Friction
by Dustin Glazier and Heather Williams, June 1998
Table of Contents
Here are the quantities you can know:
F Force
m Mass
a Acceleration
g Gravity
Mu Coefficient of Friction
Mu_{s} Coefficient of Static Friction
Mu_{k} Coefficient of Kinetic Friction
F_{N} Normal force
These quantities are defined and explained on other pages except for the Coefficient
of friction, Static friction, Kinetic friction, and the Normal force
which is explained below
New Quantities
The formulas that you already know about dynamics are:

F = ma

W = mg
Defining Friction
You already have a good idea about Force. Friction is what stops that force. There
we're done. Just kidding. For example, if you're pushing a rock up a hill, there is
something that makes that more difficult than pushing that same rock down hill. Well,
that FORCE is friction and it is caused by both gravity and the resistence of
the
material that the object is being pushed along. The gravity, as you well know
mulitplied by the mass of the object gives its weight. This weight, in terms of
friction, is called the normal force or F_{N}. The resistence of the material is
called the coefficient of friction or Mu. There are two different types of coeffecients
of friction, static (Mu_{s}) and kinetic (Mu_{k}). If you're sliding a
box it will require a force greater than its static coefficient of friction because you
need to get the box moving. After the box starts to move it requires a certain force
greater than the kinetic coefficient to maintain motion.
Formulas
So now we have all the formulas we need for determining friction:

