Ch 4 Forces & Newton’s Laws · 2018. 10. 22. · 90.0 km/h. At that speed the forces resisting motion, including friction and air resistance, total 400 N. (Air resistance is analogous

Ch 4 Forces & Newton’s Laws

M1M2

M3

Pulley Clamps

Lab Table

Aluminum Support Rods

M3

T1 T2

M3 g

Protractor

Wooden Block

T1 T2

M3 g

T1 cos T2 cos

T1

T2 cos

cos

Horizontal Force Components

Equation

T1 sin T2 sin M3 g

T2 cos

cos

sin T2 sin M3 g

Vertical Components Equation

T1sin T2 sin

M3 g

T1 cos T2 cos

Static Equilibrium

Horizontal forces balance.

T1 cos T2 cos

Vertical forces balance.

T1 sin T2 sin M g

A mass, (M 23 kg ), is

suspended by two ropes from a

ceil ing. Rope 1 makes an angle

of 60 deg from the ceil ing,

while Rope 2 makes an angle of

40 deg . What are the

tensions T1 and T2?

T1

T2 cos

cos

T1 sin T2 sin M g

T2 cos

cos

sin T2 sin M g

1.

TL is the tension of the

horizontal cable on the

left, attached to the wall.

TR is the tension of the

angled cable on the right,

attached to the ceil ing.

1.

TL is the tension of the

horizontal cable on the

left, attached to the wall.

TR is the tension of the

angled cable on the right,

attached to the ceil ing.

Fy 0 TR sin 37 deg( ) 625 N Fx 0 TR cos 37 deg( ) TL

Setting up a T1 T2 Problem

Find the tensions T1 for the rope

attached to the left, and T2 for the

rope attached to the right.

Fx 0 T1 cos 43.0deg( ) T2 cos 55 deg( )

Fy 0 T1 sin 43.0deg( ) T2 sin 55 deg( ) 43.8kg9.8 N

kg

This horizontal equation substitutes

into the vertical equation, enab ling us to

solve for T2.

T1

T2 cos 55 deg( )

cos 43.0deg( )

T2 cos 55 deg( )

cos 43.0deg( )

sin 43.0deg( ) T2 sin 55 deg( ) 43.8kg9.8 N

kg

0.574

0.731T2 0.682

0.819T2 429.24N

0.535T2 0.819T2 429.24N

1.354T2 429.24N

T2429.24N

1.354

T1

T2 cos 55 deg( )

cos 43.0deg( )

T2 317N

T1317N( ) cos 55 deg( )

cos 43.0deg( )

T1 249N

4.1 Concepts of Force & Mass

•Force is a push or

pull.

•Mass is a measure of

the quantity of matter.

O-Stax Ch 4 Page 168 Prob 9.

Suppose two children push horizontally, but in exactly opposite directions, on a

third child in a wagon. The first child exerts a force of 75.0 N, the second a force of

90.0 N, friction is 12.0 N, and the mass of the third child plus wagon is 23.0 kg.

(a) What is the system of interest if the acceleration of the child in the wagon is to

be calculated? (b) Draw a free-body diagram, including all forces acting on the

system. (c) Calculate the acceleration. (d) What would the acceleration be if

friction were 15.0 N?


A powerful motorcycle can produce an acceleration of 3.50 m/s2 while traveling at

90.0 km/h. At that speed the forces resisting motion, including friction and air

resistance, total 400 N. (Air resistance is analogous to air friction. It always

opposes the motion of an object.) What is the magnitude of the force the

motorcycle exerts backward on the ground to produce its acceleration if the mass

of the motorcycle with rider is 245 kg?


The rocket sled shown in Figure 4.33 accelerates at a rate of 49.0 m/s2 . Its

passenger has a mass of 75.0 kg. (a) Calculate the horizontal component of the

force the seat exerts against his body. Compare this with his weight by using a

ratio. (b) Calculate the direction and magnitude of the total force the seat exerts

against his body.

Figure 4.33


Repeat the previous problem for the situation in which the rocket sled decelerates

at a rate of 201 m/s2 . In this problem, the forces are exerted by the seat and

restraining belts.

