Unit 2
Newton’s Laws
• Newton’s Laws of Motion
1) Law of inertia: An object will remain in its
current state of motion unless acted upon by
an external force.
2) ΣF = ma
3) For every action (force) there is an equal and
opposite reaction (force). (Forces always
come in pairs!)
ΣF = ma
or
Fnet = ma
• Types of forces
1) Field
a) Gravity
b) Electromagnetic
2) Contact
a)
b)
c)
d)
e)
Tension
Normal
Friction
Applied
Resistance
• Gravity
• Fg = mg
• Also known as “weight”
• Tension
• T = mg
for a suspended mass
• T < mg
for a falling mass
• T > mg
for a mass accelerating upward
• Normal
• Normal means perpendicular
• N = mg
on a horizontal surface*
• N = mgcosθ on a slope of angle θ*
• *Applied forces that are not parallel to the
surface will affect the normal force
• Normal
• ΣF = ma
• mg + Fsinθ – N = ma
• But a = 0
• N = mg + Fsinθ
• Friction
• f = μN
• When two surfaces slide relative to each other
it is kinetic friction μk
• When two surfaces are stationary relative to
each other it is static friction μs
• μ is almost always less than 1
• For a given pair of surfaces, μk < μs
• Applied
• Any external push or pull not already covered
• Resistance
• Usually air or water resistance
• A VERY common topic for differential
equations
• A ball falls and is subject to a resistance force
FR = –bv. Develop an equation for the velocity of
the ball as a function of time. At t = 0, v = 0
F  ma
mg  bv  ma
dv
mg  bv  m
dt
dt
dv

m mg  bv
dt
dv
 m   mg  bv
1
1
t
ln mg  bv   C
m
b
 bt
 ln mg  bv   C
m
 bt
 ln mg  bv   ln C
m
• A ball falls and is subject to a resistance force
FR = –bv. Develop an equation for the velocity of
the ball as a function of time. At t = 0, v = 0
 bt
 ln mg  bv   ln C
m
 bt
 ln C mg  bv 
m
e
 bt / m
 C mg  bv 
Cebt / m  mg  bv
Ceb 0 / m  mg  b0
C  mg
bv  mg  mge
 bt / m
mg mg bt / m
v

e
b
b
mg
v
1  e bt / m
b


• Terminal velocity
• Constant velocity of a falling object
• What would be the terminal velocity in the
previous example?
• A ball falls and is subject to a resistance force
FR = –bv. Develop an equation for the velocity of
the ball as a function of time. At t = 0, v = 0
F  ma
mg  bv  ma
a0
mg  bvT  0
mg
vT 
b
• Free Body Diagrams
• They show all external forces acting on an object
• A vector represents each force.
– The head points in the direction of the force
– The tail begins at the point of origin of the force
– The length of the vector should be representative of the
magnitude of the force
– When asked to draw a free body diagram on a test,
include only forces, not components
– Label appropriately, using symbols from the problem
• Draw a free body diagram for
mass m2 in the diagram
N
T
f
m2 g
• Components of forces
• Always chose a coordinate system to minimize
components
• Components of forces
F
θ
• Components of forces
N
f
θ
F
Fx  max
mg sin   F  f  max
Fy  ma y
mg cos   N  ma y
mg
• Applications of Newton’s third law
F
m1
m2
a
• Determine the force of m2 on m1
F  Fm2  m1a
Fm1  m2 a
F  m1  m2 a
F
a
m1  m2
 F 