F = ma

W = mg

F = Mu F_{N}

F = F_{A}  F_{F}
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General Problem Solving Strategy:
 Read the problem.
 Go through the problem and figure out what is given or implied
Make a list, and identify the quantities you know.
 Find any formula that will allow you to calculate
anything that you don't know, and apply it.
 Add what you just found in the last step to your list of knowns.
 Check to see if you have found the answer. If not, repeat the
previous two steps until you are done.
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Example problem 1
Since sliding is such a great example, if you have a box that has a mass of 50 kg, what
is it's normal force?
Here's what we got:
M = 50 kg
G = 9.8 m/s/s
F_{N} = ?
Use the formula F_{N} = mg,
So it would look like F_{N} = (50 kg)(9.8 m/s/s). Therefore F_{N} = 490
Newtons.
Now if it takes 156.8 newtons of force to get the box moving, then what is it's static
coefficient of friction?
Here's what we got now:
F = 156.8 N
F_{N} = 490 N
Mu = ?
We take the formula F = Mu (F_{N})
With 156.8 N = Mu (490 N) we can to this with simple math.
156.8/490 = Mu = .32, which is the static coefficient of friction, and the answer.
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Example problem 2
You push a 100 kg rock down the road. If the kinetic coefficient of the rock and the
pavement is .25, what is the force required to keep the rock moving?
Here's what we know:
m = 100 kg
g = 9.8 m/s/s
Mu_{k} = .25
Obviously we need to find the F_{N} for:
F_{N} = mg
Since we know m and g, F_{N} = (100 kg)(9.8 m/s/s) = 980 N. Now that we have
F_{N} we can use
F = Mu (F_{N})
Which when plugged in with what we know is
F = (.25)(980 N)
So F = 245 N that need to be continually pressed to keep it moving.
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Sample Problems
The answers to each problem follow it in parentheses. They also link to a solution to
the problem. Try the problem, check your answer, and go to the solution if you do not
understand.
1.
What is the force of friction between a block of ice that
weighs 930 N and the ground if m = .12?
acceleration?
(111.6 N)
2.
What is the coefficient of static friction if it takes 34
N of force to move a box that weighs 67 N?
(.51)
3.
A box takes 350 N to start moving when the coefficient of
static friction is .35. What is the weight of the box?
(1000 N)
4.
A car has a mass of 1020 Kg and has a coefficient of
friction between the ground and its tires of .85. What force of friction can it exert on
the ground? What is the maximum acceleration of this car? In what minimum distance could
it stop from 27 m/s?
(8500 N, 8.3 m/s/s, 43.8 m)
5.
Clarice moves a 800 gram set of weights by applying a
force of 1.2 N. What is the coefficient of friction?
(.15)
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6.
A car has a coefficient of friction between the ground and
its tires of .85. What is the mass of the car if it takes 9620 N of force to make it
slide along the ground?
(1155 kg)
7.
A 5.0 Kg block has a coefficient of friction of .15 on a
flat surface. What is its acceleration if you exert a force of 15 N sideways on it when
it is at rest? (Find the friction force first)
(1.47 m/s/s)
8.
A 10 Kg block is at rest on a level surface. It
accelerates from rest to 51.2 m/s in 8 seconds when you exert a force of 100 N on it
sideways. What is the acceleration of the block? What is the force of friction between
the surface and the block, and what is the coefficient of friction?
(6.4 m/s/s, 36 N, .56)
9.
A 120 Kg log sled accelerates at 1.4 m/s/s when a horse
pulls on it. What force must the horse exert if the coefficient of friction between the
ground and the sled is .28?
(497 N)
10.
(E.C.  Super Stud Problem) You exert a force of 24 N
sideways on an object and it accelerates from 0  12 m/s over a distance of 5.2 m. You
know that the coefficient of friction between the object and the ground is .58, so what
is its mass?
(1.23 kg)
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Solutions to Sample Problems
1.
What is the force of friction between a block of ice that
weighs 930 N and the ground if Mu = .12?
(111.6 N)
Here is what you start with:
W = 930 N
Mu = .12
Once again, our formula for Mu is:
F = Mu (F_{N})
It's pretty easy to plugin the stuff we know:
F = (.12)(930 N)
We then do the math to get 111.6 N.
Go to: Problem Formulas Table of Contents
2.
What is the coefficient of static friction if it takes 34
N of force to move a box that weighs 67 N?
(.51)
Here is what you start with:
F = 3.0 s
W = 67 N
Once again, our formula for Mu is:
F = Mu (F_{N})
We know that F_{N} is equal to W so we can plug what we know into the formula and
get:
34 N = Mu_{S} (67 N)
Divide the 34 by 67 and you end up with .507462 which happens to round to the answer of
.51.
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3.
A box takes 350 N to start moving when the coefficient of
static friction is .35. What is the weight of the box? (1000 N)
Here is what you start with:
F = 350
Mu = .35
Hey, since F_{N} = W, surprisingly enough we use the formula:
F = Mu (F_{N})
which we plugin the numbers and get
350 = .35 (F_{N})
Divide 350 by .35 to get 1000 N.
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4.
A car has a mass of 1020 Kg and has a coefficient of
friction between the ground and its tires of .85. What force of friction can it exert on
the ground? What is the maximum acceleration of this car? In what minimum distance could
it stop from 27 m/s? (8500 N, 8.3 m/s/s, 43.8 m)
Here is what you start with:
m = 1020 Kg
Mu = .85
Use the formula
W = mg
with g = 9.8 m/s/s as the force of gravity. We now know the weight to be 9996 N. We can
then use the formula
F = Mu (F_{N}) with F_{N} equal to the weight.
Plugged in it looks like:
F = (.85)(9996 N) and equals 8496.6 N but with significant digits it ends up being 8500
N. In order to find the acceleration we need a formula that uses the Force we found and
acceleration. Obviously
F = ma
seems best. We plugin the numbers and get:
8496.6 N = (1020 Kg) a
Solve for a and get 8.33 m/s/s as the acceleration
But we're not done yet! We need to solve for the distance. Since this deals with good old
velocity, acceleration and distance we need a formula that uses only what we know:
V_{o} = 37 m/s
V = 0 (It comes to a complete stop)
a = 8.33 m/s/s
Now we can use:
x = V^{2}  V_{O}^{2}

