Forces

Homework:Physics: Handout #1 – 9, 12, 14, 15,

but all are highly recommended.

Forces:

A force is a _____________ or a ____________ on an object by another object.

push pull

Forces can be applied either through direct contact, as shown in the picture below (a), (b), and (c).

Forces shown on the right are called “action at a distance”, where no physical contact is required between the objects to cause a force.

A force is tied to the motion of an object through ________________ .inertia

Inertia is the ability of an object to ________________ a change in its motion.

resist

Inertia is measured as the__________ of an object. The standard unit is the kilogram, kg.

mass

The more massive an object is, the more it is able to resist a change in its motion. It is easier to push a 10 kg object into motion than it is a 3000 kg car.

All of this begins the topic of _______________, or why objects move.dynamics

Newton’s Laws of Motion:

The relationship between motion and force was stated clearly by Isaac Newton in his 3 laws. Each law relates a different aspect of motion.

Newton’s First Law:

If the total, or net, force on an object is zero, then an object will not accelerate.

First variation: If an object is at rest, it will continue to remain at rest until acted upon by some external agent.

Example: A book placed on a desk will remain on the desk until someone removes it. There are forces acting on the book: gravity pulls down on it and the table pushed upwards on it. These forces balance out, and the object does not move.

Newton’s First Law: continued

Second variation: If an object is moving, it will continue to move with a constant velocity. Since velocity is a vector, this means the speed (how fast) and the direction remain unchanged.

Example: Take a look at a hockey game. When the puck (small black projectile) is hit, the puck travels at nearly the same speed and in a straight line across the ice. All real objects are subject to friction for anything sliding on a surface, and an icy surface is as close to frictionless as we can get.

Newton’s Second Law:

If the net force on an object is not zero, then an object will accelerate. The acceleration of the object is directly proportional to the total force acting upon it, and the acceleration is inversely proportional to the mass of the object.

As an equation: amFFnet

Here, m is the mass of the object, measured in kg.

a is the acceleration of the object, measured in m/s2. This is a vector quantity

F refers to the amount of force. Fnet or F is the sum of the forces acting on an object. Force is a vector, so it will have magnitude and direction. The net force and the acceleration point in the same direction.

Units: Force is measured in units called newtons, and this unit is represented by the letter N.

Newtons can be written in term of fundamental units through the statement of Newton’s Second Law:

22

1111s

mkg

s

mkgN

Example #1: What is the acceleration of a 2.00 kg mass if a force of 4.00 N acts upon it?

kg

N

m

Fa

00.2

00.4 200.2

sm

Example #2: An object with a mass of 3.00 kg moves from rest to a speed of 20.0 m/s in a time of 5.00 s. a. What is the acceleration of the object?

Start with: atvv o and solve for a:

200.400.5

00.20s

msm

o

st

vva

b. What is the net force acting on the object?

200.400.3s

mnet kgmaF N0.12

Both force and acceleration are vectors. The net force points in the same direction as the acceleration of the object.

Example #2: c. Let the object’s acceleration point to the right. If there is a drag force of 30.0 N opposing the motion, what other applied force must there be on the object to give the desired acceleration?

A picture helps!

dragappliednet FFF

appliedF

dragF

For one dimension, the vector signs are dropped.It is understood all vectors point along the same line.Just use a (+) sign for the vectors pointing right and a (–) sign for the vectors pointing left.The variables are assumed to be only the magnitude of the vectors.

dragappliednet FFNF 0.12

NNFFF dragnetapplied 0.300.12

NFapplied 0.42

Newton’s Third Law:

If one object (object A) exerts a force on another object (object B), then object B exerts the same force back on object A. The force is equal in magnitude, but opposite in direction.

Alternate interpretation: For every action there is an equal but opposite reaction.

Example: If you drive your fist against a wooden wall as hard as you can, you will possibly leave a mark on the wall with your fist. The mark is made by the force you exerted against the wall. The wall also exerts an equal force on your fist. You perceive this as the severe pain felt in your knuckles.

Drawing:

object Aobject B

FB on A

FA on BAonBBonA FF

Example #3: A student pushes with a 30.0 lb force towards the South on a heavy box. There is a friction (drag) force between the floor and the box that prevents the movement of the box. What is the magnitude and direction of the friction force?

Since the box does not move, the net force must be zero.

appliedF

dragF

dragappliednet FFF

0

northlbFF applieddrag 0.30

Example #4: A baseball is thrown at a speed of 40.0 m/s towards the right. The ball is hit by a baseball bat, and then travels at 50.0 m/s towards the left. The ball is in contact with the bat for 2.00 ms and the ball has a mass of 200 grams. What is the net force acting on the ball?