Figure 4.33


The weight of an astronaut plus his space suit on the Moon is only 250 N. How

much do they weigh on Earth? What is the mass on the Moon? On Earth?

4.2 Newton’s 1st Law of Motion

•Objects at rest tend to

stay at rest.

•Objects in motion tend

to stay in motion.

4.2 Newton’s 1st Law of Motion

• The object is only compelled

to change its state (whether

stationary or moving at

constant speed) if it is acted

upon by an outside net

force.

Inertia & Mass

• Inertia is the tendency of an object to remain at rest or in motion.

• Inertia depends on mass.

Non-Inertial Frame of Reference

• Objects show evidence of acceleration due to an unknown force.

• Newton’s first Law of Inertia does not appear to hold true.

Inertial Frame of Reference

• Newton’s Law of Inertia holds true.

• Zero acceleration is assumed, whether the object is at rest or moving at constant velocity.

Inertial Frame of Reference

• Zero acceleration

• Van is parked.

• Van is moving at

constant velocity.

4.3 Newton’s 2nd Law

• Acceleration of a mass is the result of an unbalanced force.

• It takes a force to cause an acceleration.

• A net force of zero results in the object staying at rest or staying at constant speed.

Units & the 2nd Law of Motion

N kgm

s2

F ma

Free-Body Diagrams (FBD)

Free-Body Diagrams (FBD)

Page 128 Problem 11

Page 128 Problem 11

• A net force causes acceleration of the block.

• Since the block is moving along the horizontal, it is accelerating along the horizontal.

Page 128 Problem 11

•Therefore, only the horizontal components of the force vectors are taken into consideration.

F m a

a F

m

45.0N cos 65°( ) 25.0N

5.00kg

a 1.20m

s2

A negative

acceleration is

a deceleration.

4.4 Vector Nature of 2nd Law

4.5 Newton’s 3rd Law of Motion

4.5 Newton’s 3rd Law of Motion

• The force of Object A on Object B has the same magnitude as the force of Object B on Object A.

• The directions of the forces are opposite.

Newton’s 3rd Law

4.6 Universal Gravitation

FG M1 M2

r2

G 6.671011

N m

2

kg2




Mass of the Earth

F m gG ME m( )

RE2

gG ME

RE2 ME

g RE2

G

Mass of the Earth

ME

g RE2

G

ME

9.80m

s2

6.38106

m 2

6.671011

N m

2

kg2

5.981024

kg

4.7 Weight

• Gravitational force that Earth exerts on an object.

• For a given mass, the object’s weight varies with the mass of the planet or moon.

Relation Between

Mass & Weight

• An object’s mass remains

constant in different

gravitational fields.

• It’s weight, being a force,

DOES change in the presence

of different gravitational fields.

4.8 Normal Force

• A normal force is the perpendicular force to the surface in question.

• In coefficient of friction problems, this is the pressing force of the surface upon which an object rests or slides.

4.8 Normal Force

Static Friction

PHYSICS LAB: Electronic Worksheet

Determination of the

Coefficients of Static

& Kinetic Friction

Copyright © 2010, by R. Faucher

4.9 Static and Kinetic Frictional Forces

Orientation Effects on m

FN Fs Fk

data0 1 2

0

1

2

3

4

2.5

3.5

4.3

7.1

12

FN data0

N Normal Force against block.

Fs data1

N1 Static Force against block.

Fk data2

N2 Kinetic Force against block.

Friction as a Function of Normal Force

Normal Force (Weight of Wood Block)

Sta

tic

& K

inet

ic F

rict

ion

5 .5

0

Fs

N

Fk

newton

best_fits

best_fitk

120 FN

newton

Fs slopes slopes

slopek slopek

Friction as a Function of Normal Force


Sta

tic

& K

inet

ic F

rict

ion

5 .5

0

Fs

N

Fk

newton

best_fits

best_fitk

120 FN

newton

Fs slopes slopes

slopek slopek

0 5 10 150

1

2

3

4

5

6Friction as a Function of Normal Force


Sta

tic

& K

inet

ic F

rict

ion

Fs

N

Fk

newton

best_fits

best_fitk

FN

newton

slopes 0.465

slopek 0.331

What information do the slopes

of Fs

FN

and Fk

FN

tel l us from

the straight-l ine graph?


