Coefficient of friction force

In summary, the block experiences static friction in the negative direction of the normal force on the incline, due to the fact that there is no net force acting on the block.
  • #1
GingerBread27
108
0
The coefficients of friction between a 1kg mass and a surface are ms=.54 and mk=.46. Assuming the only other force acting on this block is that due to gravity, what is the force of friction (give magnitude and direction) on block:

-if the block is at rest and the surface is horizontal?
would this just be f=(ms)(N), and then N would equal (mass of block *gravity)?
 
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  • #2
I believe so, yes: [itex]F_{F}=\mu_{s}F_{N}[/itex], where [itex]\mu_{s}[/itex] is the coefficient of static friction.
 
  • #3
ok so the Force of friction would be F=(.54)(1kg)(9.8m/s^2)=5.3 N to the left?

also, if the block is at rest and the surface is inclined at 17.5 degrees, would the force then be f=(Ms)(mgcos(theta)), where theta equals 17.5?
 
  • #4
GingerBread27 said:
The coefficients of friction between a 1kg mass and a surface are ms=.54 and mk=.46. Assuming the only other force acting on this block is that due to gravity, what is the force of friction (give magnitude and direction) on block:

-if the block is at rest and the surface is horizontal?
would this just be f=(ms)(N), and then N would equal (mass of block *gravity)?

Is there any force on the block other than gravity ? If there isn't, your answer would NOT be correct. Where did you get "left" from ?
 
  • #5
Well we're told to give magnitude and direction, but since the block isn't moving I guess there is no direction, so all I would give is the magnitude.
 
  • #6
if the block is at rest and there are no forces other than gravity, then there would be no friction because its not acting against anything
 
  • #7
Always draw a free-body diagram for problems like this.

If there is any net (resultant) force on the body, it will accelerate in the direction of that force. For a body to be stationary, there must be NO net force.
 
  • #8
static friction

Sirus said:
I believe so, yes: [itex]F_{F}=\mu_{s}F_{N}[/itex], where [itex]\mu_{s}[/itex] is the coefficient of static friction.

The correct expression is
[itex]F_{static} \leq \mu_{s}F_{N}[/itex].
Interpretation: "The static fricition force is whatever it needs to be to prevent sliding between the surfaces... up to some maximum strength [itex]F_{static,max} = \mu_{s}F_{N}[/itex]."

Generally, [itex]F_{static} [/itex] is less than [itex]\mu_{s}F_{N}[/itex] unless "the object is just about to start sliding", when it is equal to [itex]\mu_{s}F_{N}[/itex].
 
  • #9
GingerBread27 said:
The coefficients of friction between a 1kg mass and a surface are ms=.54 and mk=.46. Assuming the only other force acting on this block is that due to gravity, what is the force of friction (give magnitude and direction) on block:

-if the block is at rest and the surface is horizontal?
would this just be f=(ms)(N), and then N would equal (mass of block *gravity)?


Static friction is a limit, it does not have a definite value: [tex]F_s\leq{\mu_sN}[/tex].
Its value is just enough to keep the body in rest till the resultant of the other forces would exceed [tex]\mu_sN[/tex].

If the body is in rest on a horizontal surface than there are two forces normal to the support, one is gravity and the other is the normal force. These two forces cancel each other. The friction would counteract against a horizontal pull or push, but there is no such force now. So the force of static friction is zero.

If the body is in rest on an incline the normal force is [tex]F_N=mg\cos(\phi ) \mbox{ and the force along the incline is } F_i = mg\sin(\phi)-F_s [/tex]
If the body doesn't move [tex] F_i=0 \mbox{ so } F_s=mg\sin(\phi)[/tex].
 

What is the coefficient of friction force?

The coefficient of friction force is a dimensionless quantity that measures the amount of friction between two surfaces. It is represented by the symbol μ (mu) and is determined by the nature of the surfaces in contact and the force pressing them together.

How is the coefficient of friction force calculated?

The coefficient of friction force is calculated by dividing the force of friction (F) by the normal force (N) between two surfaces. This can be expressed in the equation μ = F/N. The resulting value is a ratio of the two forces and has no units.

Why is the coefficient of friction force important?

The coefficient of friction force is important because it helps determine how much force is needed to move an object over a surface. It also plays a role in determining the amount of wear and tear on surfaces in contact and can impact the efficiency of machines and equipment.

What factors affect the coefficient of friction force?

The coefficient of friction force is affected by several factors, including the roughness or smoothness of the surfaces in contact, the materials of the surfaces, the amount of force pressing the surfaces together, and the presence of any lubricants or contaminants.

Can the coefficient of friction force change?

Yes, the coefficient of friction force can change depending on the factors mentioned above. For example, if a lubricant is introduced between two surfaces, it can reduce the coefficient of friction force and make it easier for the surfaces to slide over each other. Additionally, as surfaces wear or become smoother over time, the coefficient of friction force may decrease.

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