Theory of Friction

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  • #1
izmeh

Main Question or Discussion Point

I'm currently enrolled in a "into to physics" couse at college.
I'm not doing so hot in the course, mainly because I can't understand the instructor. His native language is not english. His strategy of teaching doesn't work for me either. He tells you how to solve something, but he doesn't tell you why it needs to be done this way, and never refers to the book.
With that minor explanation out of the way...
We are currently in a section covering the theory behind physics. I'm having problems in actually working with the "formulas" or what not. I understand the working of the idea of friction, but can't follow the book on the usage of formulas.

Is anyone willing to give me a break down on that?
 

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  • #2
chroot
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Hi izmeh,

Your question is probably a little too vague to be answered. Why don't you post some of the problems you're having trouble solving? We'll help you figure out how to solve them.

- Warren
 
  • #3
izmeh
I didn't want to bust in with just asking for help on my homework...
where should i post a question at?
 
  • #4
chroot
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Well, the best place to ask for help on specific homework problems is in our Homework Help forum. On the other hand, it sounds like you're having problems with the general concepts you need to solve your problems. It might be helpful for you to post three or four of those related problems for us here. We'll try to help you understand the best way to approach them.

- Warren
 
  • #5
izmeh
I don't understand how the force of friction can be determiined if it is dimensionless.

Well, this is the first 2 problems of my assignment.

1. A baseball player slides into third base with an initial speed of 7.9m/s. If the coefficient of kinetic friction between the player and the ground is 0.41, how far does the player slide before coming to rest?

2. Hopping into your Porsche, you floor it and accelerate at 12m/s² without spinning the tires. Determine the minimum coefficient of static friction between the tires and the road needed to make this possible.

As cheesy as those sound, they are from my college txt book.
 
  • #6
NateTG
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Ok, first off friction, and the friction constant:

Someone (I unfortunately don't remember who) was testing friction and discovered that the force of friction (both static and sliding) is proportional to the normal force so:

Ffriction=Fnormal * μ
or
Ffriction/Fnormal=μ

Dimensional analysis makes it clear that μ should be dimensionless.

μ depends on the material that both surfaces are made of. (Your text should have a table of coefficients.)

One theory for the mechanism of friction is essentially that when you rub two materials against each other, the surface molecules on the materials displace each other electromagnetically and then bounce back into place heating the materials.
 
  • #7
chroot
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Originally posted by izmeh
I don't understand how the force of friction can be determiined if it is dimensionless.
Well, the force due to friction is not dimensionless -- it has units of force, e.g. newtons.

The coefficient of friction is dimensionless, since all it does is relate two forces. Take a look at the "equation" for friction forces:

F = [mu](s or k) N

where N is the normal force pressing the two bodies together and F is the resulting perpendicular force.

N is in newtons, and so is F. Therefore, [mu] is dimensionless, and simply serves as a proportionality constant between the two forces.
1. A baseball player slides into third base with an initial speed of 7.9m/s. If the coefficient of kinetic friction between the player and the ground is 0.41, how far does the player slide before coming to rest?
The first thing to do in most problems is to find the forces experienced by the moving body(ies). The force in this case is just

F = [mu]k m g

Keep in mind that you don't have to actually find a number here -- you don't know what the mass is, so you can't! You can leave it in this symbolic form and use it just fine, however.

Now, we can find the acceleration, according to Newton's second law:

F = m a

Substituting the expression for F above, this is

[mu]k m g = m a

The m's cancel, leaving us with

a = [mu]k g

Now that you have the acceleration, you can just use your knowledge of kinematics to find out how far he slides. The baseball player has initial velocity v0, and accelerates at -[mu]k g. (The minus sign to reflect the fact that the baseball player is slowing down. In other words, every second, the baseball player's velocity decreases by [mu]k g. You can therefore find his velocity at any time t with this equation:

v(t) = v0 - a t
= v0 - [mu]k g t

To find out when the baseball player stops, you just need to find out when his velocity is zero:

0 = v0 - [mu]k g t
t = v0 / ([mu]k g)

Now you know how long the baseball player is in motion, and you know his velocity at all times. To find the distance he slides, you'll need to use this equation:

s(t) = v0 t + 1/2 a t2

where s(t) is the position of the object at time t, v0 is its initial velocity, and a is its acceleration. t is the value you found above, namely v0 / ([mu]k g).

Solve this equation for s, and you'll have your answer.

Hopefully these equations are familiar to you -- if they are not, let me know and I'll help you derive them. It is important to resist the temptation to approach a physics problem with a bag of equations you don't really understand. Truly understanding what you're doing is far more valuable than just having memorized the right equation.

The second problem sounds very similar. Can you solve the first one?

- Warren
 
  • #8
russ_watters
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Originally posted by chroot
Well, the force due to friction is not dimensionless -- it has units of force, e.g. newtons.

The coefficient of friction is dimensionless, since all it does is relate two forces.
To expand, izmeh, a coefficient MUST be dimensionless by definition. The purpose is to relate other numbers that have the same dimensions. For example, in the friction equation, one force times the coefficient yields another force. If you are having trouble with how that works, consider the dimensions to be just another variable to be manipulated algebraically like any other variable. In this case, when you divide one force by another, the dimensions cancel each other out(say for example newtons: N/N=1). So the resulting coefficient has no dimensions.
 