Initial information: Let right be positive and left be negative.vo = +40.0 m/s, v = – 50.0 m/s, t = 2.00 x 10-3 s, m = 0.200 kg

atvv o

2000,45

1000.2

0.400.503 s

msm

sm

o

st

vva

2000,45200.0s

mnet kgmaF N9000

A horse is hitched to a wagon. Which statement is correct?

Example #5

A. The force that the horse exerts on the wagon is greater than the force that the wagon exerts on the horse.

B. The force that the horse exerts on the wagon is less than the force that the wagon exerts on the horse.

C. The force that the horse exerts on the wagon is just as strong as the force that the wagon exerts on the horse.

D. The answer depends on the velocity of horse and wagon.

E. The answer depends on the acceleration of horse and wagon.

A horse is hitched to a wagon. Which statement is correct?A. The force that the horse exerts on the wagon is greater than the force that the wagon exerts on the horse .

B. The force that the horse exerts on the wagon is less than the force that the wagon exerts on the horse.

C. The force that the horse exerts on the wagon is just as strong as the force that the wagon exerts on the horse.

D. The answer depends on the velocity of horse and wagon.

E. The answer depends on the acceleration of horse and wagon.

Example #5

You are standing at rest and begin to walk forward. What force pushes you forward?

A. the force of your feet on the ground

B. the force of your acceleration

C. the force of your velocity

D. the force of your momentum

E. the force of the ground on your feet

Example #6

You are standing at rest and begin to walk forward. What force pushes you forward?

A. the force of your feet on the ground

B. the force of your acceleration

C. the force of your velocity

D. the force of your momentum

E. the force of the ground on your feet

Example #6

Table Cloth Pull: An example of the Law of Inertia.

Tension, Gravity, and Normal Force

Homework:

Physics Day 1: #1 – 7, 10

Physics Day 2: #8, 9, 11 – 14,

15 a,b,c.

Definitions: Forces are any kind of push or pull on a system or on an object. There are some special types of forces that we will define today. The first is the weight of an object.

The weight of an object is the force of ___________ acting on the object by Earth.

gravity

The amount of force is directly proportional to the ____________ of the object.

mass

The equation for weight is as follows: mgw

w = weight of the object. This is a force measured in N (newtons).

m = mass of the object, in kg (kilograms).

g = acceleration of gravity, 9.80 m/s2.

The direction of the force is downwards, towards the center of Earth.

Example #1: What is the weight of a 55.0 kg person?

280.90.55s

mkgmgw N539

Example #2: In the British system, mass is measured in slugs and force is measured in pounds. Weight is still measured as mg, with one slug mass weighing 32.2 pounds. Convert the mass of the person above into slugs.Conversion Factor: 1.0000 kg mass weighs 2.2046 lbs

2

121

32.2 fts

w lbm

g slugs77.3

Definition: The __________ force on an object is defined as the force exerted on an object by a surface. The term “normal” is another term used in mathematics for perpendicular. The normal force exerted by a surface is always perpendicular to that surface.

The normal force is represented by the letter _________ . Do not confuse this with newtons of force!

n

Problem Solving: Always draw a picture, and label all the forces acting on the object. This is referred to as drawing a “Free Body Diagram”.

Only draw the forces acting on the object. By Newton’s third law, the forces on an object result also in forces exerted by the object on the surroundings. We are only interested in the forces on the object by the environment.

normal

Example #3: A 1.4 kg book is placed on a flat surface at rest. What is the normal force acting on the book?

Start with a labeled picture:

m = 1.4 kg

weight = mg

normal = nNext apply Newton’s Second Law:

amFnet

For one dimension, define up as positive and down as negative. The forces may be written as follows:

and

The variables show the magnitude, or size of the vector. The ± sign gives the direction.

The net force becomes:

mgnwnFnet

Since the object is at rest, the acceleration of the object is zero.

0 amFnet

Combine the above equations and solve for n.

mgnFnet 0

280.94.1s

mkgmgn N7.13

Example #4: {Classic Physics Problem!!!} A person stands in an elevator on a scale. (Don’t ask why…..) When the elevator is at rest, the scale reads 588 N. What will the scale read if the elevator accelerates upwards at 2.20 m/s2?

Solution: The scale is designed to measure a force. The scale does not measure the weight of the person, rather the force exerted upwards on the person to support them. In other words, the scale measures the normal force acting on the person.

