Coefficient of Friction of Fixed Atwood Pulley

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SUMMARY

The discussion focuses on calculating the coefficient of friction in a fixed Atwood pulley system with two free-hanging masses. The equations presented include Fnet = Ft - Fg - μFn for mass m1 and Fnet = Fg - Ft - μFn for mass m2, leading to the derived formula μ = (Ft - mg - m1a) / mg. It is clarified that if the pulley is massless, the coefficient of friction cannot be determined, as even minimal static friction prevents slippage. The conversation also highlights that if the pulley has mass, a minimum static friction coefficient can be calculated to prevent slippage.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with the concepts of force, tension, and friction
  • Knowledge of basic pulley mechanics
  • Ability to manipulate algebraic equations
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  • Study the dynamics of Atwood machines with massless and massive pulleys
  • Learn about the moment of inertia and its impact on pulley systems
  • Explore static and kinetic friction coefficients in mechanical systems
  • Investigate scenarios involving slipping ropes on pulleys
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Students studying physics, particularly those focusing on mechanics, engineers designing pulley systems, and educators teaching concepts of friction and motion.

DrewCrew
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Homework Statement


Hi, I was wondering how you would isolate for and calculate the coefficient of friction in an Atwood pulley, with one fixed pulley, with two masses hanging off of it on either side. There is no table or surface that the masses are touching or resting on, they are both free-hanging on either side of the pulley. There is no person applying a force to it, the pulley is free-standing. No values are given, I am just trying to make a general equation for the coefficient of friction. A general diagram of the situation is:
Atwood%20Problem%20Setup1%20FBD.png


Homework Equations


I know that, for m1, Fnet = Force of tension - Force of Gravity - Force of Friction, if it were to set "up" as the positive direction
And for m2, Fnet = Force of Gravity - Force of tension - force of friction
Force of friction = μ X Force normal
Feel free to correct me if I'm wrong for any of these equations

The Attempt at a Solution


Fnet1 = Ft - Fg - μFn
m1a = Ft - mg - μFn
In this case, would Fn be equivalent to Fg?
m1a = Ft - mg - μmg
μmg = Ft - mg - m1a
μ = (Ft - mg - m1a) / mg

I am attempting this entirely off of assumptions, as I am not aware of the actual friction forces or formulas on the pulley. Thanks for the help, it is greatly appreciated!
 
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What friction are you considering?
Most problems involving pulleys tell you to assume the pulley is massless and "is frictionless". By that, they mean it has no frictional torque from its axle. Generally you should assume there is enough friction between the pulley and rope that the rope does not slip on the pulley, but if the pulley is massless then even the tiniest static friction will achieve that, so there is no way to determine what the coefficient is.

It gets a bit more interesting if the pulley has mass, and hence a moment of inertia. In every such problem I have ever seen posed, you are still to assume the rope does not slip on the pulley. If the pulley is to accelerate then there has to be enough friction to ensure there is no slippage, so you could calculate a minimum value for the static friction coefficient.

A problem could consider the case where the pulley has mass and the static friction is not enough to prevent slipping, but I have never seen that done.
 

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