1

Unit Four:Force and Motion

John Elberfeld

JElberfeld@itt-tech.edu

518 872 2082

GE253 Physics

2

Schedule

• Unit 1 – Measurements and Problem Solving• Unit 2 – Kinematics• Unit 3 – Motion in Two Dimensions• Unit 4 – Force and Motion• Unit 5 – Work and Energy• Unit 6 – Linear Momentum and Collisions• Unit 7 – Solids and Fluids• Unit 8 – Temperature and Kinetic Theory• Unit 9 – Sound• Unit 10 – Reflection and Refraction of Light• Unit 11 – Final

3

Chapter 4 Objectives• Relate force and motion and explain what is

meant by a net or unbalanced force.• State and explain Newton's first law of

motion and describe inertia and its relationship to mass

• State and explain Newton's second law of motion, apply it to physical situations, and distinguish between weight and mass.

4

Chapter 4 Objectives

• State and explain Newton's third law of motion and identify action-reaction force pairs.

• Apply Newton's second law in analyzing various situations, use free-body diagrams, and understand the concept of translational equilibrium

• Explain the causes of friction and how friction is described by using coefficients of friction.

5

Reading Assignment

• Read and study Chapters 4 in College Physics: Wilson and Buffa

• Expect a quiz on any of the material covered in the course so far

6

Written Assignments

• Do all the exercises on the handout.• You must show all your work, and carry

through the units in all calculations• Use the proper number of significant

figures and, when reasonable, scientific notation

7

Introduction

• In the previous week, you explored the motion of objects when the velocity and acceleration of the object are known.

• In real life, you have to know what actions we need to perform to cause and control this motion

8

Introduction

• For example, when pushing an object, you have to know what to push, where to push it, and how hard to push it.

• You also want to understand what happens to an object when you push it.

• In other words, you want to understand the relation between forces and the motion of objects that forces act upon.

9

Force

• A force is a push or a pull. • For example, when you try to lift a heavy object,

you feel the force of gravity pulling the object down.

• Springs and magnets also exert forces, as do your muscles.

• Force is also associated with motion because when you exert force on an object, it may move.

• Force is a vector quantity because it has direction.

• When you exert force, it is in a particular direction.

10

Net Force

• This brings us to the concept of net force or total force.

• If the forces acting on an object are equal and in opposite directions, they will cancel each other out.

• In such a case, the total force or net force is zero.

11

Net Forces

12

Net Forces

• If two forces of equal magnitude but opposite direction are applied to the same object, the vector resultant, or net force acting on the crate, in the x direction, is zero. – The forces are balanced

• If the forces are not equal in magnitude, the resultant is not zero– A net force accelerates an object

in the direction of the force

13

Classes of Forces

• There are two classes of forces: • The action-at-a-distance forces are

gravity, magnetism, and electrostatic forces.

• Contact forces include friction and the so-called normal forces.– For example, the force exerted by a

table on an object resting on it is a contact force.

14

Thought Experiment

• Force is associated with motion, but the relationship between the two may not be clear at first.

• The diagram shows an experiment performed by Galileo in which a ball rolls down and then up an inclined plane, always reaching the same level.

• Therefore, if the surface is horizontal and perfectly frictionless, the ball will continue to move without stopping.

• However, the ball will stop if another force (like friction) acts upon it.

15

Thought Experiment

• How far would the ball roll on a frictionless horizontal surface?

16

Newton’s First Law

• Based upon Galileo’s experiment and other observations, Newton derived the first law of motion.

• An object at rest stays at rest unless acted upon by a net external force.

• An object moving at a constant velocity will continue to move at that velocity unless acted on by a net external force.

17

Common Sense

• Newton’s law is in sharp contrast to the science of ancient Greece.

• According to Aristotle, an ancient Greek scientist, the natural state of an object is to be at rest.

• An object moves only when a force acts on it.

18

Inertia

• All objects have mass.

• This means Newton’s first law can be considered a property of mass called inertia.