When the person is at rest, the normal force equals the weight of the person. When the person is accelerated in the elevator, the net force on the person is no longer zero. For an upward acceleration, the normal force must be larger than the weight of the person. Let’s find out why.

Before anything, determine the mass of the person:

At rest, the person weighs 588 N. This equals mg, so the mass is:

kgN

g

wm

sm

0.6080.9

588

2

Start by drawing a complete picture and then apply Newton’s Second Law.

elev

ator

pers

on,

m

weight = mg

normal = nDraw in a coordinate axis and choose a positive direction.

up = positive

down = negative

Apply Newton’s 2nd Law:

mgnmaFnet

mgman

gamn

Substitute numbers: Solve for the new normal force for the accelerated elevator.

22 80.920.20.60s

ms

mkggamn

720.n N

The person appears to weigh more when the elevator accelerates upwards.

Example #5: What is the motion of the elevator if the person’s weight appears to be 360 N?

Just follow the same reasoning as above and show again:

Solve this expression for a.

2

360.9.80

60.0m

s

n Na g

m kg 280.3

sm

Since up was chosen as the positive direction, a negative acceleration means that the elevator is accelerating downwards.

mgnmaFnet

Example #6: What would the scale read if the elevator was in freefall?

Just follow the same reasoning as above and show again:

gamn

Solve this expression for n. Note that a = – g.

0 ggmgamn

For freefall, the scale would read zero. The person would feel weightless.

Definition: A __________ force is a ____________ force, usually applied through a string or rope.

Ropes are to be treated as ideal, meaning that the mass of the ropes can be ignored and the ropes do not stretch or sag when used.

tension pulling

The force of tension is then the same throughout the rope.

Example #7: Two masses rest on a frictionless horizontal surface. The mass on the left is 60.0 kg and the mass on the right is 40.0 kg. The masses are tied together with a light cord and a 25.0 N force is applied towards the right on the 40.0 kg mass. Find the acceleration of the system and the tension force in the cord.

m1 = 60.0 kgm2 = 40.0 kg

cord

applied forceFa = 25.0N

Solution: First treat the two masses as a single object. They move the same: same velocity and acceleration.

mtotal = m1 +m2

applied forceFa = 25.0N

Apply Newton’s 2nd Law: appliedtotalnet FamF

kg

N

mm

F

m

Fa applied

total

applied

100

0.25

21

2250.0

sm

Now isolate one mass to solve for the force of tension in the cord. If the mass on the left (m1) is chosen, the solution is as follows:

m1

T = tension force

Apply Newton’s 2nd Law and solve for the tension force.

TamFnet 1

2250.00.601 smkgamT N0.15

An alternate solution would be to choose the mass on the right (m2).

m2T = tension force

applied forceFa = 25.0N

Apply Newton’s 2nd Law and solve for the tension force. Let the positive direction be towards the right.

TFamF appliednet 2

2250.00.400.252 sm

applied kgNamFT

NNT 0.100.25 N0.15

Example #8: A mass (m1 = 10.0 kg) hangs from the ceiling of an elevator from a light cord. A second mass (m2 = 20.0 kg) is suspended from the first mass by another cord. If the elevator accelerates upwards at a rate of 4.70 m/s2, determine the tension in each cord.

elev

ator

cord with tension T1

cord with tension T2

m1

m2

up = positive

down = negative

Solution: First treat the two masses as a single object and solve for the tension in the top cord. Start with a simple picture.

T1

mtot

mtotg

Apply Newton’s 2nd Law and solve for the tension force. Let the positive direction be upwards.

gmTamF tottotnet 1

gmamT tottot 1

22 80.970.40.301 sm

smkgT

NT 4351

Now isolate one mass to solve for the force of tension in the lower cord. If the mass on the bottom (m2) is chosen, the solution is as follows:

T2

m2

m2g

Apply Newton’s 2nd Law and solve for the tension force. Let the positive direction be upwards.

gmTamFnet 222

gmamT 222

22 80.970.40.202 sm

smkgT

2 290.T N

If the mass on the top (m1) is chosen, the solution is as follows:

T1

m1

m1gT2

Apply Newton’s 2nd Law and solve for the tension force. Let the positive direction be upwards.

gmTTamFnet 1211

gmamTT 1112

22 80.970.40.104352 sm

smkgNT

2 290.T N

Q4.1

v

Motor

Cable

Elevator

An elevator is being lifted at a constant speed by a steel cable attached to an electric motor. There is no air resistance, nor is there any friction between the elevator and the walls of the elevator shaft.