F  Fm2  m1 
 m1  m2 
 F 

Fm2  F  m1 
 m1  m2 
m1 F
Fm2  F 
m1  m2
Fm1  m2 a
F
a
m1  m2
 F 

Fm1  m2 
 m1  m2 
m2 F
Fm1 
m1  m2
m1 F
Fm2  F 
m1  m2
m2 F ?
m1 F
F 
m1  m2
m1  m2
m2 ?
m1
1 
m1  m2
m1  m2
?
m2  m1  m2  m1
m2  m1  m1  m2
Tarzan’s Tension!
Tarzan’s Tension!
• Determine the tension in the 12 m vine when 90.
kg Tarzan has swung to the point where he
makes a 30.° angle with the vertical and is
moving with a linear speed of 5.0 m/s.
T  mg cos   ma
2
v
T  mg cos   m
r
v2
T  mg cos   m
r
30°
mgcosθ
 5.02 
  967 N
T  9010 cos30  90
 12 
mg
T
Jane’s Curves!
• On the way to Nairobi one day to pick
up a few supplies, Jane was driving
around a banked curve of radius r =
150 m. Some naughty monkeys
had thrown banana peels onto the road making it
essentially frictionless. If the road was banked at
an angle of θ = 30.°, what speed would she need
to drive to maintain her position (not slip up or
down)?
Jane’s Curves!
N
F  ma
θ
N sin   ma
v2
N sin   m
r
v2
mg cos  sin   m
r
2
gr cos  sin   v
v  gr cos  sin 
θ
v
mg
mg
10150cos 30sin 30
v  25 m/s
Cheetah Dances!
• On day, after he found Tarzan’s stash
of home-brewed Serengeti Special,
and drank a couple bottles, Cheetah
put on quite the show. If he completes
two revolutions (4π radians) per second,
determine the velocity of the empty bottle in his
hand. The bottle is located 0.80 m from the
center of his body, which is the axis of rotation.
d 22r 
v 
 4 0.80  10. m/s
t
1
v  r  4r  4 0.80  10. m/s
Elephant Power!
• Tarzan’s mother-in-law came for
an extended visit and when she
was finally ready to leave, her
Jeep was stuck. Tarzan called
an elephant friend named Shep to help out. If the
powerful pachyderm was pushing with a force of
2500 N when he had the vehicle moving at 20.
m/s, what was his power output?
P  Fv
P  250020  50,000 W
• Vector Math
• Vectors are added head to tail
a
+
b
b
=
r
a
• Vector Math
• Vectors are subtracted by reversing the second
and adding
a
–
b
=
a
r
b
• Vector Math
• Vectors are multiplied two ways:
1) Dot product
2) Cross product
• Vector Math
• Fundamentally,
• A dot product is
a  b  a b cos 
• A cross product is
a  b  a b sin  n
• Dot product
• Work is determined with the dot product:
W  F d
• Dot products are scalar
• Dot product
•
•
•
•
•
•
•
Evaluate 𝑚 ∙ 𝑛 if
𝑚 = 𝑎𝑖 + 𝑏𝑗 and
𝑛 = 𝑐𝑖 + 𝑑𝑗
𝑚 ∙ 𝑛 = 𝑎𝑖 + 𝑏𝑗 ∙ 𝑐 𝑖 + 𝑑 𝑗
𝑚 ∙ 𝑛 = 𝑎𝑐 𝑖 ∙ 𝑖 + 𝑎𝑑 𝑖 ∙ 𝑗 + 𝑏𝑐 𝑗 ∙ 𝑖 + 𝑏𝑑 𝑗 ∙ 𝑗
𝑖∙𝑖=𝑗∙𝑗=1
𝑖∙𝑗 =𝑗∙𝑖=0
𝑚 ∙ 𝑛 = 𝑎𝑐 + 𝑏𝑑
• Cross product
• Angular momentum is calculated with a cross
product:
• 𝐿 =𝑟×𝑝
• The direction of the cross product
is given by the right hand rule.
• Cross products are vectors
•
•
•
•
Evaluate p × 𝑞 if
𝑝 = 𝑎𝑖 + 𝑏𝑗 𝑞 = 𝑐 𝑖 + 𝑑 𝑗
𝑝 × 𝑞 = 𝑎𝑖 + 𝑏𝑗 × 𝑐 𝑖 + 𝑑 𝑗
𝑝 × 𝑞 = 𝑎𝑐 𝑖 × 𝑖 + 𝑎𝑑 𝑖 × 𝑗 + 𝑏𝑐 𝑗 × 𝑖 +
𝑏𝑑 𝑗 × 𝑗
• 𝑖×𝑖=𝑗×𝑗 =0
• 𝑖×𝑗=− 𝑗×𝑖 =𝑘
• 𝑝 × 𝑞 = 𝑎𝑑 𝑖 × 𝑗 + 𝑏𝑐 𝑗 × 𝑖
• 𝑝 × 𝑞 = 𝑎𝑑 − 𝑏𝑐 𝑘