Question about static friction adjusting itself

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Discussion Overview

The discussion revolves around the conditions for static friction in a system involving two masses and an inclined plane. Participants explore the equilibrium states of the system, particularly focusing on scenarios where static friction may not be present.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant outlines two equilibrium conditions for a mass on an incline, considering forces acting on the mass just before it accelerates downwards and upwards.
  • Another participant asserts that tension (T) is always equal to the weight of the mass (Mg), emphasizing the importance of the mass ratio (m/M).
  • A third participant reiterates the two equilibrium conditions and introduces the concept of infinitely many equilibria between the limiting cases.
  • A further contribution questions the existence of a scenario where static friction is absent, proposing a method to determine angles at which the system remains at rest without static friction.

Areas of Agreement / Disagreement

Participants express varying interpretations of the equilibrium conditions and the role of static friction, indicating that multiple competing views remain without a consensus on the specific conditions under which static friction may be absent.

Contextual Notes

The discussion includes assumptions about the mass ratio and the behavior of the system at different angles, but these assumptions are not universally accepted or resolved among participants.

Sho Kano
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incline_wpulley.gif

In a situation like this, there are two equilibrium's,
For just about to accelerate downwards:
mgsinθ - Ffriction - T = 0
For just about to accelerate upwards:
T - Ffriction - mgsinθ = 0

Is there a situation where there is no static friction i.e. when
mgsinθ = T?
 
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Sure. For the equilibrium condition with or without friction, T is always equal to Mg. The m/M ratio is key.
 
Sho Kano said:
incline_wpulley.gif

In a situation like this, there are two equilibrium's,
For just about to accelerate downwards:
mgsinθ - Ffriction - T = 0
For just about to accelerate upwards:
T - Ffriction - mgsinθ = 0
This are the limiting cases. There are infinitely many equilibriums in between.
 
Sho Kano said:
Is there a situation where there is no static friction i.e. when
mgsinθ = T?
if m>M
first set angle to 0
then increase angle by tiny amounts and after each increase reset the system to original condition( ie bodies at rest) and check if mass 'm' is at rest
let x = 1st angle at which equilibrium is achieved (body just about to accelerate down)
on further increase , for a small interval of angle , system will stay at rest
let y = 1st angle at which system again starts moving (body just about to accelerate up)
note that on increasing angle above x , tendency of m to go down decreases and to go up increases
on decreasing angle below y , tendency of m to go down increases and to go up decreases
so there must be an angle bet x and y at which both tendency balanced so no static friction needed for equilibrium
 

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