The upward force exerted on the elevator by the cable is

A. greater than the downward force of gravity.

B. equal to the force of gravity.

C. less than the force of gravity.

D. any of the above, depending on the speed of the elevator.

A4.1

v

Motor

Cable

Elevator

An elevator is being lifted at a constant speed by a steel cable attached to an electric motor. There is no air resistance, nor is there any friction between the elevator and the walls of the elevator shaft.

The upward force exerted on the elevator by the cable is

A. greater than the downward force of gravity.

B. equal to the force of gravity.

C. less than the force of gravity.

D. any of the above, depending on the speed of the elevator.

A woman pulls on a 6.00-kg crate, which in turn is connected to a 4.00-kg crate by a light rope. The light rope remains taut. A. is subjected to the same net force and has the same acceleration.

B. is subjected to a smaller net force and has the same acceleration.

C. is subjected to the same net force and has a smaller acceleration.

D. is subjected to a smaller net force and has a smaller acceleration.

E. none of the above

Q4.9

Compared to the 6.00–kg crate, the lighter 4.00-kg crate

A woman pulls on a 6.00-kg crate, which in turn is connected to a 4.00-kg crate by a light rope. The light rope remains taut. A. is subjected to the same net force and has the same acceleration.

B. is subjected to a smaller net force and has the same acceleration.

C. is subjected to the same net force and has a smaller acceleration.

D. is subjected to a smaller net force and has a smaller acceleration.

E. none of the above

A4.9

Compared to the 6.00–kg crate, the lighter 4.00-kg crate

2 – Dimensional Forces

Homework:Handout #3,

all problems.

ConcepTest 4.17 Three BlocksThree blocks of mass 3m,

2m, and m are connected

by strings and pulled with

constant acceleration a.

What is the relationship

between the tension in

each of the strings?

1) T1 > T2 > T3

2) T1 < T2 < T3

3) T1 = T2 = T3

4) all tensions are zero

5) tensions are random

T3 T2 T13m 2m m

a

ConcepTest 4.17 Three Blocks

1) T1 > T2 > T3

2) T1 < T2 < T3

3) T1 = T2 = T3

4) all tensions are zero

5) tensions are random

T3 T2 T13m 2m m

a

TT11 pulls the wholewhole setset of

blocks along, so it must

be the largestlargest. T2 pulls

the last two masses, but

T3 only pulls the last

mass.

Forces are _________ quantities: they have both a _____________

and a ____________.

vector magnitude

direction

For Newton’s 2nd Law, both the net force and the acceleration are vectors. It is easier to resolve the vectors into components to solve 2 – dimensional problems. On any given problem, draw in a coordinate set and label which directions are positive .

Example #1: A 40.0 kg block rests on a smooth surface. A force of 400 N is applied to the block at an angle of 36.9o above the horizontal. a. Label all forces acting on the object, and resolve these forces into horizontal and vertical components.

m

mg = weight

n = normal Fa

= 36.9o

m

mg = weight

n = normal Fa

= 36.9o

The normal force has only a y – component: nx = 0, ny = +n

+x

+y

The weight has only a y – component: wx = 0, wy = –mg

The x – component of the applied force, Fa, is: Fa,x = +Facos

The y – component of the applied force, Fa, is: Fa,y = +Fasin

Ex. #1 b. Determine the size of the normal force acting on the block.

Apply Newton’s 2nd Law to the y – components and solve for the normal force.

0 yynet maF

The y – component of the acceleration is zero. The y – component of the applied force is not larger than the weight of the object, so the object will not be lifted from the surface. It will only slide parallel to the surface.

sin0 aynet FmgnF

sinaFmgn

os

m Nkgn 9.36sin40080.90.40 2

Nn 152

Ex. #1 c. What is the acceleration of the block? How fast will it be moving after traveling 10.0 meters?

Apply Newton’s 2nd Law to the x – components and solve for the acceleration.

cosaxxnet FmaF

oax kg

N

m

Fa 9.36cos

0.40

400cos 200.8

sm

Reach back to chapter 2 for a motion equation: xavv xoxx 222

mxavvs

mxoxx 0.1000.8202 2

22

sm

xv 6.12

Case 1Case 1

Case 2Case 2

ConcepTest 4.10 Normal Force

Below you see two cases: a Below you see two cases: a

physics student physics student pullingpulling or or

pushingpushing a sled with a force a sled with a force FF

which is applied at an angle which is applied at an angle . In . In

which case is the normal force which case is the normal force

greater?greater?

