Newton's Laws: Friction: Inclined Plane

  • #1
My main question:
Given an angle of incline, x, how can you find the coefficients of static and kinetic friction? Is it possible?

My homework problem is:

Someone wearing leather shoes is standing in the middle of a wooden plank. One end of the board is gradually raised until it makes an angle of 17 degrees with the floor, at which point the person begins to slide down the incline. Compute the coefficient of static friction.



-----

I used the simple formula tan x = Mu s, and it seemed to work, giving me the correct answer of 0.31, although I have no idea why. Can anyone explain to me why tan x gives the coefficient, and whether it gives the static or kinetic coefficient? I know its hard to explain without a free body diagram, but I need help..
 
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Answers and Replies

  • #2
learningphysics
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you need to draw a freebody diagram... there are 3 forces in total acting on the person... the normal force... gravity and the frictional force...

All these forces need to be divided into components along the plane and parallel to the plane...

Normal force is perpendicular to the plane... frictional force is parallel to the plane... now you need to get the components of gravity perpendicular to the plane and parallel to the plane...
 
  • #3
you need to draw a freebody diagram... there are 3 forces in total acting on the person... the normal force... gravity and the frictional force...

All these forces need to be divided into components along the plane and parallel to the plane...

Normal force is perpendicular to the plane... frictional force is parallel to the plane... now you need to get the components of gravity perpendicular to the plane and parallel to the plane...

Ohh. But how can I find any forces when all I am given is an angle?
Sum of Fx = Fa- Ff and Fy= Fn-Fg right? I'm so confused..
 
  • #4
learningphysics
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Ohh. But how can I find any forces when all I am given is an angle?
Sum of Fx = Fa- Ff and Fy= Fn-Fg right? I'm so confused..

What is Fa?

It's like the inclined plane... what is the component of gravity perpendicular to the plane? mg is the net gravitational force... what are the two components?
 
  • #5
Fg perpendicular is Fg(cos 17) right?
So would it be Fy= Fg(cos 17) - mg
 
  • #6
Fy= mg(cos 17) - mg
and Fa= Ff ?
 
  • #7
learningphysics
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Fy= mg(cos 17) - mg

the mgcos(17) is right... but why do you have the "-mg" ? You need Fnormal. or rather use downwards into the perpendicular as negative and outwards as positive hence:

Fy = Fnormal - mg(cos 17) = 0
 
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  • #8
learningphysics
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Fy= mg(cos 17) - mg
and Fa= Ff ?

But what is Fa supposed to be? I don't understand what Fa is supposed to represent.
 
  • #9
Wait so what's the main equation I'm working with.. Ff= mu * Fn ?
 
  • #10
learningphysics
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Wait so what's the main equation I'm working with.. Ff= mu * Fn ?

yes, you need to use that... but first get the x and y equations... just use Ff for now... we'll substitute mu*Fn afterwards...
 
  • #11
Oh I was saying Fa is applied force.. but I don't know why cause I guess there is no force applied, sorry
 
  • #12
learningphysics
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Oh I was saying Fa is applied force.. but I don't know why cause I guess there is no force applied, sorry

You have Ff and mgsin(17) parallel to the plane.
 
  • #13
I'm sorry, what is mg*(cos 17) ? Isn't that Fgperpendicular, which is equal and opposite to the normal force?
 
  • #14
learningphysics
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I'm sorry, what is mg*(cos 17) ? Isn't that Fgperpendicular, which is equal and opposite to the normal force?

yes, in this particular case Fnormal = mgcos(17)

So using [tex]\Sigma{F_y} = 0[/tex]

then substituting in the left side:

[tex]F_{normal} - mgcos(17) = 0[/tex]

Hence Fnormal = mgcos17.

Can you give a similar equation for [tex]\Sigma{F_x}[/tex]?
 
  • #15
so is it Fx= Ff -Fgparallel ?
 
  • #16
and Fx= 0, and Fgparallel= Fg (sin 17)
 
  • #17
so 0= Ff- mg(sin 17) therefore Ff= mg(sin 17) ?
 
  • #18
then since Ff= mu * Fn ,
mg(sin 17) = mu * mg(cos 17)
and mu = (sin 17)/(cos 17)
 
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  • #19
Ohhhh and that equals 0.305!!!!!!
 
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  • #20
Thanks soooo much I get it now. I guess I was trying to take a shortcut using that tangent thing and it messed me up. Thanks your amazing
 
  • #21
learningphysics
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so is it Fx= Ff -Fgparallel ?

Yes. so as long as the object is not sliding... [tex]\Sigma[/tex]Fx = Ff-Fgparallel = 0

let's do this for an arbitrary angle where it is not slipping...

Hence Ff- mgsin(theta) = 0

Ff = mgsin(theta).

and in the y-direction:

Fnormal - mgcos(theta) = 0
Fnormal = mgcos(theta)

Now the maximum possible static frictional force is [tex]\mu*Fnormal[/tex]

The block won't slip as long as Ff< [tex]\mu*Fnormal[/tex]

in other words the block won't slip as long as [tex]mgsin(\theta)<\mu*mgcos(\theta)[/tex]

from that we get [tex]tan(\theta)<\mu[/tex] for no slipping... or in other words [tex]tan(\theta)>=\mu[/tex] when slipping happens.

if slipping happens at 17, then tan(17) = [tex]\mu[/tex]
 
  • #22
learningphysics
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Thanks soooo much I get it now. I guess I was trying to take a shortcut using that tangent thing and it messed me up. Thanks your amazing

cool! no prob!
 
  • #23
Yes. so as long as the object is not sliding... [tex]\Sigma[/tex]Fx = Ff-Fgparallel = 0

let's do this for an arbitrary angle where it is not slipping...

Hence Ff- mgsin(theta) = 0

Ff = mgsin(theta).

and in the y-direction:

Fnormal - mgcos(theta) = 0
Fnormal = mgcos(theta)

Now the maximum possible static frictional force is [tex]\mu*Fnormal[/tex]

The block won't slip as long as Ff< [tex]\mu*Fnormal[/tex]

in other words the block won't slip as long as [tex]mgsin(\theta)<\mu*mgcos(\theta)[/tex]

from that we get [tex]tan(\theta)<\mu[/tex] for no slipping... or in other words [tex]tan(\theta)>=\mu[/tex] when slipping happens.

if slipping happens at 17, then tan(17) = [tex]\mu[/tex]

Yeah yeah I see
 
  • #24
That's awesome you make it so clear thanks
 

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