Coefficient of Friction #41 Fw 45.0 N Fa 36.0 N

ms 0.650 mk 0.420

If calculated Fa Fs , then

block wil l move.

Coefficient of Friction #41

ms

Fs

FN

Fa

Fw

Holds true if block is static or

moves at constant speed, just

l ike in the friction lab we did.

Coefficient of Friction #41 If calculated Fa Fs , then

block wil l move.

Fs ms FN ms Fw 29.3 N

36.0 N 29.3N Block moves.


mk

Fk

Fw

F m a

a F

m

Fa Fk

m


aFa Fw mk

Fw

9.8N

kg

a 3.72m

s2

Static & Kinetic Friction

ms

Fs

FN

mk

Fk

FN

4.8 The Normal Force

Definition of the Normal Force

The normal force is one component of the force that a surface

exerts on an object with which it is in contact – namely, the

component that is perpendicular

to the surface.


N 26

0N 15N 11

N

N

F

F

N 4

0N 15N 11

N

N

F

F

Apparent Weight

• This is the weight that you feel in an elevator, regardless of your true weight on Earth.

• It is what a bathroom scale would read if you were on one in an elevator. (The elevator floor provides an upward force to the scale, and from the scale to your feet.)


Apparent Weight

The apparent weight of an object is the reading of the scale.

It is equal to the normal force the scale exerts on the man.


mamgFF Ny

mamgFN

apparent

weight

true

weight

Elevator Problem #37

• Since an elevator accelerates, use F=ma.

• Draw a FBD of all the forces involved.

• The algebraic sum of the colinear forces gives the net force.


F m a

vo 0 vf 45m

s mA 57 kg FN

FN mA g mA a

FN mA a mA g

FN a g( )mA

g 9.80m

s2

t 15sec


FN a g( )mA

FN

45m

s

15 s

9.8m

s2

57 kg

FN 730N

Elevator Problem # 36a

Mp 95.0kg g 9.8m

s2

FN

a 1.80m

s2

g 9.8m

s2

a 1.80m

s2

F m a

Elevator Problem # 36a

FN Mp g( ) Mp a

FN Mp a Mp g

FN Mp a g( )

FN 1.1 103

N

Elevator Problem # 36b

• Constant speed means NO

acceleration.

• Since acceleration is the result

of an unbalanced force, and

there is no acceleration, the

forces are balanced.


• The magnitude of the normal

force, FN, is therefore the same

as the magnitude of the

person’s weight, mpg.

FN Mp g


FN Mp g

Mp g 95.0kg( ) 9.8m

s2

Mp g 931 N

Elevator Problem # 36c

Mp 95.0kg g 9.8m

s2

FN

a 1.30m

s2

F m a

Elevator Problem # 36c

FN Mp g( ) Mp a

FN Mp a Mp g

FN Mp a g( )

FN 808N

4.11 Equilibrium & the 3rd Law

•Static Equilibrium

•Dynamic Equilibrium

Static Equilibrium

Equilibrium Definition

•Static (not moving) or

•No acceleration

(moving at constant

speed).

4.11 Equilibrium Application of Newton’s Laws of Motion

Definition of Equilibrium

An object is in equilibrium when it has zero acceleration.

0xF

0yF


Reasoning Strategy

• Select an object to which the equations of equilibrium are

to be applied.

• Draw a free-body diagram for each object chosen above.

Include only forces acting on the object, not forces the object

exerts on its environment.

• Choose a set of x, y axes for each object and resolve all forces

in the free-body diagram into components that point along these

axes.

• Apply the equations and solve for the unknown quantities.


035sin35sin 21 TT

035cos35cos 21 FTT

4.12 Nonequilibrium Application of Newton’s Laws of Motion

xx maF

yy maF

When an object is accelerating, it is not in equilibrium.