1) case 11) case 1

2) case 22) case 2

3) it’s the same for both3) it’s the same for both

4) depends on the magnitude 4) depends on the magnitude of the force of the force FF

5) depends on the ice surface5) depends on the ice surface

In Case 1, the force F is pushing

downdown (in addition to in addition to mgmg), so the

normal force needs to be largerlarger.

In Case 2, the force F is pulling

upup, against gravity, so the

normal force is lessenedlessened.

Example #2: An object has a two forces acting upon it. One force is 454 N, pointing due north, and the other force is 844 N, pointing due west. What is the net force on the object and the acceleration of the object? Give the answer as magnitude and direction. What one force would counter this net force?

N, +y

E, +x

F1

F2

Ftotal

a. solve with Pythagorean theorem.

22

21 FFFtotal

NFtotal 958

NWN

N o3.28844

454tan 1

SENF oopposite 3.28@958

Example #3: Statics. A lady with a weight of 522 N balances at the center of a long rope, as shown. If the rope deflects 10.0o from the horizontal, determine the tension in the rope.

mg

T2T1

find the components of each vector.

+y

+x

oy

ox

TT

TT

0.10sin

0.10cos

11

11

oy

ox

TT

TT

0.10sin

0.10cos

22

22

This is a static system, meaning there is no acceleration. Solve this with Newton’s second law. Start with the x – component.

xxxxnet TTmaF 210

oo TT 0.10cos0.10cos0 21

therefore21 TT

Next use the y – component.

mgTTmaF yyyynet 210

mgTT oo 0.10sin0.10sin0 21

mgT o 0.10sin2 1

NNmg

Too

15030.10sin2

522

0.10sin21

Example #4: A mass of 125 kg is supported by two ropes. One rope extends horizontally, towards the left from the object to a support wall. The second rope extends towards the right of the object, at an angle of 53.1o above the horizontal. Find the tension in each rope.

tension T1

tension T2

= 53.1o

+y

+x

draw each force vector.

mg

find the components of each vector.

01

11

y

x

T

TT

oy

ox

TT

TT

1.53sin

1.53cos

22

22

This is a static system, meaning there is no acceleration. Solve this with Newton’s second law. Start with the y – component. (Random choice. You could start with the x – component.)

mgTTmaF yyyynet 210

mgT o 1.53sin00 2

os

m

o

kgmgT

1.53sin

80.9125

1.53sin

2

2 N1530

Next use the x – component.

xxxxnet TTmaF 210

oTT 1.53cos0 21

therefore NT 9201

oTT 1.53cos21

Example #5: A 615 kg mass is suspended as shown below. Determine the tension in each rope.

tension T1 tension T2

= 20.0o = 60.0o

+y

+x

draw each force vector.

= 20.0o = 60.0o

mg

find the components of each vector.

oy

ox

TT

TT

0.20sin

0.20cos

11

11

oy

ox

TT

TT

0.60sin

0.60cos

22

22

This is a static system, meaning there is no acceleration. Solve this with Newton’s second law. Start with the x – component. (Random choice. You could start with the y – component.)

xxxxnet TTmaF 210

oo TT 0.60cos0.20cos0 21

221 532089.0

0.20cos

0.60cosTTT

o

o

This is one equation with two unknowns. Use the y – component to get another equation with the same two unknowns.

Next use the y – component.

mgTTmaF yyyynet 210

mgTT oo 0.60sin0.20sin0 21

mgTT oo 0.60sin0.20sin 21

Substitute the equation from the x – direction.

mgTT oo 0.60sin0.20sin532089.0 22

mgT oo 0.60sin0.20sin532089.02

Solve for T2.

oo

mgT

0.60sin0.20sin532089.02

oo

smkg

T0.60sin0.20sin532089.0

80.9615 2

2

NT 57512

Substitute the equation from the x – direction.

21 532089.0 TT N3060

2 – Dimensional ForcesDay #2

Homework:Handout #4,

problems #1 – 3, 12

Example #6: Two masses are connected by a string over a pulley, as shown. The bottom mass (m1) is 25.0 kg and the hanging mass (m2) is 10.0 kg. What is the normal force on the bottom mass?

m1

m2

Example #6: Two masses are connected by a string over a pulley, as shown. The bottom mass (m1) is 25.0 kg and the hanging mass (m2) is 10.0 kg. What is the normal force on the bottom mass?

m1

m2

Draw in the forces and a coordinate line.

