# M1 - F = mu.R?

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Thread starter 13 years ago
#1
F = mu.R

What am i exactly subbing into the 'F', is it friction, or force in general?
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13 years ago
#2
Friction
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13 years ago
#3
friction
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13 years ago
#4
friction x resultant force, but would check with someone whos clued up about mechanics, but thats what I remember from M1.
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13 years ago
#5
(Original post by Frater)
friction x resultant force, but would check with someone whos clued up about mechanics, but thats what I remember from M1.
No, just friction.

(Incidentally, it's , not necessarily unless the body is moving or about to move.)
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Thread starter 13 years ago
#6
Ok, so i cannot say then :

mu.R = ma ... as thats Force, correct?
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13 years ago
#7
(Original post by Mos Def)
Ok, so i cannot say then :

mu.R = ma ... as thats Force, correct?
Yeah, saying that would be wrong.
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13 years ago
#8
no i would be fine, assuming all other forces are in eqm
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13 years ago
#9
friction is a force

(frictional force)
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Thread starter 13 years ago
#10
So friction = ma during equilibrium?
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13 years ago
#11
(Original post by Mos Def)
So friction = ma during equilibrium?
No. 0 = ma, there's no resultant force as there's no acceleration.
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13 years ago
#12
(Original post by Mos Def)
So friction = ma during equilibrium?
No, dont think of it that way. Friction=mu.R and
Resultant Force = ma.

If and only if your resultant force is the friction does ma = mu.R
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Thread starter 13 years ago
#13
Ahh right, cheers all
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13 years ago
#14
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
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13 years ago
#15
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.
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13 years ago
#16
(Original post by 2^1/2)
no i would be fine, assuming all other forces are in eqm
And when does that happen?!

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

Edit: wait, this is false. See later post for clarification.
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13 years ago
#17
yes we are talking about a resultant here.

(in the one direction)
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13 years ago
#18
(Original post by generalebriety)
And when does that happen?!

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 ]
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13 years ago
#19
(Original post by 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 ]
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'.
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13 years ago
#20
(Original post by bruceleej)
Dont forget you have the component of weight which acts parallel to the plane...

Ah yes...yet another dumb moment for me there...

Just ignore me
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