Download 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
Along a Line
Topics:
• Equilibrium
• 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 coordinate system!!
r
r r r
Fnet = T1 + T2 + T3 = 0
−T1 + T3 cos θ = 0
−T2 + T3 sin θ = 0
4
Special
The
First
Class
Class
of of
Problems:
Problems:
Equilibrium
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
Fnet = n + T + w + f k = 0
r
fk
T − f k − w sin θ = 0
n − w cos θ = 0
Kinetic friction
5
The Second
Acceleration
SpecialClass
Classof
ofProblems:
Problems: Find
Equilibrium
r
Important:
r Fnet
a=
- Introduce convenient co-ordinate system!!
m
- Understand what the direction of acceleration is!!
r
a
r
fk
Kinetic friction
r
r r r r
r
Fnet = n + T + w + f k = ma
T − f k − w sin θ = ma
n − w cos θ = 0
6
The Second
Acceleration
SpecialClass
Classof
ofProblems:
Problems: Find
Equilibrium
r
r Fnet
a=
m
r
a
r
fk
r
r r r r
r
Fnet = n + T + w + f k = ma
T − f k − w sin θ = ma
n − w cos θ = 0
T = 20 N
is given
m = 1kg ⇒ w = mg ≈ 10 N
θ = 300
is given
is given
n = w cos θ = 10cos 300 = 8.7 N
?
Then:
T − f k − w sin θ 20 − f k − 5
a=
=
m
1
7
Friction
µs
f s ≤ f s ,max = µ s n
Static friction:
- coefficient of static
friction (it is usually
given in the problem)
n
- normal force
Kinetic friction:
f k = µk n
µk
- coefficient of kinetic
friction (it is usually
given in the problem)
Rolling friction:
f r = µr n
µr
- coefficient of rolling
friction (it is usually
given in the problem)
Usually:
µ s > µk > µr
very small
8
Friction
Static friction:
f s ≤ f s ,max = µ s n
Find the maximum tension,
µs
- coefficient of static friction
n
- normal force
Tmax
y
Equilibrium:
r
r r r r
Fnet = n + w + T + f s = 0
x
n−w = 0
r
fs
− fs + T = 0
r
n
r
T
r
w
weight,
friction
tension
normal
n=w
f s ,max = µ s n = µ s w
f s = T ≤ f s ,max
Condition of equilibrium:
T ≤ µs w
Tmax = µ s w
9
Friction
Static friction:
f s ≤ f s ,max = µ s n
Find the maximum tension,
µs
n
- coefficient of static friction
- normal force
y
Tmax
Equilibrium:
r
r r r r
Fnet = n + w + T + f s = 0
n− w = 0
x
f s ,max = µ s n = µ s w
r
fs
f s = T ≤ f s ,max
− fs + T = 0
r
n
r
T
r
w
fs = T
Condition of equilibrium:
f s ,max = µ s w
T ≤ µs w
Tmax = µ s w
Tmax
T
10
Friction
Kinetic friction:
Find acceleration,
µk
f k = µk n
n
- coefficient of kinetic friction
- normal force
(T > Tmax )
y
r
r r r r
r
Fnet = n + w + T + f k = ma
n− w = 0
− f k + T = ma
x
r
fs
r
n
r
a
r
T
r
w
weight,
friction
tension
normal
n=w
then
f k = µk n = µk w
T − f k T − µk w T
a=
=
= − µk g
m
m
m
11
Friction
Static friction:
f s ≤ f s ,max = µ s n
Find the maximum angle,
µs
- coefficient of static friction
n
- normal force
θ max
Equilibrium:
r
r r r
Fnet = n + w + f s = 0
n − w cos θ = 0
− f s + w sin θ = 0
n = w cos θ
f s ,max = µ s n = µ s w cos θ
f s = w sin θ ≤ f s ,max
Condition of equilibrium:
cos(theta)
w sin θ ≤ µ s w sin
θ
tan θ ≤ µ s
tan θ max = µ s
12
Friction
Static friction:
f s ≤ f s ,max = µ s n
Find the maximum angle,
θ max
µs
n
- coefficient of static friction
- normal force
f s ,max = µ s n = µ s w cos θ
f s = w sin θ ≤ f s ,max
f s = w sin θ
f s ,max = µ s w cos θ
θ max
θ
Condition of equilibrium: tan θ ≤ µ
s
13
Friction
Kinetic friction:
Find acceleration
µk
f k = µk n
- coefficient of kinetic friction
n
- normal force
(θ > θ max )
r
r r r
r
Fnet = n + w + f k = ma
r
a
n − w cos θ = 0
− f k + w sin θ = ma
n = w cos θ
then
f k = µ k n = µk w cos θ
− f k + w sin θ − µ k w cos θ + w sin θ
a=
= g (sin θ − µk cos θ )
=
m
m
14
Weight and Apparent Weight
Weight – gravitational force - pulls the objects down
r
r
w = mg
m
- mass of the object (the same on all planets)
m
g = 9.8 2
s
- free-fall acceleration (different
on different planets)
How can we measure weight?
1. We can measure mass by comparing with
the known mass
munknown = mknown
2. We can measure the weight by comparing
with the known force
w = Fspring
15
Weight and Apparent Weight
Apparent Weight – reading of the scale
(or the normal force)
In equilibrium:
r
Fnet = 0
then
Fspring = w
Motion with acceleration:
r
r
Fnet = ma
Fspring − w = ma
r
a
r
r
Fnet = ma
Fspring − w = − ma
then
then
Fspring = m ( g + a )
Fspring = m ( g − a )
The man feels heavier than
normal while accelerating upward
r
a
The man feels lighter than normal
16
while accelerating upward
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
the
r
r smallest,
size of the friction forces fa to fe in these five
different situations. The box and the floor
are made of the same materials in all
situations.
A.
B.
C.
D.
E.
fc >
fb >
fb >
fa >
fa =
fd >
fc =
fc >
fc =
fb >
fe >
fd =
fd >
fd =
fc =
f b > f a.
f e > f a.
f e > f a.
f e > f b.
f d = f e.
Rank order, from largest to smallest, the
r
r to in these five
size of the friction forces
fa
fe
different situations. The box and the floor
are made of the same materials in all
situations.
A.
B.
C.
D.
E.
fc >
fb >
fb >
fa >
fa =
fd >
fc =
fc >
fc =
fb >
fe >
fd =
fd >
fd =
fc =
f b > f a.
f e > f a.
f e > f a.
f e > f b.
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?