+y

– y

T

T

m1g

m2gn

This system is at rest. The suspended mass is too small to make the bottom mass rise upwards on its own.

Solve this problem by isolating each mass and apply Newton’s 2nd law to each mass.

Start with the hanging mass, m2.

m2

T

m2g

gmTamFnet 22

Set the acceleration equal to zero and solve for T.

gmT 20

NkggmTs

m 0.9880.90.10 22

Next look at the bottom mass, m1:

m1

T

m1g

n

gmnTamFnet 11

Set the acceleration equal to zero and solve for n.

gmnT 10

NkgTgmns

m 0.9880.90.25 21

Nn 147

Example #7: Given the two masses shown below, find the acceleration of the system and the tension in the string. Ignore friction.

1m

2mFirst determine the motion of each mass:

Mass m2 falls and accelerates downwards.

Mass m1 moves and accelerates towards the right.

Let these directions define the positive direction for each mass. Now label the forces. gm2

TT

Apply Newton’s 2nd law to each mass separately.

1mT

Start with mass m1. There are two unknowns: acceleration and tension

TamFnet 1

Next look at mass m2. This mass also has the same two unknowns: acceleration and tension. This will give us a second equation to solve the two unknowns.

2m

gm2

T TgmamFnet 22

By choosing the positive directions to match the directions of the accelerations, the two objects will have the same acceleration, both in magnitude and direction.

Solve the simultaneous equations: Tam 1

Tgmam 22TgmTamam 221

Notice T’s will subtract out

gmamm 221

21

2

mm

gma

Solve for the tension by substituting into one of the initial equations:

Tam 1

21

211 mm

gmmamT

21

21

mm

gmmT

Example #8: Determine the acceleration of the system and the tension in the string. Assume the mass on the right is the heavier mass. Ignore friction.

1mm

2mM

gmTam 11

Tgmam 22TgmgmTamam 2121

gmmamm 1221

gmm

mma

21

12

Solve for the tension by substituting into one of the initial equations:

gmTam 11

gmamT 11 gmgmm

mmm 1

21

121

121

121 mm

mmgm

21

12

21

121 mm

mm

mm

mmgm

21

212

mm

gmmT

Force of Friction

Homework:Handout #4

problems #5 and complete chapter reading.

Test Friday

Definition: Friction is a form of a ____________ force. This is a force that opposes the motion of an object. In general, this force may be very complex, but the form we will study will be simple.

drag

The friction force comes in two types: ___________ and ___________ .static kinetic

____________ ____________ refers to the force that will keep an object at rest. Static means that there is no motion between the surfaces in contact. The static friction force is a variable force: The force will take on whatever value it needs to prevent an object from moving, up to a maximum amount. The maximum amount of force is proportional to the normal force between the two surfaces.

____________ ____________ refers to the force that resists an object’s motion. Kinetic friction only operates when the object is moving. For our purposes, this force is constant, regardless of the speed of the object.

Static friction

Kinetic friction

Friction always tries to oppose the motion of two contacting surfaces relative to one another.

Static friction balances any applied force, up to a maximum limit.

forces balance

limit of static friction

When the applied force exceeds the maximum static friction, the object breaks free and moves with constant kinetic friction.

The size of the maximum static friction force is given in terms of the normal force the surface exerts on the object:

nFF ss max, Fs is the static friction force between two surfaces.

Fs,max is the maximum static friction force between two surfaces.

n is the normal force between two surfaces.

The maximum static friction force can be made equal to the normal force through a proportionality constant. This constant is called the coefficient of static friction, and it is represented by the symbol s.

nF ss max,

The coefficient of static friction rates the “roughness” of a surface. The higher the value of the coefficient, the stronger the static friction force can be. The coefficient does not have any units! This value is unique to a given surface, and it is usually given in some way.

Example #1: A 400 lb block rests on an icy horizontal surface. The coefficient of static friction between the block and the ice is 0.100.a. What will be the static friction force if a horizontal force of 20.0 lb is applied to the block? b. What if the applied force is 40.0 lb? c. 50.0 lb?

SOLUTION: Start by finding the maximum static friction force available.

The coefficient has a value of s = 0.100, so the maximum static friction force is:

nF ss max,

For a block resting on a flat surface, the normal force equals the weight of the object.

mgn

Combining the above, get: mgnF sss max,

lbmgF ss 400100.0max, lb0.40

When the applied force is 20.0 lb, the static friction force will match it at 20.0 lb in the opposite direction. When the applied force is 40.0 lb, that is right at the limit that static friction can hold the object still, or move it at constant speed. When the applied force is 50.0 lb, the static friction will fail and the object will accelerate.

