Objective problem regarding friction.

[sankalpmittal]

Homework Statement

Let F , F_N and f denote the magnitudes of the contact force , normal force and the force of friction exerted by one surface on the other kept in contact. If none of these is zero , then ,

(A.) F>F_N
(B.) F>f
(C.) F_N>f
(D.) F_N-f<F<F_N+f

Homework Equations

Well , I don't know whether object is moving with respect to other or not...

In case of static friction ,

f_s ≤ μ_sF_N

In case of kinetic friction ,
f_k = μ_kF_N < μ_sF_N

The Attempt at a Solution

http://postimage.org/image/yqh9rg6pl/

OK , So obviously (A) is correct. (B) is also correct. (D) obviously is correct. (C) is also correct as it would mean that tanθ = μ <1 which is also true.

So I go for A , B , C , D. But the textbook gives the answer A,B,D. Why is it so ? Where am I going wrong ?

 

[PhanthomJay]

(C) is also correct as it would mean that tanθ = μ <1 which is also true.

is it always true that μ < 1?

 

[CAF123]

What is this contact force? Do you mean something like applied force? (This is my understanding from the attached link).

 

[kushan]

The contact force F , is the resultant of force due to friction and the normal force .
I guess the question setter is assuming that u <= 1 .

 

[ehild]

is it always true that μ < 1?

No. That is why C is not true.

ehild

 

[CAF123]

$μ$ is not restricted to lie below 1. It depends on the materials being moved relative to each other and is a ratio that depends on the force of friction between the two materials and their contact. Indeed, a ratio can be greater than 1.

 

[NoPoke]

http://mathforum.org/library/drmath/view/51493.html for a brief discussion on the question of the coefficient of friction being greater than 1

 

[sankalpmittal]

is it always true that μ < 1?

But my intuition says that μ can not be greater than 1. It can just equal 1 at most.

Look at this figure :
http://postimage.org/image/yqh9rg6pl/

μ = tanθ = f/F_N. Now as θ approaches 90 , tanθ → ∞ , so the contact force F , will not be having any vertical component. This implies that F_N = 0 , which is never true as this would mean that body will bury straight away into the surface of other object.

Also , if θ>45^o inclination of contact force F vector will be such that F_N will decrease , and so there will be net force acting downward which again causes the object to bury. This is also not possible.

What is this contact force? Do you mean something like applied force? (This is my understanding from the attached link).

No , its not applied force precisely. kushan answers your question :

The contact force F , is the resultant of force due to friction and the normal force .
I guess the question setter is assuming that u <= 1 .

Assuming ? According to your conjecture , he must assume μ≥ 1

No. That is why C is not true.

ehild

Are you answering PhanthomJay's question ?

$μ$ is not restricted to lie below 1. It depends on the materials being moved relative to each other and is a ratio that depends on the force of friction between the two materials and their contact. Indeed, a ratio can be greater than 1.

I can not understand. As I said that if μ→∞ , normal reaction will become 0 and as μ > 1 , normal reaction will decrease. Also , by googling , I can rarely see anything whose μ>1.05.

http://mathforum.org/library/drmath/view/51493.html for a brief discussion on the question of the coefficient of friction being greater than 1

No doubt this site is good for mathematics concepts but does not help me much in physics.

 

[ehild]

Are you answering PhanthomJay's question ?

Yes.

ehild

 

[PhanthomJay]

You are confusing your 'thetas'. There is one shown in the figure. It apparently is the angle that the contact force F makes with the vertical. Then you say u = tan theta. That theta in u = tan theta has nothing to due with the other angle. u is a property of the materials, independent of anything else. For some slick race car tires on dry pavement, u can have a value of well over 1.
When you place an object on an incline plane and slowly increase the angle of the incline, then the object will slip at the angle where u = tan theta. That's a whole different theta. Nothing at all to do with the contact force and it's angle with the vertical. Also, the figure is messed. F should be pointing down and to the right, which is another different issue. Bottom line is that u can be greater than 1 for some materials.

 

[NoPoke]

Mathematicians can do physics too so it is not a great idea to dismiss the answer given because a maths site doesn't help you much in general. It only has to help you in this particular instance to be of value.