• Inertia is the property of mass which determines its resistance to changes in velocity.

19

Mass is in Kilograms

• Mass is always measured in kilograms in physics

• The mass in kilograms determines how much gravity affects an object (weight) as well as how hard it is to start or stop its motion (inertia)

20

Practice

• Every object on earth experiences the force of the earth’s gravity, and every object in the solar systems experiences the force of the sun’s gravity

• All mass always has some type of force on it

• An object at rest, or at constant velocity (no change in speed OR direction), has no net force on it – all forces add up to zero

• If an object has a net force on it, the velocity will change (acceleration!)

21

Practice

• On a jet airliner that is taking off, you feel that you are being pushed back into the seat. Why? – You are at rest and will stay at rest unless a

force acts upon you. When the plane takes off, you feel the force of the plane acting on you.

• Is the opposite true when a car stops?

22

Newton’s Second Law

• When a net force acts on an object, the object accelerates.

• a = Δv/Δt• The mass of an object affects its

acceleration. • According to Newton’s second law

• F = ma • where F and a are vectors. The unit of

force is Newton (N), defined as• 1N = kg m / s2

                                   

23

Second Law

• Assuming no additional forces, like friction:

24

F = ma

•                                      

25

Weight

• The weight of an object (mass) is the force of gravity acting on the object.

• Near the earth’s surface, the acceleration due to gravity is about (-) 9.8 m/s2 downward

• So, you can calculate the weight of a 1-kg object as follows:

• Fgravity = ma = W = mg = m (-9.8m/s2)

• The weight of a 1 kg mass is:• F = ma = (1kg)(-9.8m/s2) = - 9.8kg m/s2 = -9.8N

26

Gravity

• Gravity is an example of an action-at-a-distance force.

• The universal law of gravity, discovered by Newton, is that every mass in the universe is attracted to every other mass with a gravitational force proportional to the two masses and inversely proportional to the square of the distance between the two masses.

• F = Gm1m2/d2

27

Forces are the Same

• F = Gm1m2/d2

• The force of m1 on m2 is the same as the force of m2 on m1

• The sun is so much more massive than the earth, the earth has little noticeable affect on the sun, but astronomers use the motion of stars to discover invisible planet

28

On the Earth

• Weight = mass x gravity• W = m g• W = m (- 9.8 m/s2 )• A person pulls on the earth as

much as the earth pulls on the person• The force caused by gravity (weight)

causes falling objects to accelerate at at rate of 9.8 m/s2 downward

• F = m a = W = m g• a = g = - 9.8 m/s2 on the earth’s surface

29

Practice

• A tractor pulls a loaded wagon with a mass of 275 kg on a level road with a constant force of 440 N, as shown below.

• (a) What is the acceleration of the wagon?• (b) What would the acceleration of the wagon be

if there was an opposing frictional force of 140 N?• (c) Suppose the wagon starts from rest (with

friction). How far will the wagon travel in 4.0 s?

30

Calculations

• a) F = maa = F/m = 440N/275 kg =1.60 N/kg1.60 N/kg (kg m/s2 / N) = 1.60 m/s2

• b) F = Ftractor – FFriction F = 440N – 140 N = 300 Na = F/m = 300N/275kg = 1.09 m/s2

• c) x = x0 +v0t +at2/2 x = 0 + 0 +   (1.09 m/s2) *(4s)2/2x = 8.72 m

X0

X

31

Practice

• A student weighs 588 N on the surface of the earth.

• What is her mass?

• What would be her mass on the moon?

• What would be her mass at the bottom of the ocean?

32

Calculations

• W = mg

• m = W/g = 588 N / 9.8m/s2

• m = 60 kg (about 132 pounds)– 1 kg has a weight of 2.2 pounds on the

surface of the earth– Her mass is 60 kg on the Moon, at the

bottom of the ocean, on Mars…– Mass does NOT change with location

33

Practice

• In a college homecoming competition, 15 students lift a sports car. While holding the car off the ground, each student exerts an upward force of 420 N.

