DE for modelling motion of a box

  • Context: Undergrad 
  • Thread starter Thread starter acrusera
  • Start date Start date
  • Tags Tags
    Box Modelling Motion
Click For Summary

Discussion Overview

The discussion revolves around modeling the motion of a box-like object sliding down a straight ramp using differential equations. Participants explore the forces acting on the object, including gravity and friction, and question the appropriate representation of these forces in the equations of motion.

Discussion Character

  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant presents the differential equation ma=mg-kv, questioning whether k represents the coefficient of kinetic friction and if k equals μ.
  • Another participant challenges the first by suggesting that k does not equal μ, implying a distinction between kinetic and static friction.
  • A third participant reiterates the original equation and suggests that it is applicable for a body falling under gravity with air resistance proportional to its velocity, indicating a potential misunderstanding of the scenario.
  • One participant introduces a formula for k as 6πηr, questioning its relevance to the ramp scenario.
  • A fifth participant explains that the coefficient of friction, μ, is related to the normal force and weight of the object, providing a breakdown of the forces acting on the object on the ramp.

Areas of Agreement / Disagreement

Participants express differing views on the interpretation of the variable k and its relationship to the coefficient of friction. There is no consensus on the correct application of the differential equation or the definitions of the terms involved.

Contextual Notes

Participants have not fully resolved the definitions of k and μ, nor have they clarified the assumptions underlying the modeling of the box's motion on the ramp.

acrusera
Messages
1
Reaction score
0
So if you were to slide a box-like object down a straight ramp with a single gradient (its literally just an incline no curve). And you attempt to model it accurately would this be correct?
Given the DE
ma=mg-kv
where a=dv/dt, mg=force of gravity, kv= force of friction

Anyway my main question, is k the coefficient of kinetic friction? Does k=μ?
 
Physics news on Phys.org
What is µ according to you? Coefficient of static friction? then no ,k≠ µ
 
acrusera said:
So if you were to slide a box-like object down a straight ramp with a single gradient (its literally just an incline no curve). And you attempt to model it accurately would this be correct?
Given the DE
ma=mg-kv
where a=dv/dt, mg=force of gravity, kv= force of friction

Anyway my main question, is k the coefficient of kinetic friction? Does k=μ?

I think you've got your equations mixed up.

##ma = mg - kv##

Is for a body falling under gravity with air resistance proportional to its velocity.
 
PeroK said:
Is for a body falling under gravity with air resistance proportional to its velocity.
k=##6 \pi \eta r## ??
then what about the ramp?
 
The "coefficient of friction", [itex]\mu[/itex] is the number that multiplies the normal force on an object to give the friction force. An object of mass m, has weight mg. On a ramp making angle [itex]\theta[/itex] with the horizontal, we can divide the weight into components parallel to and normal to the ramp. The component normal to the ramp is [itex]mg cos(\theta)[/itex].
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 10 ·
Replies
10
Views
6K
  • · Replies 11 ·
Replies
11
Views
4K
  • · Replies 15 ·
Replies
15
Views
6K
  • · Replies 2 ·
Replies
2
Views
4K
Replies
24
Views
3K
  • · Replies 7 ·
Replies
7
Views
7K
  • · Replies 27 ·
Replies
27
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K