What that statement is saying is that the maximal value of friction is proportional to the normal reaction , but is always less than it (μ < 1) .This max value occurs when the particle is about to slip , so if it is slipping it is safe to assume f = f(max) = μR

Reply 14

Swayum

18

To avoid confusion, my teacher taught me to always write friction as Fr instead of F and I agree with him that it's more sensible.

Fr = mu * R when the object is in limiting equilibrium.

Fr = ma when friction is the only force acting on the object. This follows as Newton's second law says

F = ma (where F is the sum of all forces acting on the object)

And if Fr is the only force, then Fr = ma.

Reply 15

generalebriety

16

2^1/2

no i would be fine, assuming all other forces are in eqm

And when does that happen?! :s-smilie:

~~Friction can't act unless there are other forces acting.~~

Edit: wait, this is false. See later post for clarification. :p:

Reply 16

2^1/2

11

yes we are talking about a resultant here.

(in the one direction)

Reply 17

fusionskd

10

generalebriety

And when does that happen?! :s-smilie:

Friction can't act unless there are other forces acting.

What if theres a mass inclined to an angle at say theta degrees....and theres nothing pulling it the opposite way, so the only way it can move is downward and horizontally, friction is the only force acting and vertically, there is a normal reaction and a weight at mgsin(theta). Atleast thats what I remember from M1...I may be wrong but that was a situation where friction was the only force acting, horizontally anyway....[it is M1 though, not exactly rocket science so they do simplify the model :rolleyes:

]

Reply 18

bruceleej

2

fusionskd

What if theres a mass inclined to an angle at say theta degrees....and theres nothing pulling it the opposite way, so the only way it can move is downward and horizontally, friction is the only force acting and vertically, there is a normal reaction and a weight at mgsin(theta). Atleast thats what I remember from M1...I may be wrong but that was a situation where friction was the only force acting, horizontally anyway....[it is M1 though, not exactly rocket science so they do simplify the model :rolleyes:

]

Dont forget you have the component of weight which acts parallel to the plane...
The only situation I can think of where friction is the only {horizontal} force acting is; 'A hockey pog on ice{after a force has been applied to it}, it'll obviously slow down because the resultant force is the friction between the pog and the ice and thus mu.R=ma'.