Nicholas J. Giordano

www.cengage.com/physics/giordano

Forces and Motion in One Dimension

Applications of Newton’s Laws • We will learn how Newton’s Laws apply in various

situations • We will begin with motion in one dimension

• Idealized case – most real motion is three-dimensional • Mathematics is easier • Basic ideas and approaches can be applied to motion

in higher dimensions

Introduction

Motion of a Spacecraft

• Assume the spacecraft travels along a straight-line path from one galaxy to another

• Look at the motion in terms of position, velocity and acceleration

• No forces acting on the spacecraft so no acceleration

Section 3.1

Insights: Problem Set Up • Start with a picture

• Include the coordinate axes you have chosen for the problem

• Determine a positive direction • Be careful with units

• Be sure they are all in the same system of measurement

• Include the units with your calculations • Check the units that they correctly match the quantity

being calculated

Section 3.1

Notation Simplification • For cases involving one-dimensional motion the

usual vector notation can be simplified • The direction is specified by the sign (+ or - ) of the

quantity • This notation corresponds to specifying the

components of the quantity along the coordinate axis

Section 3.1

Motion Example – Constant Velocity = 0

• The velocity is zero • On the graph, the line is at v = 0

• The position is constant, but generally not zero • Not moving, so no change in position • On the graph, a horizontal line

Section 3.1

Motion Example – Constant Velocity ≠ 0

• The velocity is not zero • On the graph, the line is at v = constant value

• The position is changing steadily • The same Δ x occurs each second • On the graph, an upward sloping straight line • The slope of the position-time curve is equal to the value of

the velocity Section 3.1

Motion Example – Constant Acceleration

• A constant force is applied • The acceleration is a constant

• Value depends on the total force exerted on (and mass of) the object

• The value is equal to the slope of the line on the v-t graph

• The velocity is changing • On the graph, this is an

upward sloping straight line • The position is changing

• Not the same change each second

• On the graph, this is a curved line

Section 3.1

Equations to Describe Motion with Constant Acceleration • v = vo + a t

• vo is the velocity at some initial time t = 0 • It depends on what happened prior to t = 0

• x = xo + vo t + ½ a t2 • xo is the position at some initial time t = 0

• v2 = vo² + 2 a (xf - xo) • Eliminates t from the equation

• Which equation to use depends on what information you are given in the problem and what you are asked to find

Section 3.1

Constant Acceleration Equations, Summary

Section 3.1

Note About Units • Units can still be cancelled and combined as shown

in Chapter 1 • When dealing with units of force, it is simplest to

express Newtons in terms of fundamental units before combining and/or canceling • Remember: 1 N = 1 kg m / s2

• Always include units in your calculations and check that they correctly match the quantity being calculated

Weight and Normal Forces • There are many cases where gravity plays an

important role • Weight is associated with gravity • The normal force is another force often encountered

in problems

Section 3.2

Weight • Weight is the force of gravity exerted by the Earth on

an object • Denoted by • If an object has a mass m, then

• The force of gravity is a consequence of Newton’s Law of Universal Gravitation

• The value of g is approximately the same for all locations near the surface of the Earth • g ≈ 9.8 m/s²

• The weight will be different on another planet • Since it is due to the gravitational attraction of that

planet Section 3.2

Weight, cont. • The value of g is independent of the mass of the

object • The weight of an object is proportional to the mass of

the object • g is commonly referred to as the “acceleration due

to gravity” • Weight will be measured in Newtons

• It is a force • Since weight acts vertically, it will be along the y-axis • Since the weight acts downward, Fgrav = - m g

• It acts toward the center of the Earth

Section 3.2

Normal Force • A normal force acts perpendicularly to the plane of

contact • Normal forces are common in nature

• They occur whenever the surfaces of two objects come in contact

Section 3.2

Weight and Normal Force, Example

• The weight is directed downward

• The normal force is directed upward

• The person is at rest • a = 0

• ΣF = -m g + N = m a = 0 • N = m g • In this case, the normal is

equal in magnitude to the weight and opposite in direction to the person’s weight

Section 3.2

Free Body Diagram • A free body diagram should be used for analysis

using Newton’s Second Law • It is a simplified diagram showing all the forces

acting on each object involved in the problem

Section 3.2

Problem Solving with Newton’s Laws • Recognize the objects of interest

• List all the forces acting on each object • Sketch the problem

• Start with a drawing that shows all the objects of interest • Include all the forces acting on the objects • Make a separate sketch showing all the forces acting on

each using a dot for the object • This is the free body diagram

• Forces in the free body diagram should be represented by arrows • The direction of the arrow will show the direction of the force • Include only the forces acting on that particular object

Section 3.2

Problem Solving with Newton’s Laws, cont. • Identify the relationships

• Represent unknown quantities by an appropriate letter • Many unknowns have a particular symbol associated with them