The question on that maths site was "My college physics professor stated that the coefficient of friction can be greater than one..." So it isn't just the mathematicians saying u can be >1.

The answer given on the maths site for can u>1 shows two things

1) The mathematical model is fine with u>1 ["Friction" is a macro-model of nature only!]
2) That if there is even a single example where u>1 then the theory that u is always <1 is false. [basic scientific method]

FWIW I used to think that u<1 was a rule too : but it was my mistake to extrapolate from never seeing a counter example to the non-existance of counter examples.

 

[sankalpmittal]

Please delete...

 

[sankalpmittal]

You are confusing your 'thetas'. There is one shown in the figure. It apparently is the angle that the contact force F makes with the vertical.

Yes , and that angle is known as "angle of friction."

Then you say u = tan theta. That theta in u = tan theta has nothing to due with the other angle.

What "other" angle are you mentioning ? Do you mean - "Angle of repose" ?

u is a property of the materials, independent of anything else. For some slick race car tires on dry pavement, u can have a value of well over 1.

Ok , but how can any object have μ>1 ?

When you place an object on an incline plane and slowly increase the angle of the incline, then the object will slip at the angle where u = tan theta. That's a whole different theta. Nothing at all to do with the contact force and it's angle with the vertical.

Yes , I know. I am anyways not talking about angle of repose which is a topic of friction on an inclined plane. Here I am talking about angle of friction - angle made by contact force with the vertical.

Also, the figure is messed. F should be pointing down and to the right, which is another different issue. Bottom line is that u can be greater than 1 for some materials.

I cannot fathom. Anyways , here is what my textbook says about friction :

Suppose a block is kept on ground. In this case , taking block as a system , there are two forces acting on it , force of gravity and normal reaction. If we apply force on block , and suppose it does not move , then the contact force applied by ground on block to oppose relative motion of block w.r.t ground will have two components , as shown in my figure in post #1 , one will be vertical component which will balance the force due gravity (which will always remain same) and other will be horizontal component which will provide friction force which equal applied force. Now as we apply more force - 5,6,7,8N,etc... block still does not move. Such type of friction is called static friction. This must mean that frictional force , f_s = μN is also increasing. Now here is what I 'm adding - since N or normal reaction remains same , f_s is increasing because μ is increasing. μ = f/N , but tanθ = f/N so μ=tanθ. This means that angle of friction θ , is also increasing which tilts the vector of contact force further and also increases length of contact force vector. Now at some value of f_s , the block is just to move. We call that friction , limiting friction and μ_s as coefficient of limiting friction. Limiting friction is equal to minimum applied force to move that object. In this case tanθ_s = μ_s , where θ_s is known as critical angle of friction. In general f_s≤μ_sN.

Now as block begins to move , the static friction is replaced by kinetic friction. Since f_k<f_s , so μ_k<μ_s , this must mean that angle of friction during kinetic friction must be reduced and the tilting of contact force vector will also reduce. But after this f_k will always remain same and so does μ_k except for body moving with high relative velocities.

Also , in case of static friction , if μ→∞ , θ (or angle of friction ) → 90^o , does it not imply that N or normal reaction =0 ?

This is my concept regarding kinetic and static friction. Assume that object is moving on horizontal surface and not inclined plane. Tell me where is it jumbled up ?
And in my textbook , this figure (in post 1) was only drawn to show contact force applied by ground on block.

Mathematicians can do physics too so it is not a great idea to dismiss the answer given because a maths site doesn't help you much in general. It only has to help you in this particular instance to be of value.

The question on that maths site was "My college physics professor stated that the coefficient of friction can be greater than one..." So it isn't just the mathematicians saying u can be >1.

The answer given on the maths site for can u>1 shows two things

1) The mathematical model is fine with u>1 ["Friction" is a macro-model of nature only!]
2) That if there is even a single example where u>1 then the theory that u is always <1 is false. [basic scientific method]

FWIW I used to think that u<1 was a rule too : but it was my mistake to extrapolate from never seeing a counter example to the non-existance of counter examples.

Ah ! Thanks. I went through that site , and it only made me believe that coefficients of friction can be greater than 1.

