1. Physics 111: Lecture 4, Pg 1
Physics 111: Lecture 4Physics 111: Lecture 4
Today’s AgendaToday’s Agenda
Recap of centripetal acceleration
Newton’s 3 laws
How and why do objects move?
DynamicsDynamics

2. Physics 111: Lecture 4, Pg 2
Review: Centripetal AccelerationReview: Centripetal Acceleration
UCM results in acceleration:
Magnitude: |a| = v2
/ R = ω2
R
Direction: - rr (toward center of circle)
R aa
ω
^^
Useful stuff:Useful stuff:
f = rotations / sec
T = 1 / f
ω = 2π / T = 2π f = rad/sec
v = ω R

3. Physics 111: Lecture 4, Pg 3
Isaac Newton (1643 - 1727) published Principia Mathematica in 1687

4. Physics 111: Lecture 4, Pg 4
DynamicsDynamics
In this work, he proposed three “laws” of motion:
Law 1: An object subject to no external forces is at rest or moves
with a constant velocity if viewed from an inertial reference
frame.
Law 2: For any object, FFNET = Σ FF = maa
Law 3: Forces occur in pairs: FFA ,B = - FFB ,A
(For every action there is an equal and opposite reaction.)

5. Physics 111: Lecture 4, Pg 5
Newton’s First LawNewton’s First Law
An object subject to no external forces is at rest or moves
with a constant velocity if viewed from an inertial referenceinertial reference
frameframe.
If no forces act, there is no acceleration.
The following statements can be thought of as the
definition of inertial reference frames.
An IRF is a reference frame that is not accelerating (or
rotating) with respect to the “fixed stars”.
If one IRF exists, infinitely many exist since they are
related by any arbitrary constant velocity vector!
1. Dishes
2. Monkey

6. Physics 111: Lecture 4, Pg 6
Is Urbana a good IRF?Is Urbana a good IRF?
Is Urbana accelerating?
YES!
Urbana is on the Earth.
The Earth is rotating.
What is the centripetal acceleration of Urbana?
T = 1 day = 8.64 x 104
sec,
R ~ RE = 6.4 x 106
meters .
Plug this in: aU = .034 m/s2
( ~ 1/300 g)
Close enough to 0 that we will ignore it.
Urbana is a pretty good IRF.
a
v
R
R
T
RU = = =






2 2
2
ω
π2
Ice
puck

7. Physics 111: Lecture 4, Pg 7
Newton’s Second LawNewton’s Second Law
For any object, FFNET = Σ FF = maa.
The acceleration aa of an object is proportional to the
net force FFNET acting on it.
The constant of proportionality is called “mass”, denoted
m.
» This is the definition of mass.
» The mass of an object is a constant property of that
object, and is independent of external influences.
Force has units of [M]x[L / T2
] = kg m/s2
= N (Newton)

8. Physics 111: Lecture 4, Pg 8
Newton’s Second Law...Newton’s Second Law...
What is a force?
A Force is a push or a pull.
A Force has magnitude & direction (vector).
Adding forces is like adding vectors.
FF1
FF2
aa
FF1
FF2
aa
FFNET
FFNET = maa

9. Physics 111: Lecture 4, Pg 9
Newton’s Second Law...Newton’s Second Law...
Components of FF = maa :
FX = maX
FY = maY
FZ = maZ
Suppose we know m and FX , we can solve for aX and apply
the things we learned about kinematics over the last few
weeks:
tavv
ta
2
1
tvxx
xx0x
2
xx00
+=
++=

10. Physics 111: Lecture 4, Pg 10
Example: Pushing a Box on Ice.Example: Pushing a Box on Ice.
A skater is pushing a heavy box (mass m = 100 kg) across
a sheet of ice (horizontal & frictionless). He applies a force
of 50 N in the ii direction. If the box starts at rest, what is its
speed v after being pushed a distance d = 10 m?
FF
v = 0
m a
ii