• (a) What is the mass of the car in kilograms?

• (b) What is the weight in pounds?

34

Calculations

• Total upward force = 15 x 420 N• F = 6,300 N• W = mg• m = W/g = 6300N / 9.8m/s2

• m = 643 kg• W = 643 kg (2.2 pounds/kg) on

earth’s surface• W = 1,415 pounds

35

Practice

• In an emergency stop to avoid an accident, a shoulder-strap seat belt holds a 50-kg passenger firmly in place.

• If the car were initially traveling at 80 km/h and came to a stop in 6 s along a straight, level road, what was the average force applied to the passenger by the seatbelt?  

• Suppose the time is reduced to .5 s?

36

Calculations

• Find velocity in m/s• 80 km/hr (1000m/km)(1hr/3600s)= 22.2m/s• v = v0 + at • 0 = 22.2m/s + a 6s• a = -22.2m/s / 6s = -3.7 m/s2

• F = ma = 50 kg (-3.7m/s2) = -185 N– Minus indicates the force acts in a direction

opposite of the initial velocity

• a = -22.2m/s / .5 s = -44.4 m/s2

• F = ma = 50 kg (-44.4m/s2) = -2200 N • F = -493 pounds for a fast stop!

37

Introduction

• In the previous lessons, you learned what a force is becauseyou learned the connection between force and acceleration.

• In this lesson, you will study why forces exist and where they come from.

• The law of action and reaction will also be examined here.

38

Newton’s Third Law

• Newton’s second law says F = ma

• Newton’s third law says the forces come from other objects

• Your muscles (first object) push the crate (second object), but that crate pushes back just as hard

39

Newton’s Third Law

• Newton’s third law of motion:

• For every action there is an equal and opposite reaction.

• In other words, for every force, there is an equal and opposite force.

• In notation form, the law is written as follows:

• F12 = – F21

40

Pairs of Forces

• When an object exerts force on another object, the second object exerts an equal force at the sametime on the first object.

• Objects interact in pairs; and the interacting forces come in equal and opposite pairs.

41

Free Fall

• When the briefcase is released, there is an unbalanced, or nonzero net force acting on the briefcase (its weight force,) and it accelerates downward at g or -9.8m/s2 for free fall

42

Action-Reaction

• Both skateboarders move toward the center point

43

Paired Forces

• The third law is illustrated with any kind of propulsion.

• When you walk, for example, your muscles exert a horizontal force against the ground and the ground exerts an equal and opposite horizontal force against you, which propels you in the direction you are walking.

44

Jet Propulsion

• In this case, the action-reaction pair is the force of the hot gases on the plane propelling the plane forward and the force of the plane on the hot gases ejecting the gases backward.

45

Finding Forces

• The diagram shows two blocks m1 and m2 of masses 2.5 kg and 3.5 kg, respectively, resting on a frictionless surface.

• A force of 12 N acts on m1, which tautens the massless string connecting the two masses.

• As a result, the string exerts a force on m2. • Therefore, both masses accelerate. • What is the acceleration of

the system, and what is the force that the string exerts on m2?

46

Calculations

• For the system of masses

• F = ma

• 12 N = (2.5kg + 3.5kg) a

• a = 12 N / (6 kg) = 2.0 m/s2

• For the second mass

• F = ma

• F = 3.5 kg 2.0 m/s2 = 7 N

47

Action and Reaction

• m2 pulls back on m1 with the same force a m1 pulls forward on m2

• The net force on m2 = 7N and the net force on m1 = 12N – 7N = 5 N– If m1 were alone, only 5N would be

needed to accelerate it at 2.0m/s2

F = ma = 5N = 2.5 kg 2 m/s

12N-7N7N

48

Force

• A car traveling at 72.0 km/h along a straight, level road comes uniformly to a stop at a distance of 40.0 m.

• If the car weighs 8800 N, what is the breaking force?