Definition: Kinetic friction refers to the force that opposes a moving object’s motion. The kinetic friction force is proportional to the normal force between the object and the surface. The kinetic friction force is constant in magnitude. This force is independent of the speed of the object. It is also independent of the surface area of the contact area.

nFk Fk is the kinetic friction force between two surfaces.

n is the normal force between two surfaces.

The kinetic friction force can be made equal to the normal force through a proportionality constant. This constant is called the coefficient of kinetic friction, and it is represented by the symbol k.

nF kk

Example #2: A 40.0 kg box is made to slide on a horizontal surface that has a coefficient of kinetic friction of 0.510. How much force must be applied to the object to make it accelerate at 3.00 m/s2 horizontally?

appliedFkF

mg

n

a

Since the block never leaves the surface, the vertical forces balance.

mgn

Use the definition of kinetic friction:

mgnF kkk

Now use Newton’s 2nd law in the horizontal direction:

kappliednet FFmaF

mgFma kapplied

Solve for the applied force:

mgmaF kapplied

gamF kapplied

NFapplied 320

1. (Review) A force of 90.0 N pulls horizontally towards the right on an object with a mass of 20.0 kg along a surface with a coefficient of kinetic friction of 0.300. (a) What is the acceleration of this block?

appliedFkF

mg

n

a

Since the block never leaves the surface, the vertical forces balance.

mgn

Use the definition of kinetic friction:

mgnF kkk

Now use Newton’s 2nd law in the horizontal direction:

kappliednet FFmaF

mgFma kapplied

Solve for the acceleration: gm

Fa k

applied

256.1s

m

(b) Now use kinematics to get the final velocity: xavv o 222

xavv o 22

smv 59.5

2. A force of 90.0 N pulls the same object towards the right at an angle of 16.7 degrees above the horizontal. If the mass is still 20.0 kg and the coefficient of kinetic friction is still 0.300, (a) determine the normal force and kinetic friction force acting on the object.

mg

n

oF

cosoF

sinoFkF

There is no upward acceleration:

mgFnFma oyy sin0

sinoFmgn

The normal force is smaller than the weight of the mass since the applied force is also lifting upwards. A smaller normal force corresponds to a smaller friction force.

sinokkk FmgnF

sinokkk FmgF

(b) Determine the horizontal acceleration.

koxx FFFma cos

sincos okkox FmgFma

gm

Fa kk

ox sincos

276.1s

mxa

A couple of algebraic steps later…….

Force of Friction Day #2

3. A 1.20 kg book is pushed against a vertical wall by a horizontal applied force. The coefficient of static friction between the book and the wall is 0.425. What is the amount of applied force needed to hold the book up without sliding?

The forces balance in each direction:

nFo mgFfr

Use the definition of friction to combine the two equations:

ossfr FnFmg

so

mgF

425.0

80.920.1 2smkg

NFo 7.27

4. Two masses are tied together, with one suspended over a pulley as shown. Let m1 = 20.0 kg and m2 = 5.00 kg. Let the coefficient of kinetic friction between m1 and the surface be 0.150. (a) Determine the acceleration of the system, assuming m2 is initially moving downwards.

gm1

gm2

nT

T

kF

Mass m1 moves and accelerates towards the right.

Mass m2 moves and accelerates downwards.

Let these define the positive direction for the motion of each of the two masses.

Friction

Positive

Tension

m2g

Positive

Isolate the first mass. The vertical forces balance.

gmn 1 gmnF kkk 1

Solve the acceleration in the horizontal direction:

gmTFTam kk 11

Isolate the second mass and solve the acceleration :

Tgmam 22

Solve the simultaneous equations.

gmTam k 11

Tgmam 22TgmgmTamam k 2121

gmmamm k 1221

g

mm

mma k

21

12

280.9

00.50.20

0.20150.000.5s

m

kgkg

kgkga

2784.0s

ma

(b) What is the tension in the string?

Pick one equation and solve for T: Tgmam 22

agmamgmT 222

22 784.080.900.5s

ms

mkgT

NT 08.45

g

mm

mma k

21

12

280.9

00.50.20

0.20300.000.5s

m

kgkg

kgkga

2392.0s

ma

(c) What would be the acceleration if the coefficient of kinetic friction increased to 0.300? What does this mean physically?

