Does Static Friction Depend on Contact Area and Force Application Angle?

[jamesadrian]

I have some questions I have labeled below:

I believe that the coefficient of static friction, (let's call it Csf), equals the maximum force of static friction (before it slips), Fsmax, divided by the force bringing the two surfaces together (a force in the direction of the normal to their surfaces), Fn. In other words.

Csf = Fsmax/Fn

If the normal force, Fn, is constant (and we know that the coefficient, Csf, is constant), the static frictional force would seem to be independent of the contact area between the two objects.Q 1. Is this correct so far?Now please consider a round cylinder, C, somehow rigidly fixed in space (perhaps clamped to something at one end), and a flat band (like a belt) wrapped part way around C. A downward force, Fw, is provided by a weight on one side of C (let's say the right side as we look at the end of C) and the band leaves contact with C at zero degrees (the horizontal). The band is pressed against C with a holding force, Fh1, at a point. This point might be 20 degrees counterclockwise from zero. There is traction between the stationary cylinder and the band.Q 2. Is the static frictional force constant in this setting also?I believe this to be true because while the diameter of cylinder C may affect the amount of contact area between the S and C, it seems to me that the diameter does not affect any of the forces.Q 3. Is the holding force, Fh1, required to maintain traction, less than Fw?Now let's imagine moving the point at which the holding force is applied from 20 degrees to 40 degrees counterclockwise from zero and call this possibly different holding force Fh2. If the answer to Q 3 is yes, it seems that

A1(Fh1) = Fw and

A2(Fh2) = Fw
where A1 and A2 are both constant.

It also seems to me that A1 is the square of A2, meaning that the ratio of Fw to the holding force is an exponential funtion of the number of degrees (or radians) separating the holding force from zero.Q 4. Is this true?The above conjectures should lead me to an equation for the necessary holding force in terms of Fw, Csf, and the number of degrees or radians the band is wrapped around cylinder C. I need some help with that.Q 5. What is the full formula?Thank you for your help.

Jim Adrian

 

[Valerie Park]

Q 1. Yes, this is correct.Q 2. Yes, the static frictional force should remain constant in this setting.Q 3. Yes, the holding force, Fh1, is required to maintain traction and is usually less than Fw.Q 4. Yes, this is true.Q 5. The full formula for the necessary holding force is: Fh = Fw / (Csf * cos(θ)), where θ is the number of degrees or radians the band is wrapped around cylinder C.