# Friction

by Dustin Glazier and Heather Williams, June 1998

Here are the quantities you can know:
```
```
• ```F Force

```
• ```m Mass

```
• ```a Acceleration

```
• ```g Gravity

```
• ```Mu Coefficient of Friction

```
• ```Mus Coefficient of Static Friction

```
• ```Muk Coefficient of Kinetic Friction

```
• ```FN 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

```
```
1. ```F = ma

```
2. ```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 FN. The resistence of the material is called the coefficient of friction or Mu. There are two different types of coeffecients of friction, static (Mus) and kinetic (Muk). 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:
```
```
1. ```F = ma

```
2. ```W = mg

```
3. ```F = Mu FN

```
4. ```F = FA - FF

```

```Go back to: Table of Contents

```

## General Problem Solving Strategy:

2. Go through the problem and figure out what is given or implied
Make a list, and identify the quantities you know.
3. Find any formula that will allow you to calculate
anything that you don't know, and apply it.
4. Add what you just found in the last step to your list of knowns.
5. Check to see if you have found the answer. If not, repeat the
previous two steps until you are done.

## 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

```
• ```FN = ?

```
```Use the formula FN = mg,

So it would look like FN = (50 kg)(9.8 m/s/s). Therefore FN = 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

```
• ```FN = 490 N

```
• ```Mu = ?

```
```We take the formula F = Mu (FN)

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.

```

## 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

```
• ```Muk = .25

```
```Obviously we need to find the FN for:

FN = mg

Since we know m and g, FN = (100 kg)(9.8 m/s/s) = 980 N. Now that we have
FN we can use

F = Mu (FN)

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.

```

## Sample Problems

```

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) Go back to: Table of Contents

## 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)

## 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)
```
```
• ```W = 930 N

```
• ```Mu = .12

```
```Once again, our formula for Mu is:

F = Mu (FN)

It's pretty easy to plug-in the stuff we know:

F = (.12)(930 N)

We then do the math to get 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)
```
```
• ```F = 3.0 s

```
• ```W = 67 N

```
```Once again, our formula for Mu is:

F = Mu (FN)

We know that FN is equal to W so we can plug what we know into the formula and
get:

34 N = MuS (67 N)

Divide the 34 by 67 and you end up with .507462 which happens to round to the answer of
.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)
```
```
• ```F = 350

```
• ```Mu = .35

```
```Hey, since FN = W, surprisingly enough we use the formula:

F = Mu (FN)

which we plug-in the numbers and get

350 = .35 (FN)

Divide 350 by .35 to get 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)
```
```
• ```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 (FN) with FN 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 plug-in 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:

```
```
```
• ```Vo = 37 m/s

```
• ```V = 0 (It comes to a complete stop)

```
• ```a = 8.33 m/s/s

```
```Now we can use:

x = V2 - VO2

-----------------

2a

x = 43.7575 or with sig figs, 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)
```
```
• ```m = 800 grams or .8 Kg

```
• ```F = 1.2 N

```
• ```g = 9.8 m/s/s

```
```We first need to get the FN or Weight; we must convert the mass to kilograms
and then get FN = 7.84 N for the next step.

As usual when we look at the formulas we see that:

F = Mu (FN)

seems to be the only one that will uses Mu

We plug-in 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.

```

## 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)
```
```
• ```Mu = .85

```
• ```F = 9620 N

```
```We have to find the weight first so we use:

F = Mu (FN) and that equals

9620 N = (.85)(FN)

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

```

## 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)
```
```
• ```m = 5.0 kg

```
• ```F = 15 N

```
• ```Mu = .15

```
```So we can use the formula:

FN = mg or 15 N = (5.0 kg)(9.8 m/s/s)

So the FN = 49 N

Then we plug that into:

F = Mu (FN) 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

```

## 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)
```
```
• ```m = 10 Kg

```
• ```Vo = 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 = Vo + 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

FA = 100 N and FN = 64 N so the FF = 36 N

Now we plug that back into the formula:

FF = Mu (FN) we use our stuff

36 N = Mu (64 N)

Divide 36 by 64 and get .5625 as the coefficient of friction.

```

## 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)
```
```
• ```m = 120 Kg

```
• ```Mu = .28

```
• ```a = 1.4 m/s/s

```
```First we need to find FN and we use FN = 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 = FA - FF, with the values plugged in we get:

168 = FA - 329.28

solve this little problem and get 497.28 or 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)
```
```
• ```VO = 0 m/s

```
• ```V = 12 m/s

```
• ```Mu = .58

```
• ```x = 5.2 m

```
• ```FA = 24 N

```
```Letís use x = V2 - VO2

-------------

2a

we plug in what we know and get:

5.2 m = 122 - 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 = FA - FF but we donít know the FF

So FF = Mu * FN we plug stuff in and get:

Since F = ma and FF = Mu * m * g we can put it into

F = FA - FF so it looks like:

ma = FA - (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!!!!!