Chapter 6. Dynamics I: Motion Along a Line

Chapter 6. Dynamics I: Motion
Along a Line
This 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.

Chapter 6. Dynamics I: Motion
Topics:
• Equilibrium
Along a Line
• Using Newton’s Second Law
• 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!!
r r r r
F = T + T + T
=
net
1 2 3
0
− T T cosθ
0
1 + 3 =
− T T sinθ
0
2 + 3 =
4

Special The First Class Class of of Problems: Equilibrium
1. Static Equilibrium: no motion (velocity = 0, then acceleration = 0 )
2. Dynamical Equilibrium: no acceleration (velocity = constant)
Dynamic Equilibrium:
Convenient coordinate
system!!
r
r
r
r
r
F = n + T + w + f
=
net
k
0
k
T − f − w sin θ
=
0
f r n − w cosθ
=
0
k
Kinetic friction
5

The Second Special Class of Problems: Find Equilibrium Acceleration
r
Important:
r F net
- Introduce convenient co-ordinate system!! a =
m
- Understand what the direction of acceleration is!!
a r
f r k
Kinetic friction
r
r r
r r
r
F = n + T + w + f =
ma
net
T − f − w sinθ
=
ma
k
k
n − w cosθ
=
0
6

The Second Special Class of Problems: Find Equilibrium Acceleration
a r
f r k
r
r r
r r
r
F = n + T + w + f =
ma
net
T − f − w sinθ
=
ma
n − w cosθ
=
0
T = 20N
is given
m = 1kg ⇒ w = mg ≈ 10N
is given
k
k
r
a =
r
F net
m
θ =
0
30
is given
Then:
= θ
= =
0
n = w cos = 10cos 30 =
8.7N
T − fk
− w sinθ
20 − fk
−
5
a
= =
m
1
?
7

Friction
Static friction:
f ≤ f = µ n
s
s s,max
s
µ - coefficient of static
friction (it is usually
given in the problem)
n
- normal force
Kinetic friction:
k k
f
= µ n
k
µ - coefficient of kinetic
friction (it is usually
given in the problem)
Rolling friction:
r r
f
= µ n
r
µ - coefficient of rolling
friction (it is usually
given in the problem)
µ > µ >
µ
Usually: s k r
very small
8

Friction
Static friction:
fs ≤ fs,max
= µ
sn
s
n
µ - coefficient of static friction
- normal force
Find the maximum tension,
T max
y
Equilibrium:
r
r
r
r
r
F = n + w + T + f
=
net
n
− w
=
0
s
0
x
n r
T
r
weight,
friction
tension
− f
+ T
=
s
n = w
f = T ≤
f
0
s s,max
f r w r
s
f = µ n =
µ
w
s,max
s s
normal
Condition of equilibrium:
T
≤
µ w
s
T
max
=
µ w
s
9

Friction
Static friction:
Find the maximum tension,
Equilibrium:
r
r
r
r
r
F = n + w + T + f
=
0
net
n
− w
=
0
− f
+ T
=
s
0
s
fs ≤ fs,max
= µ
sn
s
n
T max
f = µ n =
µ
w
s,max
s s
f = T ≤
f
s s,max
µ - coefficient of static friction
- normal force
fs
= T
y
f r s
x
n r
w r
T r
Condition of equilibrium:
T ≤ µ w
T
max
s
= µ w
s
f
s,max
= µ w
s
T max
T
10

Kinetic friction:
k k
f
µ n
Friction
= k
µ - coefficient of kinetic friction
n
- normal force
Find acceleration,
( T > T )
max
y
weight,
r
r r r
r
r
F = n + w + T + f =
ma
net
n
− w
=
0
− f + T =
ma
k
k
f r s
x
n r
w r
a r
T
r
friction
tension
n
= w then fk = µ kn =
µ
kw
normal
a
T −
fk
T −
µ
kw T
= = = −
µ
k
g
m m m
11

Friction
Static friction:
fs ≤ fs,max
= µ
sn
s
n
µ - coefficient of static friction
- normal force
Find the maximum angle, θ max
Equilibrium:
r r
r
r
F = n + w + f
=
net
n − w cos θ
=
0
s
0
− f + w sinθ
=
0
s
n = w cosθ
f = w sinθ
≤
f
s
s,max
f = µ n =
µ w cosθ
s,max
s s
Condition of equilibrium:
w sin
θ ≤
µ w sin cos(theta)
θ
tan
θ ≤
µ
tan
θ =
µ
max
s
s
s
12

Friction
Static friction:
fs ≤ fs,max
= µ
sn
s
n
µ - coefficient of static friction
- normal force
Find the maximum angle, θ max
f = µ n =
µ w cosθ
s,max
s s
f = w sinθ
≤
f
s
s,max
f = w sinθ
s
f
s,max
=
µ w cos
θ
s
θ max
θ
Condition of equilibrium: tan ≤
s
θ µ
13

Kinetic friction:
f
k
µ
Friction
=
kn
k
µ - coefficient of kinetic friction
n
- normal force
Find acceleration
( θ
>
θ
)
max
r r r r
r
F = n + w + f =
ma
net
n − w cosθ
=
0
− f + w sinθ
=
ma
k
k
a r
n = w cosθ
then fk = µ kn =
µ kw
cos
θ
− fk
+ w sinθ − µ kw cosθ +
w sinθ
a
= = = g(sinθ −
µ k
cos θ
)
m
m
14

Weight and Apparent Weight
Weight – gravitational force - pulls the objects down
r r m - mass of the object (the same on all planets)
w = mg
How can we measure weight?
g
= 9.8 m - free-fall acceleration (different
2
s on different planets)
1. We can measure mass by comparing with
the known mass
m
unknown
=
m
known
2. We can measure the weight by comparing
with the known force
w = F spring
15

Weight and Apparent Weight
Apparent Weight – reading of the scale
(or the normal force)
In equilibrium:
r
F
net
=
0
then
r r
Fnet
= ma
F − w =
ma
spring
then
F = m( g +
a)
spring
Fspring
=
w
Motion with acceleration:
a r
r r
Fnet
= ma
F −
w = −
ma
spring
then
F = m( g −
a)
spring
a r
The man feels heavier than
normal while accelerating upward
The man feels lighter than normal
while accelerating upward
16

Chapter 6. Summary Slides

General Strategy

General Strategy

Important Concepts

Important Concepts

Applications

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.
D. zero.

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.
D. zero.

Rank order, from largest to r smallest, r the
size of the friction forces f a to fe in these five
different situations. The box and the floor
are made of the same materials in all
situations.
A. f c > f d > f e > f b > f a .
B. f b > f c = f d = f e > f a .
C. f b > f c > f d > f e > f a .
D. f a > f c = f d = f e > f b .
E. f a = f b > f c = f d = f e .

Rank order, from largest to smallest, the
size of the friction forces r to in these five
different situations. The box and the floor
are made of the same materials in all
situations.
f a
r
fe
A. f c > f d > f e > f b > f a .
B. f b > f c = f d = f e > f a .
C. f b > f c > f d > f e > f a .
D. f a > f c = f d = f e > f b .
E. f a = f b > f c = f d = f e .

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

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