Integrating F.dx: Constant μ, m & g Affect Force?

  • Context: Undergrad 
  • Thread starter Thread starter Hesperides
  • Start date Start date
  • Tags Tags
    Integrating
Click For Summary

Discussion Overview

The discussion revolves around the integration of the friction force in the context of a physics problem, specifically examining how constants such as the coefficient of friction (μ), mass (m), and gravitational acceleration (g) influence the integration process. The scope includes mathematical reasoning and conceptual clarification regarding the behavior of friction in this scenario.

Discussion Character

  • Mathematical reasoning, Conceptual clarification

Main Points Raised

  • One participant questions why the friction force does not require integration, suggesting that μ, m, and g are constants, which leads to the conclusion that integrating x^0 results in a straightforward outcome.
  • Another participant agrees, stating that the friction force remains constant along the path, and that integration effectively sums these constant forces over distance, although they acknowledge that in reality, friction may not be entirely constant.
  • A later reply introduces the concept of a line integral, noting that constants can be factored out of the integral, simplifying the expression to just the displacement.

Areas of Agreement / Disagreement

Participants generally agree on the notion that the friction force can be treated as constant for the purposes of this integration, but there is an acknowledgment that real-world scenarios may introduce variability in friction.

Contextual Notes

Participants do not fully explore the implications of varying friction in different scenarios, such as static versus kinetic friction, nor do they address potential limitations in the assumptions made regarding the constants involved.

Hesperides
Messages
2
Reaction score
0
I'm sure this is an easy question to answer - per the attached image, can someone please explain why the friction force has not required/been affected by integration?

My calculus is very rusty but the only reason I can think of is that μ, m and g are all constants (for the purposes of the question) and so x^1 drops in as a result of integrating x^0 - is that correct?
 

Attachments

  • Work Fdx.PNG
    Work Fdx.PNG
    7.8 KB · Views: 453
Physics news on Phys.org
Yup. The force of friction is the same at every point along the path for the reasons that you stated. Since integration is just adding up all of those amounts of friction at each point, it is the same as multiplying it by [itex]x[/itex]. In reality, as you can probably guess, the force of friction would not actually be totally constant, but the fluctuations are very small (unless you allow the block to stop at some point at which point you would need to worry about static friction).
 
Great! Thanks Drew
 
It's a line integral over a distance x2 - x1...

so the constants may be moved outside the integral leaving just the displacement.
 

Similar threads

  • · Replies 17 ·
Replies
17
Views
3K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 14 ·
Replies
14
Views
3K
  • · Replies 14 ·
Replies
14
Views
2K
  • · Replies 11 ·
Replies
11
Views
5K
  • · Replies 9 ·
Replies
9
Views
4K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 59 ·
2
Replies
59
Views
5K