The acceleration is along the x axis so 0ya


Force x component y component

1T

2T

D

R

0.30cos1T

0.30cos2T

0

0

D

R

0.30sin1T

0.30sin2T


00.30sin0.30sin 21 TTFy

21 TT

x

x

ma

RDTTF

0.30cos0.30cos 21


TTT 21

N 1053.10.30cos2

5

DRmaT x

Two-Body Tension

Calculate the tension in each of the

three support wires. M=5.0kg

from University Physics by H. Benson

Fx mA mB ax

ax

Fx

mA mB

Fx Arepresents the magnitude

of the Normal force FN .

ms

Fs

FN

2-Body

Double Incline Problem

FBD Method for finding ax

Fx m1

m1 ax Fx m2

m2 ax

T m1 g sin 1 m1 ax m2 g sin 2 T m2 ax

UAM Method for finding au

Fu mT

mT au

m2 g sin 2 m1 g sin 1 m1 m2 au

Atwood’s Machine

Lab: Find tan theta for block:

1. Just about to slide down.

2. Moving down at constant speed

Chapter 4

Forces and Newton’s

Laws of Motion

4.1 The Concepts of Force and Mass

A force is a push or a pull.

Contact forces arise from physical

contact .

Action-at-a-distance forces do not

require contact and include gravity

and electrical forces.


Arrows are used to represent forces. The length of the arrow

is proportional to the magnitude of the force.

15 N

5 N


Mass is a measure of the amount

of “stuff” contained in an object.

4.2 Newton’s First Law of Motion

An object continues in a state of rest

or in a state of motion at a constant

speed along a straight line, unless

compelled to change that state by a

net force.

The net force is the vector sum of all

of the forces acting on an object.

Newton’s First Law


The net force on an object is the vector sum of

all forces acting on that object.

The SI unit of force is the Newton (N).

Individual Forces Net Force

10 N 4 N 6 N


Individual Forces Net Force

3 N

4 N

5 N

64


Inertia is the natural tendency of an

object to remain at rest in motion at

a constant speed along a straight line.

The mass of an object is a quantitative

measure of inertia.

SI Unit of Mass: kilogram (kg)


An inertial reference frame is one in

which Newton’s law of inertia is valid.

All accelerating reference frames are

noninertial.

4.3 Newton’s Second Law of Motion

F

Mathematically, the net force is

written as

where the Greek letter sigma

denotes the vector sum.


Newton’s Second Law

When a net external force acts on an object

of mass m, the acceleration that results is

directly proportional to the net force and has

a magnitude that is inversely proportional to

the mass. The direction of the acceleration is

the same as the direction of the net force.

m

Fa

aF

m


SI Unit for Force

22 s

mkg

s

mkg

This combination of units is called a newton (N).



A free-body-diagram is a diagram that

represents the object and the forces that

act on it.


The net force in this case is:

275 N + 395 N – 560 N = +110 N

and is directed along the + x axis of the coordinate system.


If the mass of the car is 1850 kg then, by

Newton’s second law, the acceleration is

2sm059.0kg 1850

N110

m

Fa

4.4 The Vector Nature of Newton’s Second Law

xx maFyy maF

The direction of force and acceleration vectors

can be taken into account by using x and y

components.

aF

m

is equivalent to




+17 N

+(15 N) cos67

0 N

+(15 N) sin67

+23 N +14 N

The net force on the raft can be calculated

in the following way:

P

A


2sm 018.0kg 1300

N 23

m

Fa

x

x

2sm 011.0kg 1300

N 14

m

Fa

y

y

4.5 Newton’s Third Law of Motion

Newton’s Third Law of Motion

Whenever one body exerts a force on a

second body, the second body exerts an

oppositely directed force of equal

magnitude on the first body.


Suppose that the magnitude of the force is 36 N. If the mass

of the spacecraft is 11,000 kg and the mass of the astronaut

is 92 kg, what are the accelerations?


.astronaut On the

. spacecraft On the

PF

PF

2sm0033.0kg 11,000

N 36

s

sm

Pa

2sm39.0kg 92

N 36

A

Am

Pa

4.6 Types of Forces: An Overview

In nature there are two general types of forces,

fundamental and nonfundamental.

Fundamental Forces

1. Gravitational force

2. Strong Nuclear force

3. Electroweak force

4.6 Types of Forces: An Overview

Examples of nonfundamental forces:

friction

tension in a rope

normal or support forces

4.7 The Gravitational Force

Newton’s Law of Universal Gravitation

Every particle in the universe exerts an attractive force on every

other particle.

A particle is a piece of matter, small enough in size to be

regarded as a mathematical point.

