Static friction vs. Normal force on an incline

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Discussion Overview

The discussion revolves around the relationship between static friction and normal force on an incline, particularly focusing on determining the angle at which static friction equals the normal force and the conditions under which wood chips will slide down a steel slide. The scope includes theoretical reasoning and mathematical modeling related to frictional forces and inclined planes.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant questions the possibility of equal static friction and normal force without considering mass, suggesting that they can only be equal if the coefficient of friction is one.
  • Another participant describes a scenario involving wood chips of varying mass and seeks to find the minimum angle for the normal force to overcome static friction.
  • There is a discussion about the forces acting on the chips, with emphasis on the role of gravity and the need to resolve forces into components.
  • Participants explore the relationship between the angle of the incline and the forces involved, leading to the formulation of equations involving sine and cosine functions.
  • One participant confirms that the normal force can be expressed as N = mgcos(θ), while another derives the condition for sliding based on the balance of forces.
  • There is a calculation of the angle required for the chips to slide, resulting in a value of approximately 30.96 degrees, which is noted to be independent of mass.

Areas of Agreement / Disagreement

Participants express differing views on the relationship between static friction and normal force, particularly regarding the role of mass. While some agree on the mathematical relationships derived, there is no consensus on the initial question of equal forces without considering mass.

Contextual Notes

Limitations include assumptions about the coefficient of friction and the specific conditions under which the forces are analyzed. The discussion does not resolve the initial question of equal static friction and normal force without mass considerations.

Who May Find This Useful

This discussion may be useful for individuals interested in physics, particularly those studying mechanics, friction, and inclined planes, as well as those looking to refresh their understanding of these concepts.

Alex75
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I have two surfaces that have a coifficent of friction of .6. Disregarding mass (if possible) at what angle would the force of static friction and the normal force be equal
 
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Alex75 said:
Disregarding mass (if possible) at what angle would the force of static friction and the normal force be equal

Not possible. The normal force depends on the mass. Since the frictional force is equal to the coefficient of friction times the normal force they can never be equal unless the coefficient of friction is one and that would be true of any angle.
 
This problem relates to woods chips sliding down a steel slide they range in mass from one gram to 500 grams what would the minimum angle be so that the normal force applied to every chip would overcome the force of static friction on each individual chip
 
I see the situation. The problem is asking at what angle will the chips begin to slide over one another. and that will occur when the force that is causing them to slide is equal to the frictional force that is preventing them from sliding. The normal force is not the force which will overcome the friction. The normal force is that which determines the frictional force. What force is acting such as to cause the chips down the slide?
 
The acceration of gravity acting soon an incline
 
*upon
 
And what would that be in the case of a slide at an angle Θ?
 
Mgsin°=n
 
Where m is mass
G is acceleration of gravity
° is the angle of the steel slide
N is normal force applied to object
 
  • #10
according to that if the slide is at 0 deg then the normal force is 0? Is that OK with you?
 
  • #11
No I am trying to determine the minimum angle the slide needs to be placed at so the chips accelerate down the slide. I.e. when the force from gravity is enough to overcome the static friction of the chips
 
  • #12
I was questioning your formula for the normal force.
 
  • #13
MgsinΘ is not the normal force so what is the normal force?
 
  • #14
That was the equation I found when reshuching frictional forces
 
  • #15
May I ask your physics background?
 
  • #16
Ff=un
Ff is frictional force
U is the coifficent of friction
N is normal force
 
  • #17
None really outside college physics when I was a senior I have always had a love for science and physics though which is way I am attempting to educate myself on what I still need to learn and refresh what I already know
 
  • #18
Do you remember the idea of the resolving (splitting) of a force ( i.e., a vector) into components? That is such that if you add the components you get the original force (vector) back>
 
  • #19
Vaguely
 
  • #20
I believe that was what we ended up doing in physics class to determine at what angle a car's tires would loose friction between the road surface
 
  • #21
similarly here. The problem here is to find the angle of the slide and of the stack of chips such that the chips will just begin to slide over one another . The weight of the chips i.e. the force of gravity must be split into two components .1, one normal to the slide or chip surface pushing the chips together, and 2 one that is moving the chips down the slide. MgsinΘ is that component of the force of gravity that is acting to force the chips down the slide. Can you find or write down the normal force in terms of the mass and angle?
 
  • #22
Wouldn't the normal force be N=mgcos°
 
  • #23
Correct. now you want the frictional force to be just overcome and that will occur when the frictional force just equals the force moving the chips down the slide.
 
  • #24
So mgsin°=umgcos° would be the equation I would have to evaluate and isolate °?
 
  • #25
correct
 
  • #26
OK I got that far on my own here's where the five years since when I last took physics becomes noticeable how do you isolate the angle from both triginomic functions
 
  • #27
μ = sinΘ/cosΘ = tanΘ
 
  • #28
So then °=tan^-1(u) ?
 
  • #29
Yes and can you find it?
 
  • #30
30.96°
 

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