Force of Friction
I am having a little trouble understanding the force of friction. It is given by the formula
F_{f} = [itex]\mu[/itex]F_{N} Now taking a mass on an inclined plane with angle θ into consideration, I want to find out what the value for μ. I know what F_{N} is I can easily calculate that but I am having trouble understanding F_{f}. If the mass has a force acting on it which goes up the inclined plane (The mass is attached to a string which goes trough a pulley located at the top of the inclined plane. The direction of the force applied on the mass will be parallel to the inclined place going up the plane) assuming that the force applied is just enough to overcome the sum of static friction and the parallel component of gravity on the mass, how can I find F_{f} using the mass and the force acting on the mass? I am sorry if this is badly worded. please ask for clarification if needed 
Re: Force of Friction
If the forces are 'just enough' to start things moving, then you can set the sum of forces equal to zero.

Re: Force of Friction
The direction of the static frictional force is equal in magnitude and opposite to the direction to the net force from the other applied forces. Thus, if the weight hanging over the pulley is greater than the component of the mass weight along the plane, the tangential static frictional force will be in the direction down the incline. If the weight hanging over the pulley is less than the component of the mass weight along the plane, the tangential static frictional force will be in the direction up the incline. Its magnitude will be equal to the difference of the two forces. The mass will only begin to slide if the magnitude of the static frictional force calculated in this way is greater or equal to the normal force times the coefficient of friction.

Re: Force of Friction
@Chestermiller, yes I know all that. sadly I have two unknowns and only one equation. I want to find the coefficient of friction but all I have is the normal force, I don't know the total amount of the force of friction being applied.
@Doc Al, That will work but depending on the masses I accuracy of the masses I am using I will end up with a big error percentage. Is there any any way to reduce the error. I should have mentioned this in my first post but I am trying to do this experimentally. The smallest mass I will have access to is 1 gram. 
Re: Force of Friction
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Re: Force of Friction
Are you trying to find the coefficient of static or kinetic friction?

Re: Force of Friction
would a 30 gram mass significantly reduce the error? and I am trying to measure static friction

Re: Force of Friction
If you are trying to determine the coefficient of static friction, try a series of increasing masses on the wire over the pulley until the mass on the slide just starts to move. The coefficient of friction will be determined by the mass that just causes the body on the slide to start to move: μ = (m_{pulley}  m_{incline} sin θ )/m_{incline} cosθ

Re: Force of Friction
yes that's exactly what I am trying to do. but I didn't really know how to get either the coefficient or the force of friction. can you tell me how you got that formula? or its name so I can look it up. And what is m?

Re: Force of Friction
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Re: Force of Friction
The parameter m in the equation is mass. I got this equation by doing a free body diagram on the mass, and solving for the friction force F_{f} necessary to hold the mass in equilibrium, for various values of the mass on the pulley. Your trouble is that the static friction law isn't quite the way you stated it. It should really read:
μF_{n}≥ F_{f} where the greater than sign applies to the situation before the mass starts sliding, and the equal sign applies just at the point where the mass starts sliding. So if, for a given mass on the pulley, you want to find the coefficient of friction necessary to just prevent the mass on the incline from sliding, you do a force balance on the mass on the incline, and, for the frictional force, you substitute μF_{n}. You then solve for μ. 
Re: Force of Friction
I did my own calculations by making a free body diagram and I got the formula the same as yours. thanks for the help both of you.

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