49

Calculations

• F = ma – need both m and a

• v2-v02=2a(Δx) now this is very useful

• 72km/hr (1000m/km)(1hr/3600s)=20m/s• 0 - (20m/s)2 = 2 a 40m• a = 5 m/s2

• W = 8800 N = m 9.8m/s2

• m = 898kg• F = ma = 898kg 5m/s = 4490 N

50

Normal Force

• The diagram shows a block resting on a table.

• The table is exerting an upward force on the block that is perpendicular to the surface of the table.

• This is called the normal force because normal means perpendicular.

• The weight of the block is equal to the normal force because the block is in equilibrium.

51

Normal Force

• The normal force is a contact force and is the force of the table pressing up on the block

• It is paired with the force of the block pressing down on the table due to gravity

•                  

52

Inclined Plane Problems

• Set up a new axis pair that is perpendicular and parallel to the plane

W=mg

ΘΘ

mgsinΘ

mgcosΘ

53

Inclined Plane

• Geometry – the angle of inclination is the same as the angle between gravity and the perpendicular to the plane (normal)

W=mg

ΘΘ

mgsinΘ

mgcosΘ

54

Angle Details

Θ

Θ

mgcosΘ

Φ

Φ90

90

55

Slide Down the Plane

• With no friction, a block will slide with constantacceleration

W=mg

ΘΘ

mgsinΘ

mgcosΘ

N = mgcosΘ

56

Normal Force

• N = mgcosΘ• The plane balances the component

of the weight perpendicular to the plane with N

57

Typical Problem

• If m = 35 kg, find acceleration down the 25º

plane

W=mg

ΘΘ

mgsinΘ

mgcosΘ

58

Typical Problem • If m = 35 kg, find

acceleration down a 25º plane

35 kg 9.8m/s2

ΘΘ

35 kg 9.8m/s2 sin25º

35 kg 9.8m/s2 cos25º

59

Find Acceleration

• F = ma• 35 kg 9.8m/s2 sin25º = 35 kg a• a = 35 kg 9.8m/s2 sin25º / 35 kg • a = 4.14 m/s2

• Acceleration is constant• Velocity is increasing at a constant rate

while the object slides down the plane• Acceleration is independent of the mass

of the object AND is less than g!

60

Frictionless Inclined Plane

• Two masses m1 and m2 are connected by a light string running over a light pulley.

• The pulley offers negligible friction. • Mass m1 (Mass = 5.0 kg) is on a

frictionless 20° inclined plane. • Mass m2 (Mass = 1.5 kg)

is freely suspended. • What is the acceleration

of the masses?

61

Force Diagrams

62

Forces

• F = ma – What is the total force used to accelerate BOTH objects?

• F = W – m1gsinΘ = m2g – m1gsinΘ

• F = 1.5kg9.8m/s2 – 5kg 9.8m/s2 sin20º• F = -2.06N • a = F/m = -2.06N/(1.5kg+5kg) =

-.32m/s2

• The mass on the plane will slide down the plane and drag the mass on the pulley up

63

Components of Force

• A force of 10.0 N is applied at an angle of 30° to the horizontal on a 1.25-kg block at rest on a frictionless surface.

• (a) What is the magnitude of the resulting acceleration of the block?

• (b) What is the magnitude of the normal force?• The space diagram displays all the forces acting

on the block and the free-body diagram displays the block as a point mass.

• The x-axis is along the table surface because motion isalong that direction.

64

Calculations

• W = mg = 1.25kg (–9.8m/s2) = -12.25N

• Fx = 10N cos30º = 8.66 N

• Fy= 10N sin30º = -5N (down)

• N = 12.25N + 5N = 17.25N (vertical)• F = ma (horizontal)• a = F / m • a = 8.66N / 1.25 kg• a = 6.93 m/s2

65

Equilibrium

66

Equilibrium

• Equilibrium is the state where the object is at rest or at constant velocity because the net force acting on it is zero.

• Therefore, in equilibrium, in the diagram, the woman is in a state of static translational equilibrium because she is at rest.

