• Support PF! Buy your school textbooks, materials and every day products Here!

Gragging a board across surfaces with different frictions

  • Thread starter ph123
  • Start date
  • #1
41
0
"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 (mu1), and in region 2, the coefficient is (mu2). The positive direction is from the region with mu1 to the region with mu2. 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. Express the net work in terms of M, L, g, mu1 and mu2"

In my force formula, x is equal to the amount of the board in region 2 (with mu2).

F = mu1(mg(1-(x/L))) + mu2(mg(x/L)

I know I'm supposed to integrate this with respect to L (or am I). Unfortunately, I suck at calculus. This is what I got, sorta.

((1/2)((mu1*g)/L)L^2) + ((1/2)((mu2*g)/L)L^2

Again, I suck at calc. Any thoughts?
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,258
618
You should be integrating with respect to x (not L as you said). x going from 0 -> L. And you are doing pretty well. But you forgot to integrate the 1 in the first term of your force expression.
 
  • #3
41
0
((1/2)(x-((mu1*g)/L))L^2) + (1/2)((mu2*g)/L)L^2

the 1 should just become an x, no?
 
  • #4
Dick
Science Advisor
Homework Helper
26,258
618
Evaluate the x between 0 and L. Just like you did the x^2 terms. And your x should be inside of the m*g*mu. And OUTSIDE of the 1/2. Do it again. Carefully this time.
 
Last edited:
  • #5
41
0
well, justguessing, but:

[((mu1+mu1)Mg)/L]*(1/2)L^2
 
  • #6
Dick
Science Advisor
Homework Helper
26,258
618
I don't think you're guessing. You can cancel one of the L's too. So you just average the two mu's. Makes sense, right?
 
  • #7
41
0
yeah, the board spends the same amount of time in each region so that does make sense.
 

Related Threads on Gragging a board across surfaces with different frictions

Replies
3
Views
252
Replies
3
Views
2K
Replies
10
Views
3K
Replies
1
Views
1K
Replies
4
Views
756
Replies
2
Views
6K
Replies
5
Views
1K
Replies
1
Views
2K
Replies
1
Views
2K
Top