• Generally, you will use Newton’s Laws to solve for the unknowns

• Solve • Use the information in the free body diagram to write

Newton’s Second Law • You may need to include some algebraic steps to actually

solve for the unknown • Check

• Consider what your answer means and check that it makes sense

Section 3.2

Apparent Weight

• The normal force is not always equal to the weight

• One example is moving upward in an accelerating elevator

• Letting upward be positive: • ΣF = m a = N – mg • So N = m a + m g

• If the elevator moved downward, N = m g – m a

• In this case, the normal force is called the object’s apparent weight

Section 3.2

Weight and Mass • Weight and mass are closely related • They are not the same • Mass is an intrinsic property of the object • Weight depends on the location of the object • Example:

• Assume mass on Earth is 10.0 kg • The weight on the surface of the Earth would be 98.0 N

• Fgrav = m g • The mass on the Moon would be 10.0 kg • The weight on the Moon would be ~ 16.3 N

• gMoon ~ 1/6 gEarth

Section 3.2

Forces and Interactions • Newton recognized that two objects could exert

forces on each other even when they are not in direct contact

• This is known as action-at-a-distance • Gravitational force is an example • Applies to many other forces

• Contact on a microscopic scale • Contact forces are a result of electric forces between

atoms that are in very close proximity • The atom-atom interaction is described by quantum

mechanics, not Newtonian Mechanics Section 3.2

More About Mass • The force of gravity depends on the mass of the

object • This is called gravitational mass

• The mass of an object also determines how the object will move in response to forces • This is called inertial mass

• Physicists believe that the inertial mass is precisely equal to the gravitational mass • Why will be addressed later

Section 3.2

Adding Friction • Friction is another force commonly encountered in

problems • Friction can be

• Kinetic • Related to moving

• Static • When objects are at rest

Section 3.3

An Object Coming to Rest • Aristotle’s idea was that rest was the natural state of

terrestrial objects • Newton’s view is that an object comes to rest

because a force acts on it • This force is often due to a phenomena called friction

Section 3.3

Friction

• The force of friction opposes the motion

• The magnitude of the frictional force is related to the magnitude of the normal force

• Force of kinetic friction • Ffriction = μk N • μk is called the coefficient

of kinetic friction • A pure number with no

units

Section 3.3

Motion with Friction • Set up the problem as usual, including the force of

friction • For example, as seen in fig. 3.11:

• ΣF = Ffriction = -μk N = m a • From the y-direction, Fgrav = N = m g • Therefore, -μk m g = m a and a = - μk g • Once you have found the acceleration, other quantities

involved with motion can also be found • See Table 3.2 for values for coefficients of friction for

various materials • The values depend on the smoothness of the surfaces

Section 3.3

Static Friction

• In many situations, the relevant surfaces are not slipping (moving) with respect to each other

• This situation involves static friction

• The amount of the push can vary without the object moving

Section 3.3

Static Friction, cont. • |Ffriction | ≤ μs N

• μs is called the coefficient of static friction • Static indicates that the two surfaces are not moving relative

to each other

• If the push is increased, the force of static friction also increases and again cancels the force of the push

• The magnitude of the static friction has an upper limit of μs N

Section 3.3

Kinetic Friction Compared to Static Friction • Consider both the kinetic and static friction cases

• Use the different coefficients of friction • The force of kinetic friction is just Ffriction = μk N • The force of static friction varies by |Ffriction | ≤ μs N • For a given combination of surfaces, generally μs >

μk • It is more difficult to start something moving than it is

to keep it moving once started

Section 3.3

Friction and Walking

• The person “pushes” off during each step

• The bottoms of his shoes exert a force on the ground • This is

• If the shoes do not slip, the force is due to static friction • The shoes do not move

relative to the ground

Section 3.3

Friction and Walking, cont.

• Newton’s Third Law tells us there is a reaction force •

• This force propels the person as he moves

• If the surface was so slippery that there was no frictional force, the person would slip

Section 3.3

Friction and Rolling

• The car’s tire does not slip

• There is a frictional force between the tire and road

• There is a reaction force on the tire

• This is the force that propels the car forward

Section 3.3

Friction and Motion • The force of friction makes walking and running

possible • Friction plays a key role in rolling motion

Section 3.3

Free Fall

• A specific type of motion • Only gravity acts on the

object when it is in free fall • There is actually some

air drag, but it is generally considered negligible

• We will analyze the motion in terms of acceleration, velocity, and position

Section 3.4

Free Fall – Acceleration • Take t = 0 to be the instant after the ball leaves the

thrower’s hand • Before then, there is an additional force from her hand

acting on the ball • This would not be an example of free fall since a force other than

gravity would be acting on the object • Choose a coordinate system that measures position as

the height y above the ground • Using Newton’s Second Law:

• The negative sign means gravity is directed downward

Section 3.4

Free Fall – Velocity and Position – Equations • Find the velocity and position as functions of

time:

• Note the acceleration is constant (and equal to g) • The velocity and acceleration are not always in

the same direction

Section 3.4

Free Fall – Velocity and Position – Graphs

• The motion can be expressed graphically as well

• The velocity can be positive, negative, or zero

• The slope of the v-t graph is constant and negative for the entire motion

• The constant negative acceleration is shown in graph C

Section 3.4

Free Fall – Final Notes • When the ball starts and stops at the same location

(assume the ground), the ball’s speed just before it hits the ground is equal to its initial speed • The velocities are in opposite directions

• The time spent on the way up is equal to the time spent falling back down

• Shows the symmetry in the problem

Section 3.4

Transmitting Forces • Strings exert a force on the objects they are

connected to • Also applies to cables, ropes, etc.