• An object moving at constant speed is in translational equilibrium.

67

Practice

• Jane and John, with masses of 50 kg and 60 kg, respectively, stand on a frictionless surface 10 m apart. John pulls on a rope that connects him to Jane, giving Jane an acceleration of 0.92 m/s2.

• At what rate will John accelerate? • What is the direction

of his acceleration?

68

Calculations

• F = ma = 50kg .92m/s2 = 46 N on the Jane• This is the same force that is exerted back

on John• F = ma = 46N = 60kg a• a = 46N / 60 kg = .767 m/s2 = boy’s

acceleration• The more massive object has less

acceleration, but the forces are the same

69

Practice

• A boy pulls a box of mass 30 kg with a force of 25 N at angle of 30º from the horizontal

• (a) Ignoring friction of the box, what is the acceleration of the box?

• (b) What is the normal force exerted on the box by the ground?

70

Calculations

• Fx = 25N cos(30º) = 21.7 N

• Fy = 25N sin(30º) = 12.5 N

• N = 12.5N – 30kg9.8m/s2 = -282N– Not enough to overcome weight, so

moves in horizontal direction• F = ma• a = 21.7N/30kg• a = .723 m/s2

294N

282N

21.7N

12.5N

71

Practice

• The Atwood machine consists of two masses suspended from a fixed pulley. If m1 = 0.5 kg and m2 = 1.0 kg,

• (a) What is the acceleration of the system

• (b) What is the magnitude of the tension in the string?

72

Atwood Machine

73

Atwood Machine

• The tension is the same at both ends

• Acceleration is thesame on bothblocks but oppositein direction T

W=m1g

m1

m2

T

W=m2g

74

Calculations

• F = ma – Total force accelerating both objects

• F = m2g-m1g = (m1+m2) a• 1kg 9.8m/s2 - .5kg 9.8m/s2 = (1.5kg) a• a = (4.9 kg m/s2 ) / 1.5 kg• a = 3.266 m/s2

• T = mg + ma • T = .5kg 9.8m/s2 + .5kg 3.267 m/s2 • T = 6.534 N on string to m1

75

Check

• If T = 6.54 N, Fm1 = T - W = 6.54 N - 4.9 N

• Fm1 = 1.634N UP

• a = 1.634 N/ .5kg • a = 3.27 m/s2 UP

• Fm2 = T - W

• Fm2 = 6.534 N – 9.8 N

• a = 3.27 N/ 1kg DOWN• a = 3.27 m/s2 DOWN

T

W=m1g

m1

m2

T

W=m2g

76

Equal Masses

• If the masses are the same:• m1g = m2g• There is no net force• There is no acceleration• There is NO MOTION• The blocks do not

change position• Starting heights

do not matter

T

W=m1g

m1

m2

T

W=m2g

77

Friction

• Friction is the force that dictates many of our daily activities.

• It is because of the force of friction that you take small steps on a slippery surface and press down on a piece of paper when writing with a pencil.

• It is friction that prevents you from moving heavy objects and hurts your joints in old age.

• This lesson covers the various types of friction and what determines the force of friction.

78

Friction Opposes Motion

• When you slide a file cabinet across the floor, the force of friction acts opposite to the motion of the heavy object.

• The force of friction ALWAYS points in the direction opposite to the direction of the external force.

• The attraction between the molecules of the two surfaces gives rise to the force of friction.

79

Friction

• The attraction between the molecules of the two surfaces gives rise to the force of friction.

• In addition, any surface roughness can contribute to friction as the sliding object must go up (and down).

