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Newton’s Second Law:
Motion Along a Line
Readings: Chapter 6 (2nd edition)
1
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
2
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!!
Fnet  T1  T2  T3  0
T1  T3 cos   0
T2  T3 sin  0
3
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!!
Fnet  n  T  w  f k  0
T  f k  w sin   0
n  w cos  0
fk
Kinetic friction
4
The Second
Acceleration
SpecialClass
Classof
ofProblems:
Problems:Find
Equilibrium
Important:
- Introduce convenient co-ordinate system!!
- Understand what the direction of acceleration is!!
Fnet
a
m
a
fk
Kinetic friction
Fnet  n  T  w  f k  ma
T  f k  w sin   ma
n  w cos  0
5
The Second
Acceleration
SpecialClass
Classof
ofProblems:
Problems:Find
Equilibrium
Fnet
a
m
a
Fnet  n  T  w  f k  ma
fk
T  f k  w sin   ma
n  w cos  0
T  20 N
is given
m  1kg  w  mg  10 N
  300
is given
n  w cos   10cos 300  8.7 N
is given
Then:
T  f k  w sin  20  f k  5
a

m
1
?
6
Friction
s
f s  f s ,max   s n
Static friction:
- coefficient of static
friction (it is usually
given in the problem)
n
Kinetic friction:
f k  k n
k
Rolling friction:
f r  r n
r
Usually:
- normal force
- coefficient of kinetic
friction (it is usually
given in the problem)
- coefficient of rolling
friction (it is usually
given in the problem)
 s  k  r
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very small
Friction
Static friction:
f s  f s ,max   s n
Find the maximum tension,
s
n
- coefficient of static friction
- normal force
Tmax
y
Equilibrium:
x
Fnet  n  w  T  f s  0
n
T
n w  0
fs
 fs  T  0
weight,
friction
tension
normal
nw
w
f s ,max   s n   s w
f s  T  f s ,max
Condition of equilibrium:
T  s w
Tmax   s w
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Friction
Static friction:
f s  f s ,max   s n
Find the maximum tension,
Equilibrium:
Fnet  n  w  T  f s  0
n w  0
 fs  T  0
s
n
- coefficient of static friction
- normal force
y
Tmax
x
f s ,max   s n   s w
n
T
f s  T  f s ,max
fs
w
fs  T
Condition of equilibrium:
f s ,max   s w
T  s w
Tmax   s w
Tmax
T
9
Friction
Kinetic friction:
Find acceleration,
f k  k n
k
n
- coefficient of kinetic friction
- normal force
(T  Tmax )
y
Fnet  n  w  T  f k  ma
x
n w  0
 f k  T  ma
a
n
T
fs
w
weight,
friction
tension
normal
n  w then
f k  k n  k w
T  f k T  k w T
a

  k g
m
m
m
10
Friction
Static friction:
f s  f s ,max   s n
Find the maximum angle,
s
n
- coefficient of static friction
- normal force
max
Equilibrium:
Fnet  n  w  f s  0
n  w cos  0
 f s  w sin   0
n  w cos
f s ,max   s n   s w cos 
f s  w sin   f s ,max
Condition of equilibrium:
w sin   s w sin 
tan   s
tan  max   s
11
Friction
Static friction:
f s  f s ,max   s n
Find the maximum angle,
max
s
n
- coefficient of static friction
- normal force
f s ,max   s n   s w cos 
f s  w sin   f s ,max
f s  w sin 
f s ,max   s w cos 
max

Condition of equilibrium: tan   
s
12
Friction
Kinetic friction:
Find acceleration
k
n
f k  k n
- coefficient of kinetic friction
- normal force
(  max )
Fnet  n  w  f k  ma
a
n  w cos  0
 f k  w sin   ma
n  w cos
then
f k  k n  k w cos
 f k  w sin   k w cos   w sin
a

 g(sin   k cos  )
m
m
13
Weight and Apparent Weight
Weight – gravitational force - pulls the objects down
w  mg
m
- mass of the object (the same on all planets)
g  9.8
m
s2
- 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
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Weight and Apparent Weight
Apparent Weight – reading of the scale
(or the normal force)
In equilibrium:
Fnet  0
then
Fspring  w
Motion with acceleration:
Fnet  ma
Fspring  w  ma
a
Fnet  ma
Fspring  w   ma
then
then
Fspring  m ( g  a )
Fspring  m ( g  a )
The man feels heavier than
normal while accelerating upward
a
The man feels lighter than normal
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while accelerating upward