Understanding Coefficients of Static Friction for Roofing: A Roofer's Dilemma

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Homework Help Overview

The problem involves determining the minimum coefficient of static friction required for a roofer to remain stationary on a sloped wooden roof inclined at 44 degrees. The original poster expresses uncertainty about their calculations and the implications of the coefficient of friction in relation to the angle of inclination.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the relationship between the angle of inclination and the coefficient of static friction, with some suggesting that the mass of the roofer is irrelevant to the calculation. Questions arise regarding the interpretation of coefficients greater than 1 and the conditions under which sliding occurs.

Discussion Status

The discussion is active, with participants providing insights and clarifications about the coefficient of friction and its implications. There is an exploration of different interpretations regarding the limits of the coefficient of friction, and some participants offer examples to illustrate their points.

Contextual Notes

Some participants reference a textbook that states the coefficient of friction typically ranges from 0.0 to 1.0, prompting further discussion about the validity of this claim in various contexts. The original poster is also considering the implications of kinetic friction in relation to the current problem.

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Homework Statement



A wooden roof is sloped at a 44 deg angle. What is the minimum coefficient of static friction between the sole of a roofer's shoe and the roof if the roofer is not to slip?


Homework Equations


coefficient of static friciton = friction force / Normal force



The Attempt at a Solution



Well the mass isn't given so initially I thought it just wanted an equation. But after some workings here is what i got.
coefficient of static friction = Fg sin44deg/ Fg cos44deg = 0.9657

I am not sure if this is the minimum or even the right answer.
From my understanding, at the angle of inclination of 44deg an object which is on the incline at rest ...between the incline and the bottom of the object (in this case the roofers shoes) should have atleast a coeff of static friction of 0.9657 to stop from sliding.

can someone elaborate to help me understand?

For the same problem( with same weight) if the sole of the shoes were higher quality the coeff of static friction could be 0.98 right?

One problem I have is, when I raised the angle to 50deg I get a value over 1 which is not right.

I want to follow this up with the concept of kinetic friction. But I will when I solve the above problem.

Thank you for taking the time to help me.
 

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You are correct. I'd quibble with your presentation though. He will slide when the tangential force mg*sin(44) exceeds the static frictional force mu*mg*cos(44). So yes, you need mu>tan(44). But writing Fg just looks funny. What is that? Ohhhh. Is Fg=m*g equal gravitational force? In that case no problems.
 
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The mass isn't given because it doesn't matter. Your answer is correct. The tangent of theta will give the coefficient of static friction...i.e tan(44deg).
 
One problem I have is, when I raised the angle to 50deg I get a value over 1 which is not right.

Why do you think that is not right? One can certainly have a coefficient of friction that is greater than 1.
 
stewartcs said:
Why do you think that is not right? One can certainly have a coefficient of friction that is greater than 1.

Thank you for your answers. The reason why I think so is because this paragraph in a book i was reading.

The coefficient of friction is in the range of 0.0 to 1.0. The value of 0.0 is only
possible if you have a surface that has absolutely no friction at all. The maximum
possible force due to friction, FF, is FN, the normal force. Among other things, this means that if you rely solely on the force of friction to keep you in
place, you can’t push a car with any force greater than your weight. That’s
the maximum value, and it’s possible only if μ = 1.0. (If you dig yourself in
when pushing the car, that’s a different question, because you’re not just relying
on the force of friction to keep you in place.)


I'm not sure if this only has to do with respect to a flat surface.

So let me get this straight.

If mu(=tan44)*FN is equal to the frictional force required and the guy will not move.

If mu(>tan44)*FN is greater than the frictional force required then the guy still will not move.

If mu(<tan44)*FN is less than the frictional force required and the guy will slide down?

Thank you.
 
I think you are right. And, yes, a coefficient of friction>1 is more properly referred to as 'super glue' and not friction.
 
The coefficient of friction is in the range of 0.0 to 1.0.

I would assume your book is speaking in a general sense. If it is stating that the coefficient of friction cannot be greater than 1.0, then it is absolutely wrong. While it is true that it can be no less than 0.0, the coefficient may certainly be greater than 1.0.

The coefficient of friction is simply the ratio for Frictional force to the Normal force, mu = Ff/Fn. Thus from simple inspection of the formula, one can see that if Ff is greater than Fn, the coefficient will be greater than 1.0.

Here is a link that gives some common coefficients of friction for various materials.

http://www.engineershandbook.com/Tables/frictioncoefficients.htm

You'll notice that quite a few exceed 1.0.

Your book should give a bit more information on friction, if it doesn't you might want to get another book. Specifically, you should investigate the microscopic properties between two objects when they are pressed together.

CS
 
So let me get this straight.

If mu(=tan44)*FN is equal to the frictional force required and the guy will not move.

If mu(>tan44)*FN is greater than the frictional force required then the guy still will not move.

If mu(<tan44)*FN is less than the frictional force required and the guy will slide down?

It depends on how the problem is stated. If it says that he is on the VERGE of sliding at 44deg, then when theta = 44deg, the static frictional force will be at its MAXIMUM value.

That then means if theta > 44deg, he WILL slide.

If theta < 44deg, he will NOT slide.

Intuitively this should make sense. The steeper the slope, the easier it is to slide.

CS
 
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I can assure you that I recently attended a horse pulling contest where I watched a team of horses weighing about 3300 pounds pulling a load of 3900 pounds. Their coefficient of friction is thus about 1.08. Many pieces of construction equipment exceed 1.0. Why do you think that might be? HINT: Think about how they contact the earth.
 

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