PHYS 20 LESSONS Unit 3: Dynamics Lesson 7: Friction.

PHYS 20 LESSONS

Unit 3: Dynamics

Lesson 7: Friction

Reading Segment #1:

FRICTION

To prepare for this section, please read:

Unit 3: p.24

Friction

Definition:

Friction is the force thatresists the sliding of onesurface over another.

In other words, any time an object is slid across a surface, friction will resist that motion.

fF

To draw a proper force diagram involving friction, use the following two rules:

1. Friction force always acts in the opposite direction to the motion.

2. The friction force is always parallel to the surface.

The Direction of Friction

Imagine a box is moving to the right across a horizontal surface, as shown below:

v

What would be the direction of the friction force?

Notice that the friction force:

acts opposite to the motion (i.e. velocity)

is parallel to the surface

v

fF

Imagine a box is sliding down a slope, as shown:

What would be the direction of the friction force in this case?

v

Notice, again, that the friction force:

acts opposite to the motion (i.e. velocity)

is parallel to the surface

fF

v

Example

A person is pushing a rock against a vertical wall, but because its weight is so great, the rock slides down the wall at a constant speed.

Sketch a complete force diagram of the rock, assuming the person’s force is horizontal.

Describe the relationship between the forces.

First, draw a free body diagram of the rock.

Do not draw any other objects, since they are not part of the system.

Since it is helpful to see the wall, however, we draw it as a dotted line.

v

v

FpersonThe person is exerting a horizontal force on the rock, acting towards the wall.

This force will cause the rock to push into the wall. Based on Newton’s 3rd Law, the wall exert an equal but opposite force on the rock.

v

Fperson Fwall

Newton’s 1st law:

These two forces are equal but opposite, and so there is no horizontal acceleration.

v

Fperson Fwall

The force of gravity acts downward, to pull the rock down to the surface of the Earth.

v

Fperson Fwall

Fg

The friction force acts opposite to the velocity and parallel to the surface (i.e. wall)

v

Fperson Fwall

Fg

Ff

Newton’s 1st law:

Since there is no vertical acceleration (i.e. constant velocity), the vertical forces (Fg and Ff) are equal but opposite.

v

Fperson Fwall

Fg

Ff

Equation:

The magnitude of the friction force on an object is calculated using

Nf FF SI Units: Newtons (N)

where

FN is the magnitude of the normal force (N)

is the coefficient of friction (no units)

The Magnitude of Friction

Thus, the magnitude of the friction force depends on only two things:

1. Normal force

The greater the normal force, the greater the friction force. This is a direct relationship.

i.e.

Nf FF

That is, the harder the two surfaces are pressed together, the greater the friction.

If I wanted to increase the normal force between my shoes and the ground, how would I do it?

FN

By adding a backpack, I increase my weight (Fg). Thus, the normal force will be greater, which increases the friction force between the shoes and the ground.

FN

Fg

FN

Fg

If you press down harder on the surface, then the normal force will increase. This also increases the force of friction (i.e. it is harder to slide your shoe back and forth along the ground).

FN FN

The magnitude of the friction force also depends on:

2. Coefficient of friction ()

The coefficient is a number that indicates how rough the two surfaces are.

The greater the coefficient of friction, the rougher the two surfaces are (and thus, the greater the friction force).

Rubber on concrete:

= 1.02

Rubber on ice:

= 0.005

Rubber on concrete has a higher coefficient of friction, and so, it experiences a greater friction force (i.e. better traction)

If friction depends only on normal force and coefficient of friction (roughness), all other factors have no effect.

This is quite surprising, as we will see in the next couple of pages.

Speed has no effect on friction force!

Ff low speed

Ff high speed

Same friction force

The only difference here is that the faster vehicle will have more air resistance.

Surface area has no effect on friction force!

You may be wondering why drag cars have wide tires? Look at Eric Peterson’s explanation by clicking on this link:

Same friction force

Narrow tire

Wide tire

Wide Tires on Drag Cars - Explained

Example

The truck below has rear wheel drive.

What would you do to ensure that the rear tires have the greatest friction force (i.e. traction)?

To increase the normal force acting on the rear wheel, place additional weight at the back of the truck (over the rear wheels).

Greater FN means a greater Ff.

Increased FN

Greater weight over the rear wheels

To increase the coefficient of friction, change the tire material. Choose a rubber that grips the ground better, and you will increase the friction.

