How Does Friction Affect a Stationary 1kg Mass on a Horizontal Surface?

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Homework Help Overview

The discussion revolves around the effects of friction on a stationary 1kg mass placed on a horizontal surface, with specific coefficients of static and kinetic friction provided. Participants explore the calculation of the force of friction under different conditions, including when the surface is horizontal and inclined.

Discussion Character

  • Exploratory, Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of the static friction formula and the normal force, questioning whether friction exists when the block is at rest and if any other forces are acting on it. Some explore the implications of an inclined surface on the frictional force.

Discussion Status

There is an active exploration of the concepts of static friction and the conditions under which it applies. Participants have raised questions about the direction of friction and the necessity of a net force for the block to remain stationary. Guidance has been offered regarding the interpretation of static friction as a variable force that adjusts to prevent motion.

Contextual Notes

Participants note the importance of considering all forces acting on the block, including gravity and the normal force, and question the assumptions made about the presence of external forces. The discussion reflects a range of interpretations regarding the nature of static friction and its role in maintaining equilibrium.

GingerBread27
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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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I believe so, yes: [itex]F_{F}=\mu_{s}F_{N}[/itex], where [itex]\mu_{s}[/itex] is the coefficient of static friction.
 
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?
 
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 ?
 
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.
 
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
 
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.
 
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].
 
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].
 

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