Newton's Laws: Friction: Inclined Plane

In summary, to find the coefficient of static friction for a person wearing leather shoes standing on a wooden plank at an incline of 17 degrees, one can use the formula tan x = Mu s. This will give the correct answer, but it may be difficult to understand without a free body diagram. The forces involved are the normal force, gravity, and frictional force, which must be divided into components along and parallel to the incline. The normal force is equal and opposite to the perpendicular component of gravity, while the frictional force is equal and opposite to the parallel component of gravity. Using the equations for force in the x and y directions, one can solve for the maximum possible static frictional force, which is equal to the
  • #1
johnsonandrew
88
0
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.



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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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  • #2
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
learningphysics said:
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
johnsonandrew said:
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
johnsonandrew said:
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
johnsonandrew said:
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
johnsonandrew said:
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
johnsonandrew said:
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
johnsonandrew said:
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
johnsonandrew said:
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
johnsonandrew said:
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
learningphysics said:
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
 
  • #25
johnsonandrew said:
That's awesome you make it so clear thanks

you're welcome.
 
  • #26
learningphysics said:
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]

Oh but I have a question, what happens if it IS sliding? Doesn't Fx still = Ff-Fgparallel?
 
  • #27
It does... but when it slides [tex]\Sigma[/tex]Fx is not 0 and the frictional force changes... it depends on the coefficient of kinetic friction.

We assume the object is not sliding and found the "required" static frictional force to keep it from sliding... the required static frictional force is mgsin(theta).

But we aren't sure yet if this static frictional force is possible... we need to check that it is less than [tex]\mu_s*F_{normal}[/tex]...

if it isn't... then it is impossible for the block to be prevented from sliding...

When the block slides frictional force changes to [tex]\mu_k*Fnormal[/tex].

You may in the future have a problem where you need to determine if the block will slide or not... the way to do it is to see what the "required frictional force is" to prevent sliding... then to see if it is less than [tex]\mu_s*Fnormal[/tex]... if it isn't, then the block slides... then you may be asked to calculate the acceleration of the sliding block... so then you'd use frictional force = [tex]\mu_k*Fnormal[/tex]...
 
  • #28
learningphysics said:
It does... but when it slides [tex]\Sigma[/tex]Fx is not 0 and the frictional force changes... it depends on the coefficient of kinetic friction.

We assume the object is not sliding and found the "required" static frictional force to keep it from sliding... the required static frictional force is mgsin(theta).

But we aren't sure yet if this static frictional force is possible... we need to check that it is less than [tex]\mu_s*F_{normal}[/tex]...

if it isn't... then it is impossible for the block to be prevented from sliding...

When the block slides frictional force changes to [tex]\mu_k*Fnormal[/tex].

You may in the future have a problem where you need to determine if the block will slide or not... the way to do it is to see what the "required frictional force is" to prevent sliding... then to see if it is less than [tex]\mu_s*Fnormal[/tex]... if it isn't, then the block slides... then you may be asked to calculate the acceleration of the sliding block... so then you'd use frictional force = [tex]\mu_k*Fnormal[/tex]...

Ohhh alright that makes sense. Well thanks again for all your help!
 

1. What are Newton's Laws?

Newton's Laws are three fundamental laws of physics that describe the behavior of objects in motion. They were developed by Sir Isaac Newton in the 17th century and are still used today to explain the motion of objects.

2. What is friction?

Friction is a force that opposes the motion of an object. It is caused by the contact between two surfaces and can be either static (when the object is not moving) or kinetic (when the object is moving).

3. How does friction affect an object on an inclined plane?

On an inclined plane, friction acts in the opposite direction of the object's motion. This means that it will slow down or prevent the object from sliding down the incline. Friction also increases the normal force on the object, making it harder to move up the incline.

4. What is the relationship between friction and the angle of an inclined plane?

The amount of friction on an inclined plane is directly proportional to the angle of the incline. This means that the steeper the incline, the greater the friction force will be.

5. How does the weight of an object affect its motion on an inclined plane?

The weight of an object affects its motion on an inclined plane by increasing the force of gravity pulling it down the incline. This can cause the object to move faster and experience more friction, making it harder to move up the incline.

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