Choose a “tackier” tire to increase

Example

A 37 kg object is being pulled forward by a 400 N force across a level surface. The coefficient of friction between the object and the surface is 0.68.

Using force diagram analysis, determine the object's acceleration.

Try this example on your own first.Then, check out the solution.

Force Diagram:

Fg

FN

Fapp = 400 N

Ff37 kg

a

To analyze this motion, it is best to deal with the vertical and horizontal forces separately.

Vertical:

Fg

FN

37 kg

Newton’s 1st Law (balanced forces)

Be certain to state the principles first.

Vertical:

Fg

FN

37 kg

Newton’s 1st Law (balanced forces)

gN FF

Vertical:

Fg

FN

37 kg

Newton’s 1st Law (balanced forces)

gN FF

gm

= (37 kg) (9.81 N/kg)

= 362.97 N

Then,

Nf FF

= (0.68) (362.97 N)

= 246.82 N

Fapp = 400 N

Ff = 246.82 N 37 kg

a

Ref: Forward +Backward

Newton’s 2nd Law (unbalanced forces)

amFnet

For 2nd law, include a

reference system. Note that friction is negative.

Horizontal:

Fapp = 400 N

Ff = 246.82 N 37 kg

a

amFnet

m

FF

m

Fa fappnet

kg37

N82.246N400

= 4.1 m/s2 forward

Check:

Answered question

Proper units

2 sig digs

Reading Segment #2:

STATIC AND KINETIC FRICTION

To prepare for this section, please read:

Unit 3: p.24 - 25

There are two types of friction we need to consider in Physics 20:

1. Kinetic friction- when the object is sliding

2. Static Friction- when the object is not moving

Kinetic and Static Friction

This is the type of friction we have been talking about up until now. It exists only when the object is moving, and it always acts to resist the motion.

1. Kinetic Friction

This is a constant value, determined by the formula:

Nkkf FF

where k is the coefficient of kinetic friction

This type of friction happens when the object is not moving.

2. Static Friction

Static friction always resists the intended motion of the object.

Consider when a person pushes a box to the left, but it does not move.

What is the direction of the static friction force?

Static friction always goes opposite to the intended motion.

So, it is directed towards the right.

Intended motion

Ff (s)

Consider when an object is at rest on an incline.

What is the direction of the static friction force?

Gravity tries to pull the object down the slope. This then is the direction of the intended motion.

Static friction acts in the opposite direction to the intended motion, so it acts up the slope.

Intended motionFf (s)

Equation:

The maximum static friction possible between two surfaces is

Nssf FF max SI Units: Newtons (N)

where

FN is the magnitude of the normal force (N)

s is the coefficient of static friction

However, static friction is a reaction force.

As a result, the greater the force applied to the stationary object, the greater the static friction force.

So, most of the time, static friction is less than its maximum value

i.e.

Nssf FF

Fapp

If the applied force is small on a stationary object ...

FappFf (s)

... then the static friction force is also small (equal but opposite)

If the applied force is small on a stationary object ...

Fapp

If the applied force is large on a stationary object ...

FappFf (s)

... then the static friction force is also large (equal but opposite)

If the applied force is large on a stationary object ...

However, eventually the static friction force reaches a maximum value.

If the applied force grows larger than this maximum value, then the object will accelerate.

FappFf (s max)

a

The coefficient of kinetic friction is almost always less than the coefficient of static friction (k < s).

Comparing Kinetic and Static Friction

Materials s k

Steel on dry steel

Rubber on dry concrete

Rubber on dry concrete

0.41 0.38

1.0 0.7

0.7 0.5

e.g.

Thus, kinetic friction is almost always less than the maximum static friction.

Kinetic friction and static friction are best compared and understood in the following situation.

Imagine exerting an increasing amount of force on a stationary object, as shown below:

The nature of the friction force (as the applied force increase) is summarized in the graph below:

Magnitude of friction force, Ff

Time, t

Can you interpret the graph?

Magnitude of friction force, Ff

Time, t

At restWhen there is no applied force, there is no friction force.

At rest

Since static friction is a reaction force, it is equal but opposite to the applied force.

So, as the applied force increases, static friction increases.

Magnitude of friction force, Ff

Time, tStationary

Static friction

Fapp

Ff (s)

Nssf FF

At restEventually, static friction reaches a maximum value.

Magnitude of friction force, Ff

Time, tStationary

Max static friction

Fapp

Ff (s max)

Nssf FF max

If the applied force increases some more, the object begins to move (accelerate).

