Question about static friction adjusting itself

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The discussion explores the conditions for static friction in a system involving two masses and an inclined plane. It identifies two equilibrium states: one just before downward acceleration and another just before upward acceleration, both governed by the relationship mgsinθ = T. It posits that static friction can be absent when mgsinθ equals T, particularly when the mass ratio m/M is considered. The conversation highlights that there are multiple equilibrium points between the two limiting cases, with specific angles x and y where the system transitions between states of motion. Ultimately, an angle exists between x and y where equilibrium can be achieved without static friction.
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
 
For simple comparison, I think the same thought process can be followed as a block slides down a hill, - for block down hill, simple starting PE of mgh to final max KE 0.5mv^2 - comparing PE1 to max KE2 would result in finding the work friction did through the process. efficiency is just 100*KE2/PE1. If a mousetrap car travels along a flat surface, a starting PE of 0.5 k th^2 can be measured and maximum velocity of the car can also be measured. If energy efficiency is defined by...

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