Calculating Work by Friction using Integrals

[henry3369]

Homework Statement

A uniform board of length L and mass M lies near a boundary that separates two regions. In region 1, the coefficient of kinetic friction between the board and the surface is μ1, and in region 2, the coefficient is μ2. The positive direction is shown in the figure.

Find the net work W done by friction in pulling the board directly from region 1 to region 2. Assume that the board moves at constant velocity.

http://imgur.com/tFIDobO

Homework Equations

W = Fs

The Attempt at a Solution

Force of friction in region 1: μ1Mg
Force of friction in region 2: μ2Mg
Average force = μ1Mg + μ2Mg/2 = (Mg/2) x (μ1 + μ2)
W = Average force x distance traveled = (-gML/2) x (μ1 + μ2)

So I was able to get the correct answer using average force, but I was wondering how I would approach this if I wanted to use an integral instead. I know the integral of force finds the work done, but I'm not sure what exactly force would be. Is a way to integrate this if an equation is not already given for force?

 

[BvU]

You want to integrate $\displaystyle \int \vec F \cdot \vec {ds}$ where now $\vec F =\vec F(s)$. So I would say you can't do this without an expression for F.

If the contact between board and surface is uniform, you could write $F = -\Bigl ( \mu_2 mgs/L + \mu_1 mg(L-s)/L \Bigr )$ where s runs from 0 to L

 

[haruspex]

If the contact between board and surface is uniform

Yes, that's an important caveat for this method. But more generally, you could model the board as a series of boards of varying lengths each contacting the ground at its mid point and experiencing a normal force in proportion to its weight. I believe this gives the same answer.