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

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The discussion centers on the force of friction acting on a stationary 1kg mass on a horizontal surface, with coefficients of static and kinetic friction given as ms=0.54 and mk=0.46. When at rest on a horizontal surface, the force of static friction is zero because there are no external horizontal forces acting on the block. The correct expression for static friction is that it can vary up to a maximum value, defined as F_static ≤ μ_sN, where N is the normal force. If the surface is inclined, the normal force changes, and static friction must counteract any component of gravitational force along the incline to prevent motion. Overall, static friction only acts when there are external forces attempting to move the block.
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: F_{F}=\mu_{s}F_{N}, where \mu_{s} 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: F_{F}=\mu_{s}F_{N}, where \mu_{s} is the coefficient of static friction.

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

Generally, F_{static} is less than \mu_{s}F_{N} unless "the object is just about to start sliding", when it is equal to \mu_{s}F_{N}.
 
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: F_s\leq{\mu_sN}.
Its value is just enough to keep the body in rest till the resultant of the other forces would exceed \mu_sN.

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 F_N=mg\cos(\phi ) \mbox{ and the force along the incline is } F_i = mg\sin(\phi)-F_s
If the body doesn't move F_i=0 \mbox{ so } F_s=mg\sin(\phi).
 
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