Why Are Coefficients of Friction Dimensionless?

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Coefficients of friction are dimensionless because they represent a ratio of the frictional force to the normal force acting between two surfaces. The equation F = μ * Fn shows that when calculating μ, the units of force (Newtons) cancel out, resulting in a dimensionless value. This means that μ is simply a comparative measure of how much frictional force exists relative to the normal force. The discussion confirms that understanding the relationship between these forces clarifies why coefficients of friction do not have dimensions. Thus, the dimensionless nature of friction coefficients is rooted in their definition as ratios of forces.
k-rod AP 2010
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Homework Statement



Why are both types of friction dimensionless?


Homework Equations





The Attempt at a Solution



i said it was because the coefficients of friction are just ratios relating the normal force/frictional force of specific types of surfaces sliding past each other.

F=μ*Fn so μ=Fn in N/F in N so the Newtons cancel and leaves a dimensionless number

and the Frictional force F, is the force applied to resist motion, gained from μ*{Fn} where normal force is (mass*gravity)

would this be correct?
 
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yes, \mu is dimensionless since its a ratio of two forces
 
ok that's what I thought thanks for clearing that up
 

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Thread 'A cylinder connected to a hanging mass'

Thread 'A cylinder connected to a hanging mass'

Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
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