Work done by friction on a variable slope

[Bill M]

Hello,

I'm trying to figure out a method of calculating the work done by friction on an object sliding down a surface with a variable slope, assuming an equation can be determined to fit the line along which the object travels and we have a known coefficient of friction for the surface.

The sticking point for me is that the force of friction is obviously going to vary with the slope of the path of travel since it's dependent upon the normal force. I studied calc and physics extensively, but that was over 10 years ago. I've broken out the calculus book and looked at the section on line integrals in a vector field, but I don't know that the friction could be treated as a vector field opposing the motion of the object like gravity could, for example.

So, I suppose I'm a little lost where to begin. Any hints would be appreciated.

 

[tiny-tim]

Hello Bill!

Hint: assume the hill is in straight sections …

what is the work done (as a function of h) for anyone section?

 

[Bill M]

If I were doing it for a straight surface, I'd use:

W = mgfd, where m is mass of the object, d is the distance traveled, and f is an adjusted coefficient of friction based on slope (or angle theta).

My problem now is that, first, distance is not so easily calculated (though I could use an arc length formula for a more complicated path) but that the frictional force (mu*N = mu*mgcos(theta)) varies with theta (or slope). So I'm not making the connection with how to solve this. I'm sure it involves integrating over the curve traveled somehow, but everything I've tried on paper has lead to erroneus answers.

 

[Skrambles]

Have you tried using conservation of energy? If you can figure out the object's potential and kinetic energy at two points, the difference will be the work done.

 

[nasu]

You can find an expression for the friction force as a function of position on the curve and then do the line integral to calculate the work.
Energy conservation does not work here. Energy is not conserved if there is friction. (I mean mechanical energy, of course)

 

[Dyson]

For the friction, the curve integral of a certain closed contour is not zero. So it is none sense to define a field of it. The curve integral makes sense only when you have defined a definite path.

 

[Bill M]

Okay, see if this looks right:

The sum of forces acting along the ground on the object would be:

\sum F = w(\mu cos\theta + sin\theta )

where mu is the coefficient of kinetic friction and w is weight.

This is equivalent to:
\sum F = w(\frac{\mu +m}{\sqrt{m^{2}+1}})

where m is the slope.

So, I'd need to solve:

\oint w(\frac{\mu +m}{\sqrt{m^{2}+1}}) ds

over the curve represented by some function y = f(x) where x is between 0 and some end point.

Since m is just dy/dx and given the definition of arc length as the square root of (dx/dx)^2 + (dy/dx)^2 , this should work out to the integral:

\int_{xo}^{xf}w(\frac{\mu +\frac{dy}{dx}}{\sqrt{\frac{dy}{dx}^{2}+1}})\sqrt{1+(\frac{dy}{dx})^{2}} dx

Does this look right?

 

[Dyson]

I haven't found any errors.

 

[Skrambles]

You can find an expression for the friction force as a function of position on the curve and then do the line integral to calculate the work.
Energy conservation does not work here. Energy is not conserved if there is friction. (I mean mechanical energy, of course)

That's not entirely true, friction is not a conservative force (it always does negative work), but energy is ALWAYS conserved. Hence the 1st law of thermodynamics. If you knew the change in velocity with respect to position, you could find the work done by friction with a bit of algebra.

 

[tiny-tim]

Hi Bill!

The sum of forces acting along the ground on the object …

Let's not go overboard on this …

you're only trying to find the work done by the friction …

so you can forget about the component of the weight (you know the work done by that anyway, it's w times height).

Just take your value for the friction, multiply it by distance travelled, and express that in terms of height and/or horizontal distance …

what do you get? ​

 

[nasu]

You can find an expression for the friction force as a function of position on the curve and then do the line integral to calculate the work.
Energy conservation does not work here. Energy is not conserved if there is friction. (I mean mechanical energy, of course)

That's not entirely true, friction is not a conservative force (it always does negative work), but energy is ALWAYS conserved. Hence the 1st law of thermodynamics. If you knew the change in velocity with respect to position, you could find the work done by friction with a bit of algebra.

What is not entirely true? I made it clear (I hope) that I mean mechanical energy is not conserved.
Of course, IF you know how the speed changes along the slope you could use this info to calculate the work. But how would you know this before hand?
Anyway this has nothing to do with conservation of (mechanical) energy.
You would use the work-energy theorem and not conservation:
Delta KE = Work_total

 

[Bill M]

Hi Bill! Let's not go overboard on this …

you're only trying to find the work done by the friction …

so you can forget about the component of the weight (you know the work done by that anyway, it's w times height).

Just take your value for the friction, multiply it by distance travelled, and express that in terms of height and/or horizontal distance …

what do you get? ​

You're right that I originally asked only about the work done by friction. I originally was only asking about this component because I already knew how to find the work done by gravity (quite easy, as you point out). My end goal (though unstated) was to find the total work done and that's why I went ahead and added the weight in there so I could accomplish everything in one calculation.

The only thing really left out is air resistance, but I already proved that to be negligible at the speeds I'm working with.

So, I THINK that what I came up with above will give me what I'm looking for, unless someone can find an error in my logic. Again, this is probably fairly simple stuff, but like I said I'm out of practice by a decade or so. I had to read the section on line integrals in my university calc book to get what I got above. And you should have seen the headaches I got when I had to work with differential equations again a few months ago!

Anyway, thank you to everyone for the help!

 

[tiny-tim]

Hi Bill!

If the friction force is µmgcosθ and the distance (along the slope) is d, then the work done is µmg(dcosθ), which is µmg times the horizontal distance …

this will apply for any shape of the hill, so long as the object is sliding directly downhill.

 

[Bill M]

I'm envisioning a more varying slope, with perhaps several "humps".

 

[tiny-tim]

yes, that's fine, it works for any shape of hill, including humps, provided the motion is always directly uphill or downhill …

total work done by friction = µmg times the horizontal distance ​