# Magnitude of kinetic friction and static friction!

by matt@USA
Tags: friction, kinetic, magnitude, static
 P: 25 1. The problem statement, all variables and given/known data A 12g coin slides upward on a surface that is inclined at an angle of 14degrees above the horizontal. The coefficient of kinetic friction between the coin and the surface is 0.25; the coefficient of static friction is 0.40. Find the magnitude of the force of friction when the coin is sliding. Find the magnitude of the force of friction when the coin comes to rest. 2. Relevant equations I found the magnitude of friction when the coin was sliding using the equation N=mgcostheta, and then plugging N into fs=muk*N. Would the same equation not tell me my magnitude of friction after the coin has stopped? 3. The attempt at a solution
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 Quote by matt@USA 1. The problem statement, all variables and given/known data A 12g coin slides upward on a surface that is inclined at an angle of 14degrees above the horizontal. The coefficient of kinetic friction between the coin and the surface is 0.25; the coefficient of static friction is 0.40. Find the magnitude of the force of friction when the coin is sliding. Find the magnitude of the force of friction when the coin comes to rest. 2. Relevant equations I found the magnitude of friction when the coin was sliding using the equation N=mgcostheta, and then plugging N into fs=muk*N. Would the same equation not tell me my magnitude of friction after the coin has stopped?[/b]
No. Carefully examine the equation for the static friction force. Note, for example, what would be the static friction force for a block sitting at rest on a table with a coef of friction of 0.5 between block and table.
 P: 25 I meant to say the force of kinetic friction is muk*N. So with what you are essentially saying is that my force of static friction is mus*N, where N is equal to just m*g, and not m*gcostheta?
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## Magnitude of kinetic friction and static friction!

 Quote by matt@USA I meant to say the force of kinetic friction is muk*N.
yes, that is correct
 So with what you are essentially saying is that my force of static friction is mus*N, where N is equal to just m*g, and not m*gcostheta?
No, my response was probably not clear, your answer to the kinetic friction force is correct.
 Would the same equation not tell me my magnitude of friction after the coin has stopped?
, the answer is no. To help understand why that answer is 'no', consider a block resting on table with mus =0.5. The block is AT REST. The block weighs 50 N. The normal force on the block is 50 N. The static friction force acting on the block is NOT 25 N. What is the value of the static friction force in this case, using Newton 1 in the x direction?Now use Newton 1 in your actual problem to solve for the static friction force.
 P: 25 I'm not quite sure I understand what you are saying. You do realize this is on an inclined plane right? You are telling me in your example that the force N is equal to 50, but the force of static friction in not 25N. So is the equation of static friction not, fs=mu*N?
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 Quote by matt@USA I'm not quite sure I understand what you are saying. You do realize this is on an inclined plane right? You are telling me in your example that the force N is equal to 50, but the force of static friction in not 25N. So is the equation of static friction not, fs=mu*N?
That's right, the equation of static friction is NOT always fs = mus*N. It is fs = mus*N only when the object is right on the verge of moving. Otherwise, fs is ___??____ (please look it up using a web search), and its value must be determined using Newton's first law. Try using Newton's first law on the inclined plane when the coin is at rest to solve for the static friction force in that case.
 P: 25 It says if the net force of an object is = to 0, the velocity must be constant. I am starting to confuse myself. So according to this, fs=F, where F=ma. Can someone just tell me how to answer this.
 PF Patron HW Helper Sci Advisor P: 5,794 To avoid more confusion, the force of static friction is fs is less than or equal to usN. it is equal to us*N only when relative motion between the 2 surfaces is just on the verge of taking pace. Otherwise, it is less than that value, and you must use newton's first law to solve for it. back to your problem on the incline, when the coin is at rest, there are 2 forces acting on it in the direction parallel to the incline. Identify those forces, use Newton 1, and solve for the static friction force (you seem to have done that correctly in an earlier problem you posted about a backpack at rest ,on a surface with friction, being pulled by a spring ).
 P: 25 I hate to keep being a burden ... what I took out of that was that fs=mgsintheta?
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