• The mass of the cable may have to be taken into account

• Pulleys can redirect forces • Forces can be amplified

Section 3.5

Tension

• Strings exert a force on the objects they are connected to • Cables and ropes act the

same way • The strings exert force

due to their tension • The ends of the string

both exert a force of magnitude T on the supports where they are connected

• T is the tension in the string

Section 3.5

Tension Example – Elevator Cable

• Two forces are acting on the compartment • Gravity acting

downward • Tension in cable acting

upward, T • Assume an acceleration

upward • Applying Newton’s

Second Law gives T = mg + ma

Section 3.5

Tension Example – Elevator, cont. • Now consider the cable

• Assume the cable is massless • Applying Newton’s Second Law gives: TC = T

• The tension is the same for all points along the cable • This is true for all massless cables

• Tension has force units

Section 3.5

Cables with Mass

• Apply Newton’s Second Law to the cable

• To support the cable, the upper tension, T1 must be larger than the tension from the box, T2 • T1 = T2 + mcable g

• If no acceleration • Can assume a massless

cable if the mass of the cable is small compared to the other tensions present

Section 3.5

Single Pulleys

• We often need to change the direction of the force

• A simple pulley changes the direction of the force, but not the magnitude • See diagram • Assume the rope and

pulley are both massless • Assume the cable does

not slip on the pulley Section 3.5

Pulleys To Amplify Forces

• The person exerts a force of T on the rope

• The rope exerts a force of 2T on the pulley

• This force can be used to lift an object

• More complex sets of pulleys can amplify an applied force by greater factors • The distance decreases

to compensate for the increase in force

Section 3.5

Reasoning and Relationships – Problem Notes • We may need to identify important information that is

“missing” from the initial description of the problem • We need to recognize that additional information is

needed • Then make reasonable estimates of the “missing”

quantities • An approximate mathematical solution and an

approximate numerical answer are generally sufficient • The estimates of the “missing” values will vary from

case to case

Section 3.6

Reasoning and Relationships – Problem Solving Strategy • Recognize the principle

• Determine the key physics ideas central to the problem • What principles connect the quantity you want to calculate

with the quantities you know • Sketch the problem

• Show all the given information • Draw a free body diagram, if needed

• Include all the forces, velocities, etc. • Identify the relationships

• Motion equations are an example of a set of relationships • If some values are unknown, make estimates for these

values

Section 3.6

Reasoning and Relationships – Problem Solving Strategy, cont. • Solve

• An exact mathematical solution typically is not needed • Cast the problem into one that is easy to solve

mathematically • Check

• Consider what your answer means • Check to be sure the answer makes sense

Section 3.6

Parachutes, Air Drag, and Terminal Velocity • Many constant acceleration problems are good

approximations to real-life examples • Some interesting and important cases occur when

acceleration is not constant • These cases include an object moving through a

fluid • The object is subject to a resistive drag force arising

from contact with the fluid molecules

Section 3.7

Air Drag • There are situations where air drag cannot be neglected

• Air drag depends on speed, so at higher speeds it becomes more of an effect

• Also depends on the area, so large areas cause a large air drag

• An estimate of air drag can be found by using • Fdrag = ½ ρ A v2

• A more complete equation is • Fdrag = ½ CD ρ A v2 • CD is the drag coefficient and depends on the aerodynamic

shape • CD is 1 for boxy shapes and less than 1 for many

streamlined shapes

Section 3.7

Skydiving Example

• Assume a skydiver opens their chute at t = 0

• Initially, the skydiver can be considered in free fall • Velocity is small, so air

drag is negligible • As the velocity increases,

so does the air drag • Eventually, the skydiver

will reach a constant velocity • Called the terminal

velocity

Section 3.7

Terminal Velocity • Once the skydiver reaches terminal velocity, the net

force acting on her is zero • The drag force is (nearly) equal in magnitude but

opposite in direction of the force of gravity • The magnitude of the terminal velocity is

• Air drag is generally small for slowly moving objects, but it is very important in some cases

Section 3.7

Bacterium Example

• Consider the motion of a small object in a liquid

• Stokes showed that the drag force on a spherical object moving slowly through a fluid is given by

• C is a constant that

depends on the properties of the fluid

• r is the radius of the object • The negative sign indicates

the drag force is always directed opposite of the velocity

Section 3.8