The acceleration is negative. If the first mass were initially moving towards the right, then it would slow to a stop. If the system starts at rest, it would remain at rest {static friction}.

Complete Homework Set 5 Problems.

Inclined Planes

Homework: Set 6{take two days to finish…}

Introduction: An inclined plane (or incline or ramp) is a flat surface set to some angle relative to the horizontal.

= angle of the incline to the horizontal.

Motion of the object will be parallel to the surface of the incline.

0 F maF ||

The following slides include examples of inclined planes.

Johnstown, Pennsylvania

Reconstruction of Galileo’s device for determining the law of odd numbers.

Problem Solving Strategy: Since the motion is parallel to the ramp, take coordinates that are parallel and perpendicular to the ramp:

direction||

direction

This choice puts the acceleration only along one axis. This will simplify the problems.

When the cart is released on the ramp, it accelerates down the ramp. Note the water level when at rest or accelerating down the ramp.

http://www.physics.umd.edu/lecdem/services/demos/demosc4/c4-12.htm

Ex. #1: Determine the components of gravity for a mass on an inclined surface. The weight force pulls straight downwards. Determine the components parallel and perpendicular to the surface of the incline.

mg

90 cosmg

ǁ= mg sin θ

Ex. #2: A mass m slides down a ramp without friction. (a) What is the normal force acting on the mass and (b) what is the acceleration of the mass?

The perpendicular components balance:

cosmgn

The parallel components produce the acceleration:

sinmgma singa

Ex. #3: A mass m slides down a ramp with friction. (a) What is the normal force acting on the mass and (b) what is the acceleration of the mass?

The perpendicular components still balance.

cosmgn

By definition, the friction force is:

k kF n

cosmgF kk

The parallel components produce the acceleration. Take to the bottom of the ramp as the positive direction.

kFmgma sin

Substitute values and solve for the acceleration:

cossin mgmgma k

cossin kga

Note: The acceleration will be zero when:

tancossin kk

Ex. #4: A mass m is pulled to the top of a frictionless ramp by an applied force f. (a) If the mass is pulled to the top at a constant acceleration of a = 4.90 m/s2, what is the required force?Let = 36.9° and m = 80.0 kg.

Since there is no drag, the perpendicular components still balance are not needed.

Apply Newton’s 2nd law to the parallel components:

sinmgfma

Solve for f: sinmgfma

sinmgmaf

9.36sin80.990.40.80 22 sm

smkgf

Nf 863

Ex. #5: Repeat the previous problem, but this time include friction on the surface of the ramp. The coefficient of kinetic friction is 0.400.

There is an additional drag force pointing to the bottom of the ramp.

nF kk

cosmgn By definition, the friction force is:

nF kk

cosmgF kk

Now compute the net force parallel to the incline:

kFmgfma sin

kFmgmaf sin

cossin mgmgmaf k

cossin kgamf

plug in values: 1110f N

Ex. #6: A mass m is placed on a frictionless incline of angle to the horizontal. What horizontal force, F, is needed to hold the mass at rest? Values are given in the diagram.

n

mg

Balance the vertical forces:

mgn cos

Balance the horizontal forces:

Fn sin

Divide the two equations to eliminate the normal force:

mg

F

n

n

cos

sin tan

tanmgF

0.30tan80.9100 2smkgF

NF 566

Ex. #7: Solve for the acceleration of this system. Mass m1 is 30.0 kg, mass m2 is 25.0 kg, and the coefficient of kinetic friction k is 0.200. Assume m2 is initially moving downwards.

n

gm1

gm2

TT

kF

Tackle this beast with the same technique as used on the Atwood’s machine.

The perpendicular components for mass m1 balance:

cos1gmn

cos1gmF kk

Taking to the top of the ramp as the positive direction, solve for the parallel direction of incline:

kFgmTam sin11

cossin 111 gmgmTam k

Now find Fnet on mass m2.

gmTam 22

Add the two equations together:

cossin 111 gmgmTam k

Tgmam 22

cossin 11221 gmgmgmamam k

cossin 11221 mmmgamm k

2 1

1 2

sin coskm ma g

m m

2856.0s

ma

Substituting numbers gives:

Ex. #8: {Bonus type question} Solve for the acceleration of this system. Assume the ramp is frictionless.

0.30 0.60

1m

2m

Solution Hints:

sin11 gmTam

sin22 gmTam

gmmamm sinsin 1221

g

mm

mma

21

12 sinsin

2455.0s

m