chapter 6. dynamics i: motion along a line - gsu p&a · pdf filechapter 6. dynamics i:...

Chapter 6. Dynamics I: Motion Chapter 6. Dynamics I: Motion

Along a LineAlong a LineThis chapter focuses on objects that

move in a straight line, such as runners,

bicycles, cars, planes, and rockets.

Gravitational, tension, thrust, friction, and

drag forces will be essential to our

understanding.

Chapter Goal: To learn how to solve

problems about motion in a straight line.

Topics:

• Equilibrium

• Using Newton’s Second Law

• Mass, Weight, and Gravity

Chapter 6. Dynamics I: Motion Chapter 6. Dynamics I: Motion

Along a LineAlong a Line

• Mass, Weight, and Gravity

• Friction

• Drag

• More Examples of Newton’s Second Law

1) Object – as a particle

2) Identify all the forces

3) Find the net force (vector sum of all individual forces)

4) Introduce convenient co-ordinate system

5) Find the acceleration of the object (second Newton’s law)

6) With the known acceleration find kinematics of the object

3

The First Class of Problems: Equilibrium

1. Static Equilibrium: no motion (velocity = 0, then acceleration = 0 )

2. Dynamical Equilibrium: no acceleration (velocity = constant)

- In both cases acceleration = 0

- Second Newton’s Law – Net Force = 0

Static Equilibrium:Convenient co-ordinate system!!

4

Static Equilibrium:ordinate system!!

1 2 30

netF T T T= + + == + + == + + == + + =r r r rr r r rr r r rr r r r

1 3cos 0T T θθθθ− + =− + =− + =− + =

2 3sin 0T T θθθθ− + =− + =− + =− + =

Special Class of Problems: Equilibrium

1. Static Equilibrium: no motion (velocity = 0, then acceleration = 0 )

2. Dynamical Equilibrium: no acceleration (velocity = constant)

Dynamic Equilibrium: Convenient co-ordinate system!!

sin 0T f w θθθθ− − =− − =− − =− − =

0net k

F n T w f= + + + == + + + == + + + == + + + =rrrrr rr rr rr rr rr rr rr r

The First Class of Problems: Equilibrium

5

sin 0k

T f w θθθθ− − =− − =− − =− − =

Kinetic friction

kfrrrr cos 0n w θθθθ− =− =− =− =


Important:

- Introduce convenient co-ordinate system!!

- Understand what the direction of acceleration is!!

The Second Class of Problems: Find Acceleration

netF

am

====

rrrrrrrr

arrrr

6

sink

T f w maθθθθ− − =− − =− − =− − =

Kinetic friction

kfrrrr

net kF n T w f ma= + + + == + + + == + + + == + + + =

rrrrr rr rr rr rr r rr r rr r rr r r

cos 0n w θθθθ− =− =− =− =


sink

T f w maθθθθ− − =− − =− − =− − =

kfrrrr

net kF n T w f ma= + + + == + + + == + + + == + + + =


The Second Class of Problems: Find Acceleration

netF

am

====

rrrrrrrra

rrrr

7

cos 0n w θθθθ− =− =− =− =

20T N==== is given

1 10m kg w mg N==== ⇒⇒⇒⇒ = ≈= ≈= ≈= ≈ is given

030θθθθ ==== is given

Then:

0cos 10cos 30 8.7n w Nθθθθ= = == = == = == = =

sin 20 5

1

k kT f w f

am

θθθθ− − − −− − − −− − − −− − − −= == == == =

?

Friction

Static friction: ,maxs s s

f f nµµµµ≤ =≤ =≤ =≤ = sµµµµ - coefficient of static

friction (it is usually given in the problem)

n - normal force

Kinetic friction: k k

f nµµµµ==== kµµµµ - coefficient of kinetic


8

given in the problem)

Rolling friction: r r

f nµµµµ==== rµµµµ - coefficient of rolling


Usually: s k rµ µ µµ µ µµ µ µµ µ µ> >> >> >> >

very small

Friction

Static friction:,maxs s s

f f nµµµµ≤ =≤ =≤ =≤ = sµµµµ - coefficient of static friction

n - normal force

Find the maximum tension, max

T

Equilibrium:

0net s

F n w T f= + + + == + + + == + + + == + + + =rrrrr rr rr rr rr rr rr rr r

0n w− =− =− =− =

nrrrr

Trrrr

x

y

9

0s

f T− + =− + =− + =− + =

0n w− =− =− =− =

n w====

,maxs sf T f= ≤= ≤= ≤= ≤

,maxs s sf n wµ µµ µµ µµ µ= == == == =

sT wµµµµ≤≤≤≤Condition of equilibrium:

max sT wµµµµ====

weight,

friction

tension

normal

wrrrrs

frrrr

T

Friction



n - normal force

Find the maximum tension, max

T

Equilibrium:

