Work: Integral or average

  • Thread starter henry3369
  • Start date
  • #1
194
0

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?
 

Answers and Replies

  • #2
BvU
Science Advisor
Homework Helper
14,126
3,545
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
 
  • #3
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
35,576
6,449
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.
 

Related Threads on Work: Integral or average

  • Last Post
Replies
2
Views
4K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
3
Views
3K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
15
Views
3K
Replies
1
Views
4K
Replies
9
Views
256
Replies
5
Views
3K
Replies
5
Views
9K
Top