Now, the box experiences kinetic friction, which is less than the max static friction.

So, the friction force drops.

Magnitude of friction force, Ff

Time, t

Starts to move

a

FappFf (s max)

Once the object begins to move, the kinetic friction remains constant (no matter how fast it moves).

Magnitude of friction force, Ff

Time, tMoving

v

FappFf (s max)

Example

Explain why it is easier to slide a heavy object across a floor when it is already moving (as compared to when it is at rest).

v

According to Newton’s first law, an object that is in motion wants to stay in motion (due to inertia). This makes it much easier.

A stationary object wishes to stay at rest, and so, it would have to be accelerated forward. This requires much more force.

Also, since the object is moving, the object experiences kinetic friction.

Kinetic friction is always less than static friction, and so less force is required to overcome it (compared to the stationary object).

v

Ff (k)

Example

A 70.0 kg oak desk is at rest on a level oak floor. A person will push the desk with a horizontal force.

Using the table on Unit 3 p. 25, compare what happens when he pushes with aforce of:

a) 300 Nb) 500 N

Try this example on your own first.Then, check out the solution.

If Fapp < Ff (s max), then it will not move.

If Fapp > Ff (s max), then it will move and the oak desk will experience kinetic friction.

Strategy:

Determine the maximum static friction force.

Force Diagram:

Fg

FN

Fapp

Ff70.0 kg

To analyze this motion, it is best to deal with the vertical and horizontal forces separately.

Vertical:

Fg

FN

70.0 kg

Newton’s 1st Law (balanced forces)

gN FF

gm

= (70 kg) (9.81 N/kg)

= 686.7 N

s k

Dry oak on dry oak 0.5 0.3

Based on the table above, the maximum static friction would be

Nssf FF max

= (0.5) (686.7 N)

= 343.35 N

a) When Fapp = 300 N, it is less than Ff (s max).

Thus, the box will not move.

At restFapp = 300 N

Ff (s) = 300 N

Notice that static friction is a reaction force. It will be equal but opposite to the applied force.

b) When Fapp = 500 N, it is greater than Ff (s max).

Thus, the box will move and will experiencekinetic friction.

s k

Dry oak on dry oak 0.5 0.3

Nkkf FF

= (0.3) (686.7 N)

= 206.01 N

a

Fapp = 500 NFf (k) = 206 N

Ref: Forward + Backward

Newton’s 2nd law (unbalanced forces)

amFnet

m

FF

m

Fa kfappnet

kg0.70

N206N500

= 4 m/s2 forward

Check:

Answered question

Proper units

1 sig dig

Practice Problems

Try these problems in the Physics 20 Workbook:

Unit 3 p. 16 #1 - 3, 6, 7

Remember:

When you are dealing with diagonal forces,resolve them into their x- and y-components.

Then, analyze the vertical and the horizontalforces separately.

2-D Friction Questions

Example

An 18.5 kg box experiences a diagonal force of 190 N, and as a result, accelerates forward at 2.50 m/s2.

24.0

F = 190 N 2.50 m/s2

18.5 kg

If it also is moving forward, then determine:a) the magnitude of the normal forceb) the coefficient of friction

Force diagram:

24.0

F = 190 N 2.50 m/s2

18.5 kg

FN

Fg

Ff

24.0

F = 190 N

Fy

Fx

Since F is diagonal, find its x- and y-components:

F

Fy24sin

N28.7724sinN190 yF

F

Fx24cos

N57.17324cosN190 xF

2.50 m/s2

18.5 kg

FN

Fg

Ff

Fy = 77.28 N

Fx = 173.57 N

This is a complete force diagram.

We are now able to analyze its motion horizontally and vertically.

18.5 kg

FN

FgFy = 77.28 N

Newton’s 1st law (balanced forces)

a) Vertical:

Sum of up forces =

Sum of down forces

gyN FFF

mgFF yN

= (77.28 N) + (18.5 kg) (9.81 N/kg)

= 259 N

Newton’s 2nd law (unbalanced forces)

b) Horizontal:

amFnet

2.50 m/s2

18.5 kg

Ff

Fx = 173.57 NRef: Forward +

Backward

amFF fx

xf FamF

= (18.5 kg) (2.5 m/s2) 173.57 N

= 127.32 N

To determine the coefficient of friction:

Nf FF

N

f

F

F

N77.258

N32.127

= 0.492

Practice Problems

Try these problems in the Physics 20 Workbook:

Unit 3 p. 16 #4, 5