• The force of friction opposes motion.• The force of friction points opposite the

direction of motion

80

Static and Kinetic Friction

• STATIC BIG – KINETIC SMALL• It takes a big force to get a still object to

start moving – you have to overcome static friction

• Once it is moving, the same force is more than needed to over come kinetic friction

• You can apply less force and get constant velocity, or keep the same force and get constant acceleration

81

Static and Kinetic Friction

82

Coefficients of Friction

• The force of friction is proportional to the force pressing the surfaces together (the Normal force)

• FFriction = μ FNormal

• μ is the coefficient of friction

• μStatic is used if the object is stationary

• μKinetic is used if the object is moving

• μStatic is big, μKinetic is small

83

Big Small

84

Friction Force

• The force of friction is directly proportional ONLY to the Normal force AND the coefficient of friction

• FFriction does NOT depend on the surface area

• Ff = μ N where μ is the coefficient of friction

• You must know μ and N (FNormal, FN,…) to calculate friction

• Use Static Friction if the object is stationary• Use Kinetic Friction if the object is moving

85

Practice

• What is the force needed to keep a mass of 158 kg moving horizontally across a level floor if the coefficient of kinetic friction is .27? (No acceleration, force is horizontal)

F

mg

μN

N

86

Calculation

• Ffriction = μN

• N = mg = 158 kg 9.8m/s2

• N = 1,548 N

• Ffriction = μN

• Ffriction = .27 x 1,548 N

• Ffriction = 418 N

87

Practice

• The coefficient of static friction between the 40.0-kg crate and the floor is 0.650.

• What minimum horizontal force must the worker apply to move the crate?

• If the worker maintains the force once the crate starts to move and the coefficient of kinetic friction between the surfaces is 0.500, what is the magnitude of the acceleration of the crate?

88

Calculations

• W = mg• W = 40kg(- 9.8m/s2)• W = -392 N• N = +392N

• fs = μsN

• fs = .650 x 392N

• fs = 255N

• This is the force needed to start motion

89

Maintain Motion

• 255N is applied even after the crate starts to slide

• fk = μsN

• fs = .500 x 392N = 196N

• F = ma = 255N – 196N = 59 N• a = F / m = 59 N / 40 Kg = 1.48 m/s2 in the

direction of the applied force• Acceleration is constant• Velocity increases at a

steady rate

90

Terminal Velocity

• An object moving through the air meets air resistance that opposes motion

• Air resistance is very complicated, but it increases with velocity

• Eventually, an object in free fall has enough air resistance to balance the pull of gravity

• An object reaches terminal velocity when air friction balances gravity: acceleration is zero and velocity remains constant

91

No Friction vs Friction

92

Friction increases with velocity

93

Practice

• A packing crate is placed on a plane inclined at an angle of 35o from horizontal.

• If the coefficient of static friction between the crate and the plane is 0.65, will the crate slide down the plane? Justify your answer.  

94

Slide Down the Plane

W=mg

ΘΘ

mgsinΘ

mgcosΘ

N

ff=μN

95

Calculations

• Is the force down the plane greater than the force of friction?

• Is mgsinΘ > μmgcosΘ ?• Is sinΘ > μ cosΘ ?• Is sin35º > .65 cos35º • Is .573 > .532 => Yes, so crate

slides down the plane, no matter what the mass is or acceleration due to gravity

96

Slide Down the Plane

• Find μ if a = 0

W=mg

ΘΘ

mgsinΘ

mgcosΘ

N

ff=μN

97

Calculations to find μ

• If the object slides down the plane with no acceleration, find μ

• N = mgcosΘ• μ N = mgsinΘ• μ mgcosΘ = mgsinΘ• μ = mgsinΘ / mgcosΘ • μ = sinΘ / cosΘ • μ = tanΘ• Note: This is independent of the

mass, g, shape of the object, etc.

98

No Math

• The slope of an inclined plane is increased until a crate slides down the plane

• Describe the motion of the crate after it starts moving

99

Reasoning

• The slope is increased until the component of gravity down the plane is just barely greater than the force of static friction

• This same force than is much bigger the force of kinetic friction which opposes motion as soon as the crate moves

• The object will accelerate down the plane with a constant acceleration less than g

• Velocity will increase at a steady rate

100

Summary

• What is the net force acting on an object?

• What is inertia and Newton’s first law of motion?

• What is Newton’s second law of motion?

• What is Newton’s third law of motion?