discussion Status

Participants are engaged in clarifying the definitions and implications of static friction. Some suggest that the original poster's intuition about the relationship between applied force and static friction may be correct, while others note discrepancies in the wording of the problem statement. There is an ongoing exploration of the conditions under which an object remains stationary.

Contextual Notes

Participants are discussing the implications of textbook definitions versus intuitive understanding of static friction, particularly in relation to the forces acting on an object on a slope. There is mention of potential confusion arising from the phrasing of the problem statement regarding static and kinetic friction.

Black Riven
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This is mostly a theoretical question, I'm studying mechanics by myself so I have no teacher to ask this.

Homework Statement



The minimum force needed to move a stationary object is equal to the max static friction. However, I just encountered a question that included a sloped surface and an object moving upwards across it.

Eventually, mgcos(a)+Fk bring it to a halt, and the question is what should the minimal static friction constant be in order for the object to remain stationary.

The Attempt at a Solution



Fs must be strong enough to hold out against mgcos(a). The equation we are supposed to have is Fs-mgcos(a)=0
However, this confuses me. If when a force applied on the object equals to Fs(max) it causes it to move, Shouldn't the force needed to keep the object from moving be Fs>mgcos(a)?

In other words, is Fs(max) the breaking point or the last point before movement occurs?
 
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It sounds like the question should read "minimal static friction constant"
 
Black Riven said:
The minimum force needed to move a stationary object is equal to the max static friction.
Assuming a level surface, why do you say "equal to"? If an object is at rest and the max static friction force available is 20N and you push on it with a 20N force, will it move?
Eventually, mgcos(a)+Fk bring it to a halt, and the question is what should the minimal kinetic friction constant be in order for the object to remain stationary.
As noted above, change the word 'kinetic' to 'static'.
If when a force applied on the object equals to Fs(max) it causes it to move,
does it?
Shouldn't the force needed to keep the object from moving be Fs>mgcos(a)?
What does Newton's first law tell you?
 
PhanthomJay said:
Assuming a level surface, why do you say "equal to"? If an object is at rest and the max static friction force available is 20N and you push on it with a 20N force, will it move?
This is the problem then. My intuition tells me that no, it will not move because the force must be bigger than Fs(max). However, the book I'm using says
The minimal force which causes relative movement between two bodies that share contact equals to the maximal static friction
Based on your answer I assume my intuition is correct? If Fs(max)=F the object will not move, and the minute F>Fs(max) it will start moving.
If that's the case any idea why the book says what it does?
PhanthomJay said:
As noted above, change the word 'kinetic' to 'static'.
Done.
 
I think it's just saying that if the max static friction force is 20.000000000N, a force of 20.000000001N will, in theory, cause it to move. Close enough to 20, I guess.
 

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