 

[PhanthomJay]

OK I looked at this picture a bit more carefully and read your notes and its title. The title talks about the "three forces exerted by the ground on the object" , when in fact there are just two forces exerted by the ground on the object..the friction force, f , and the normal force, N. The force shown as F is the resultant of these 2 contact forces...it is not a 3rd force. The resultant contact force makes an angle theta with the vertical which is the Angle of friction at the limiting value of f =uN , OK. Not shown in the diagram is the weight force mg acting vertically down, and the applied force, P. Confusing sketch to me.

Ok , but how can any object have μ>1 ?

I think you've answered this yourself already. You found a site referencing u = 1.05.. I believe copper on copper can have u =1.3.. And slick racecar rubber on dry concrete, u = 1.7. In these cases, the friction angle theta is greater than 45 degrees.

Now as we apply more force - 5,6,7,8N,etc... block still does not move. Such type of friction is called static friction. This must mean that frictional force , f_s = μN is also increasing.

the frictional force is increasing, but it is only equal to uN at the limiting case when the object just starts to move. Otherwise, the increasing friction force will always be less than uN, and equal to the horizontal applied force.

Now here is what I 'm adding - since N or normal reaction remains same , f_s is increasing because μ is increasing

u does not increase, f_s increases. u is a constant for the materials involved, and is unchanging. μ > or = f/N ,

but tanθ = f/N so μ=tanθ

only as the object starts to move does u = tan theta, where theta is the so called angle of friction.

This means that angle of friction θ , is also increasing

the angle of the resultant contact force F with the vertical is increasing..the angle of friction is the limiting value of theta ..too many thetas...

which tilts the vector of contact force further and also increases length of contact force vector.

yes

Now at some value of f_s , the block is just to move. We call that friction , limiting friction and μ_s as coefficient of limiting friction. Limiting friction is equal to minimum applied force to move that object. In this case tanθ_s = μ_s , where θ_s is known as critical angle of friction. In general f_s≤μ_sN.

yes

Now as block begins to move , the static friction is replaced by kinetic friction. Since f_k<f_s , so μ_k<μ_s , this must mean that angle of friction during kinetic friction must be reduced and the tilting of contact force vector will also reduce. But after this f_k will always remain same and so does μ_k except for body moving with high relative velocities.

OK

Also , in case of static friction , if μ→∞ , θ (or angle of friction ) → 90^o , does it not imply that N or normal reaction =0 ?

No. N is always just the weight mg if the applied force is horizontal. It is the friction force magnitude that changes as the appied load increases.

It is confusing to talk about 'angle of friction' and 'angle of the resultant contact force' in this example. Bottom line is that u can be greater than 1, and static friction force is less than or equal to uN.

 

[sankalpmittal]

OK I looked at this picture a bit more carefully and read your notes and its title. The title talks about the "three forces exerted by the ground on the object" , when in fact there are just two forces exerted by the ground on the object..the friction force, f , and the normal force, N. The force shown as F is the resultant of these 2 contact forces...it is not a 3rd force. The resultant contact force makes an angle theta with the vertical which is the Angle of friction at the limiting value of f =uN , OK. Not shown in the diagram is the weight force mg acting vertically down, and the applied force, P. Confusing sketch to me.



I think you've answered this yourself already. You found a site referencing u = 1.05.. I believe copper on copper can have u =1.3.. And slick racecar rubber on dry concrete, u = 1.7. In these cases, the friction angle theta is greater than 45 degrees. the frictional force is increasing, but it is only equal to uN at the limiting case when the object just starts to move. Otherwise, the increasing friction force will always be less than uN, and equal to the horizontal applied force. u does not increase, f_s increases. u is a constant for the materials involved, and is unchanging. μ > or = f/N , only as the object starts to move does u = tan theta, where theta is the so called angle of friction.the angle of the resultant contact force F with the vertical is increasing..the angle of friction is the limiting value of theta ..too many thetas... yes yes OK No. N is always just the weight mg if the applied force is horizontal. It is the friction force magnitude that changes as the appied load increases.

It is confusing to talk about 'angle of friction' and 'angle of the resultant contact force' in this example. Bottom line is that u can be greater than 1, and static friction force is less than or equal to uN.

Thanks. This problem has been resolved. Thanks a ton !

This problem was brain-bashing me for more than a week ! Now it has been solved.