0net s

F n w T f= + + + == + + + == + + + == + + + =rrrrr rr rr rr rr rr rr rr r

0n w− =− =− =− =,maxs s

f T f= ≤= ≤= ≤= ≤

,maxs s sf n wµ µµ µµ µµ µ= == == == =

nrrrr

wrrrrs

frrrr

Trrrr

x

y

10

0s

f T− + =− + =− + =− + =

,maxs s

sT wµµµµ≤≤≤≤

Condition of equilibrium:

max sT wµµµµ====

wrrrrs

f

Tmax

T

,maxs sf wµµµµ====

sf T====

Friction

Kinetic friction:k k

f nµµµµ==== kµµµµ - coefficient of kinetic friction

n - normal force

Find acceleration, max

( )T T>>>>

net kF n w T f ma= + + + == + + + == + + + == + + + =


f T ma− + =− + =− + =− + =

0n w− =− =− =− =nrrrr

Trrrr

x

y

arrrr

11

kf T ma− + =− + =− + =− + =

n w====k k k

f n wµ µµ µµ µµ µ= == == == =

weight,

friction

tension

normal

wrrrrs

frrrr

T

k k

k

T f T w Ta g

m m m

µµµµµµµµ

− −− −− −− −= = = −= = = −= = = −= = = −

then

Friction



n - normal force

Find the maximum angle, max

θθθθ

Equilibrium:

0net s

F n w f= + + == + + == + + == + + =rrrrrrrr r rr rr rr r

cos 0n w θθθθ− =− =− =− =

12

sin 0s

f w θθθθ− + =− + =− + =− + =

cos 0n w θθθθ− =− =− =− =

cosn w θθθθ====

,maxsin

s sf w fθθθθ= ≤= ≤= ≤= ≤

,maxcos

s s sf n wµ µ θµ µ θµ µ θµ µ θ= == == == =

sin sins

w wθ µ θθ µ θθ µ θθ µ θ≤≤≤≤Condition of equilibrium:

tans

θ µθ µθ µθ µ≤≤≤≤

maxtan

sθ µθ µθ µθ µ====

cos(theta)

Friction



n - normal force

Find the maximum angle, max

θθθθ

,maxsin

s sf w fθθθθ= ≤= ≤= ≤= ≤

,maxcos

s s sf n wµ µ θµ µ θµ µ θµ µ θ= == == == =

sins

f w θθθθ====

13

Condition of equilibrium: tans

θ µθ µθ µθ µ≤≤≤≤

θθθθmaxθθθθ

,maxcos

s sf wµ θµ θµ θµ θ====

Friction

Kinetic friction:n - normal force

Find acceleration max

( )θ θθ θθ θθ θ>>>>

net kF n w f ma= + + == + + == + + == + + =

rrrrrrrr r r rr r rr r rr r r

cos 0n w θθθθ− =− =− =− =

k kf nµµµµ==== k

µµµµ - coefficient of kinetic friction

arrrr

14

sink

f w maθθθθ− + =− + =− + =− + =

cosn w θθθθ==== cosk k k

f n wµ µ θµ µ θµ µ θµ µ θ= == == == =then

sin cos sin(sin cos )k k

k

f w w wa g

m m

θ µ θ θθ µ θ θθ µ θ θθ µ θ θθ µ θθ µ θθ µ θθ µ θ

− + − +− + − +− + − +− + − += = = −= = = −= = = −= = = −

Weight and Apparent Weight

Weight – gravitational force - pulls the objects down

w mg====r rr rr rr r m - mass of the object (the same on all planets)

29.8

mg

s==== - free-fall acceleration (different

on different planets)

How can we measure weight?

1. We can measure mass by comparing with

15

1. We can measure mass by comparing with the known mass

2. We can measure the weight by comparing with the known force

springw F====

unknown knownm m====

Weight and Apparent Weight

Apparent Weight – reading of the scale

(or the normal force)

springF w====

In equilibrium:

0net

F ====rrrr

then

Motion with acceleration: arrrr

16

Motion with acceleration:

netF ma====rrrr rrrr

springF w ma− =− =− =− =

( )spring

F m g a= += += += +

then

The man feels heavier than normal while accelerating upward

netF ma====rrrr rrrr

springF w ma− = −− = −− = −− = −

( )spring

F m g a= −= −= −= −

then

The man feels lighter than normal while accelerating upward

arrrr

arrrr

Chapter 6. Summary SlidesChapter 6. Summary SlidesChapter 6. Summary SlidesChapter 6. Summary Slides

General Strategy

Important Concepts

Applications

An elevator that has descended from the 50th floor is coming to a halt at the 1st floor. As it does, your apparent weight is

A. less than your true weight.

B. equal to your true weight.

C. more than your true weight.C. more than your true weight.

D. zero.

Rank order, from largest to smallest, the size of the friction forces to in these five different situations. The box and the floor are made of the same materials in all situations.

rfa

rfe

A. fc > fd > fe > fb > fa.

B. fb > fc = fd = fe > fa.

C. fb > fc > fd > fe > fa.

D. fa > fc = fd = fe > fb.

E. fa = fb > fc = fd = fe.

The terminal speed of a Styrofoam ball is 15 m/s. Suppose a Styrofoam ball is shot straight down with an initial speed of30 m/s. Which velocity graph is correct?

chapter 6. dynamics i: motion along a line - gsu p&a · pdf filechapter 6. dynamics i:...

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