The force that each exerts on the other is directed along the line

joining the particles.


For two particles that have masses m1 and m2 and are

separated by a distance r, the force has a magnitude

given by

2

21

r

mmGF

2211 kgmN10673.6 G


N 104.1

m 1.2

kg 25kg 12kgmN1067.6

8

2

2211

2

21

r

mmGF



Definition of Weight

The weight of an object on or above the earth is the

gravitational force that the earth exerts on the object.

The weight always acts downwards, toward the center

of the earth.

On or above another astronomical body, the weight is the

gravitational force exerted on the object by that body.

SI Unit of Weight: newton (N)


Relation Between Mass and Weight

2r

mMGW

E

mgW

2r

MGg E


2

26

242211

2

sm 80.9

m 106.38

kg 1098.5kgmN1067.6

E

E

R

MGg

On the earth’s surface:


Definition of the Normal Force

The normal force is one component of the force that a surface

exerts on an object with which it is in contact – namely, the

component that is perpendicular

to the surface.


N 26

0N 15N 11

N

N

F

F

N 4

0N 15N 11

N

N

F

F


Apparent Weight

The apparent weight of an object is the reading of the scale.

It is equal to the normal force the scale exerts on the man.


mamgFF Ny

mamgFN

apparent

weight

true

weight


When an object is in contact with a surface there is a force

acting on that object. The component of this force that is

parallel to the surface is called the

frictional force.


When the two surfaces are

not sliding across one another

the friction is called

static friction.


The magnitude of the static frictional force can have any value

from zero up to a maximum value.

MAX

ss ff

Ns

MAX

s Ff m

10 sm is called the coefficient of static friction.


Note that the magnitude of the frictional force does

not depend on the contact area of the surfaces.


Static friction opposes the impending relative motion between

two objects.

Kinetic friction opposes the relative sliding motion motions that

actually does occur.

Nkk Ff m

10 sm is called the coefficient of kinetic friction.



The sled comes to a halt because the kinetic frictional force

opposes its motion and causes the sled to slow down.


Suppose the coefficient of kinetic friction is 0.05 and the total

mass is 40kg. What is the kinetic frictional force?

N20sm80.9kg4005.0 2

mgFf kNkk mm

4.10 The Tension Force

Cables and ropes transmit

forces through tension.

4.10 The Tension Force

A massless rope will transmit

tension undiminished from one

end to the other.

If the rope passes around a

massless, frictionless pulley, the

tension will be transmitted to

the other end of the rope

undiminished.


Definition of Equilibrium

An object is in equilibrium when it has zero acceleration.

0xF

0yF


Reasoning Strategy

• Select an object(s) to which the equations of equilibrium are

to be applied.

• Draw a free-body diagram for each object chosen above.

Include only forces acting on the object, not forces the object

exerts on its environment.

• Choose a set of x, y axes for each object and resolve all forces

in the free-body diagram into components that point along these

axes.

• Apply the equations and solve for the unknown quantities.


035sin35sin 21 TT

035cos35cos 21 FTT



N 3150W


1T

2T

W

0.10sin1T

0.80sin2T

0

0.10cos1T

0.80cos2T

W


00.80sin0.10sin 21 TTFx

00.80cos0.10cos 21 WTTFy

The first equation gives 21

0.10sin

0.80sinTT

Substitution into the second gives

00.80cos0.10cos0.10sin

0.80sin22

WTT


0.80cos0.10cos0.10sin

0.80sin2

WT

N 5822 T N 1030.3 3

1 T


xx maF

yy maF

When an object is accelerating, it is not in equilibrium.


The acceleration is along the x axis so 0ya



1T

2T

D

R

0.30cos1T

0.30cos2T

0

0

D

R

0.30sin1T

0.30sin2T


00.30sin0.30sin 21 TTFy

21 TT

x

x

ma

RDTTF

0.30cos0.30cos 21


TTT 21

N 1053.10.30cos2

5

DRmaT x

Problem 59 (Honors)

Documents

Ch 4 Forces & Newton’s Laws · 2018. 10. 22. · 90.0 km/h. At that speed the forces resisting motion, including friction and air resistance, total 400 N. (Air resistance is analogous