The coefficient of Friction changing under air resistance

  • #1

Homework Statement

The question asks me to work out the coefficient of friction for a block accelerating down a rough slope assuming their is no air resistance, which I could do easily. In the next part it then asks how the coefficient would change if their was air resistance

Homework Equations

Where F=friction
μ=coefficient of friction
R=resultant force of plane on block

The Attempt at a Solution

Well I concluded that because the surfaces are still the same the coefficient would still be the same, but the air resistance would push block down more incresing R and so slowing block down because their is more friction.
However the mark scheme said that their is less friction so coefficient must be less, which I really don't understand so I was hoping someone could explain it to me.

Any help would be much appreciated, thanks
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  • #2
The coefficient of friction is the ratio of the frictional force to the normal reaction.

Now, if there is an additional velocity dependent retarding force (air resistance), the total force impeding the motion is the sum of the frictional force and this retarding force.

Perhaps, the idea here is to define some kind of "effective" frictional force by computing the ratio of this total impeding force to the normal reaction. I don't know why one would want to do this, but that may as well be the answer to your question.

I would reason however that the coefficient of friction is a function of the two surfaces and not any additional retarding forces (since after all, air resistance is not a relative surface interaction).
  • #3
After reading your comment and re-reading the question, I see where I went wrong.
In the previous bit of the question you have to work out the coefficient from acceleration and the force acting down the slope using F=ma
Force down slope-Friction=mass x acceleration
But if you have air resistance as well
Force down slope-Friction-air Resistance=mass x acceleration. Therefore friction is smaller and so the coefficient you calculate is smaller. It wants you to assume they are different cases and not the same surface like I did
Thanks for your help
  • #4
Good thing you figured it out. To me, it still seems like a vague/weird question :tongue:

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