2a
x = 43.7575 or with sig figs, 43.8 m
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5.
Clarice moves a 800 gram set of weights by applying a
force of 1.2 N. What is the coefficient of friction?
(.15)
Here is what you start with:
m = 800 grams or .8 Kg
F = 1.2 N
g = 9.8 m/s/s
We first need to get the F_{N} or Weight; we must convert the mass to kilograms
and then get F_{N} = 7.84 N for the next step.
As usual when we look at the formulas we see that:
F = Mu (F_{N})
seems to be the only one that will uses Mu
We plugin the numbers to get:
1.2 = Mu (7.84 N)
Divide 1.2 by 7.84 to get .15306 or .15 for the coefficient of friction.
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6.
A car has a coefficient of friction between the ground and
its tires of .85. What is the mass of the car if it takes 9620 N of force to make it
slide along the ground?
(1155 Kg)
Here is what you start with:
Mu = .85
F = 9620 N
We have to find the weight first so we use:
F = Mu (F_{N}) and that equals
9620 N = (.85)(F_{N})
9620 divided by .85 = 11317.647 N
But we need this in mass so we use:
F = mg
11317.647 = (m)(9.8 m/s/s)
so the mass equals 1154.8619 or 1155 kg
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7.
A 5.0 Kg block has a coefficient of friction of .15 on a
flat surface. What is its acceleration if you exert a force of 15 N sideways on it when
it is at rest? (Find the friction force first)
(1.47 m/s/s)
Here is what you start with:
m = 5.0 kg
F = 15 N
Mu = .15
So we can use the formula:
F_{N} = mg or 15 N = (5.0 kg)(9.8 m/s/s)
So the F_{N} = 49 N
Then we plug that into:
F = Mu (F_{N}) that looks like
F = (.15)(49 N)
And get the actual force to be 7.35 N
And then we can plug that back into the formula:
F = ma or 7.35 N = (5.0)a
Divide 7.35 by 5 and get 1.47 m/s/s
Go to: Problem Formulas Table of Contents
8.
A 10 Kg block is at rest on a level surface. It accelerates
from rest to 51.2 m/s in 8 seconds when you exert a force of 100 N on it sideways. What
is the acceleration of the block? What is the force of friction between the surface and
the block, and what is the coefficient of friction?
(6.4 m/s/s 36 N, .56)
Here is what you start with:
m = 10 Kg
V_{o} = 0 m/s
V = 51.2 m/s
t = 8 seconds
F = 100 N
Find the acceleration using the our freindly neighborhood aceleration formula once
again:
V = V_{o} + at.
Plug in a couple of numbers and get:
51.2 m/s = 0 m/s + (a * 8 s)
Solve for a, and we get 6.4 m/s/s
The force of friction = Force applied ñ The normal force
F_{A} = 100 N and F_{N} = 64 N so the F_{F} = 36 N
Now we plug that back into the formula:
F_{F} = Mu (F_{N}) we use our stuff
36 N = Mu (64 N)
Divide 36 by 64 and get .5625 as the coefficient of friction.
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9.
A 120 Kg log sled accelerates at 1.4 m/s/s when a horse
pulls on it. What force must the horse exert if the coefficient of friction between the
ground and the sled is .28?
(497 N)
Here is what you start with:
m = 120 Kg
Mu = .28
a = 1.4 m/s/s
First we need to find F_{N} and we use F_{N} = mg
F = (120 Kg)(9.8 m/s/s) and F equals 1176 N
We then multiply that by the coefficient of friction (.28)to get the force of friction
which is 329.28 N. We then use the formula:
F = F_{A}  F_{F}, with the values plugged in we get:
168 = F_{A}  329.28
solve this little problem and get 497.28 or 497 N.
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10.
(E.C.  Super Stud Problem) You exert a force of 24 N
sideways on an object and it accelerates from 0  12 m/s over a distance of 5.2 m. You
know that the coefficient of friction between the object and the ground is .58, so what
is its mass? (1.23 Kg)
Here is what you start with:
V_{O} = 0 m/s
V = 12 m/s
Mu = .58
x = 5.2 m
F_{A} = 24 N
Letís use x = V^{2}  V_{O}^{2}

2a
we plug in what we know and get:
5.2 m = 12^{2}  ^{2} / 2a
solve for acceleration and get a = 13.846 m/s/s
we want to find F = ma but weíll need to use:
F = F_{A}  F_{F} but we donít know the F_{F}
So F_{F} = Mu * F_{N} we plug stuff in and get:
Since F = ma and F_{F} = Mu * m * g we can put it into
F = F_{A}  F_{F} so it looks like:
ma = F_{A}  (Mu * m * g) we can then put in what we know
m (13.846) = 24 N ñ (.58 * m * 9.8m/s/s) Wait I donít Know!
13.846 = 24/m ñ 5.684 add 5.684m to both sides and NO That doesnít Work!
6.684m * (13.846) = 24 OHHH MAN!
Ohhh dang it this just ends up being 1.23 Kg!!!!!
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