  • #9
izmeh
Well, I at least understand the process of what you have given me....however my answer does not match the back of the book....it gives 7.8m as an answer...

this is what I did
F = [mu]k Mg
F = ma
[mu]Kg = a

0 = 7.9m/s - [mu]Kgt
t = 7.9m/s/[mu]Kg = 7.9/4.018 = 1.96

then

s(t) = v0 t + 1/2 a t²
s(t) = 7.9(1.96) + 1/2 (4.018)(1.96²) = 23.20

obviously that is wrong
 
  • #10
chroot
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Originally posted by izmeh

s(t) = v0 t + 1/2 a t²
s(t) = 7.9(1.96) + 1/2 (4.018)(1.96²) = 23.20

obviously that is wrong
Ooops! The acceleration should be negative.

Try s(t) = 7.9(1.96) - 1/2 (4.018)(1.96²)

- Warren
 
  • #11
izmeh
Originally posted by russ_watters
To expand, izmeh, a coefficient MUST be dimensionless by definition. The purpose is to relate other numbers that have the same dimensions. For example, in the friction equation, one force times the coefficient yields another force. If you are having trouble with how that works, consider the dimensions to be just another variable to be manipulated algebraically like any other variable. In this case, when you divide one force by another, the dimensions cancel each other out(say for example newtons: N/N=1). So the resulting coefficient has no dimensions.
Thank you for that explination, that puts that in perspective for me...
 
  • #12
izmeh
Originally posted by chroot
Ooops! The acceleration should be negative.

Try s(t) = 7.9(1.96) - 1/2 (4.018)(1.96²)

- Warren
That's the prob, i got the answer now...greatly appreciated...
i'll post my work for the second one soon..
 
  • #13
izmeh
or first, could you give me a better explination of Static Friction?
 
  • #14
chroot
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Static friction and kinetic friction are really the same effect. They both have the same cause -- molecules on both surface make weak bonds with each other, which cause the objects to resist sliding.

The only reason we differentiate kinetic and static friction is because the coefficient is different when the objects are sliding than when they are not.

Here's a good page:

http://hyperphysics.phy-astr.gsu.edu/hbase/frict2.html

(Hyperphysics, by the way, is an excellent site for all kinds of basic physics -- you might want to bookmark it.)

http://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html

- Warren
 
  • #15
izmeh
this my attempt at the second problem

f=[mu]smg; f=ma
[mu]smg = ma
[mu]s = a/g

[mu]s = 12/9.8 = 3.77

not sure where to go with that....i'm suppose find the minimum threshold, but how?
 
  • #16
NateTG
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That is the minimum:

You can't be sure that μs isn't any higher than that because the car didn't slip, right?
OTOH if μs were lower, you'd have seen slippage.
 
  • #17
izmeh
Originally posted by NateTG
That is the minimum:

You can't be sure that μs isn't any higher than that because the car didn't slip, right?
OTOH if μs were lower, you'd have seen slippage.
the books answer is 1.2
 
  • #18
chroot
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Originally posted by izmeh
[mu]s = 12/9.8 = 3.77

not sure where to go with that....i'm suppose find the minimum threshold, but how?
Well, other than the fact that your calculator appears to be malfunctioning I think you're done.

- Warren
 
  • #19
izmeh
Originally posted by chroot
Well, other than the fact that your calculator appears to be malfunctioning I think you're done.

- Warren
crap
why didn't i figure that out....
my input was 12^9.8
 
  • #20
chroot
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If you have any more general questions, feel free to ask them here -- if you have some more difficult homework q's, you can post them in the Homework Help forum. Glad we could help!

- Warren
 
  • #21
izmeh
i really appreciate the help, i'm sure i'll be getting back to you soon.
 
  • #22
izmeh
The nex problem;

To move a large crate across a rough floor, you push down on it at an angle of 21˚. Find the force necessary to start the crate moving, given that the mass of the crate is 32Kg and the coeffecient of static friction between the crate and the floor is .57

Is it necessary to find the acceleration in this problem?

Exactly how does the angle effect it all?
 
  • #23
chroot
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Originally posted by izmeh
Is it necessary to find the acceleration in this problem?
Nope -- the only requirement for the crate to move is that the force you apply overcomes the force due to friction. That force has a definite maximum, [mu]s m g. All you need to do is apply a horizontal force of at least this magnitude, and the box will begin moving.
Exactly how does the angle effect it all?
Well, the force is a vector, right? And you can decompose a vector into components -- in this case, you'd decompose it into the horizontal and vertical components, like this:
Code:
  .        |
    .      |
      .    | Fy
     F  .  |
        o .|
 ---------->
    Fx
(I assume the angle, o, is 21 degrees from the horizontal.)

The magnitude of the horizontal component of the force, Fx, is what opposes friction. Fx = F cos(o) of course.

So you just need to solve

F cos(o) = [mu]s m g

for F.

- Warren
 
  • #24
izmeh
I'm getting

Fcos(21) = .57(32)9.8
F(.9335) = 178.752

F = 178.752/.9335 = 191.46

yet, the book has 250N
 
  • #25
chroot
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Hmm, I hate to ask this -- but did you copy all the numbers properly from the problem?

- Warren
 

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