11. Physics 111: Lecture 4, Pg 11
Example: Pushing a Box on Ice.Example: Pushing a Box on Ice.
A skater is pushing a heavy box (mass m = 100 kg) across
a sheet of ice (horizontal & frictionless). He applies a force
of 50 N in the ii direction. If the box starts at rest, what is its
speed v after being pushed a distance d = 10m ?
d
FF
v
m a
ii

12. Physics 111: Lecture 4, Pg 12
Example: Pushing a Box on Ice...Example: Pushing a Box on Ice...
Start with F = ma.
a = F / m.
Recall that v2
- v0
2
= 2a(x - x0 ) (Lecture 1)
So v2
= 2Fd / m v
Fd
m
=
2
d
FF
v
m a
ii

13. Physics 111: Lecture 4, Pg 13
Example: Pushing a Box on Ice...Example: Pushing a Box on Ice...
Plug in F = 50 N, d = 10 m, m = 100 kg:
Find v = 3.2 m/s.
d
FF
v
m a
ii
v
Fd
m
=
2

14. Physics 111: Lecture 4, Pg 14
Lecture 4,Lecture 4, Act 1Act 1
Force and accelerationForce and acceleration
A force F acting on a mass m1 results in an acceleration a1.
The same force acting on a different mass m2 results in an
acceleration a2 = 2a1.
If m1 and m2 are glued together and the same force F acts
on this combination, what is the resulting acceleration?
(a)(a) 2/3 a1 (b)(b) 3/2 a1 (c)(c) 3/4 a1
F a1
m1
F a2 = 2a1
m2
F a = ?
m1 m2

15. Physics 111: Lecture 4, Pg 15
Lecture 4,Lecture 4, Act 1Act 1
Force and accelerationForce and acceleration
Since a2 = 2a1 for the same applied force, m2 = (1/2)m1 !
m1+ m2 = 3m1 /2
(a)(a) 2/3 a1 (b)(b) 3/2 a1 (c)(c) 3/4 a1
F a = F / (m1+ m2)
m1 m2
So a = (2/3)F / m1 but F/m1 = a1
a = 2/3 a1

16. Physics 111: Lecture 4, Pg 16
ForcesForces
We will consider two kinds of forces:
Contact force:
» This is the most familiar kind.
I push on the desk.
The ground pushes on the chair...
Action at a distance:
» Gravity
» Electricity

17. Physics 111: Lecture 4, Pg 17
Contact forces:Contact forces:
Objects in contact exert forces.
Convention: FFa,b means
“the force acting on a due to b”.
So FFhead,thumb means “the force on
the head due to the thumb”.
FFhead,thumb

18. Physics 111: Lecture 4, Pg 18
Action at a distanceAction at a distance
Gravity:

19. Physics 111: Lecture 4, Pg 19
GravitationGravitation
(Courtesy of Newton)(Courtesy of Newton)
Newton found that amoon / g = 0.000278
and noticed that RE
2
/ R2
= 0.000273
This inspired him to propose the
Universal Law of Gravitation:Universal Law of Gravitation: |FMm |= GMm / R2
R RE
amoon g
where G = 6.67 x 10 -11
m3
kg-1
s-2

20. Physics 111: Lecture 4, Pg 20
Gravity...Gravity...
The magnitude of the gravitational force FF12 exerted on an
object having mass m1 by another object having mass m2
a distance R12 away is:
The direction of FF12 is attractive, and lies along the line
connecting the centers of the masses.
F G
m m
R
12
1 2
12
2
=
R12
m1 m2
FF12 FF21

21. Physics 111: Lecture 4, Pg 21
Gravity...Gravity...
Near the Earth’s surface:
R12 = RE
» Won’t change much if we stay near the Earth's surface.
» i.e. since RE >> h, RE + h ~ RE.
RE
m
M
h 2
E
E
g
R
mM
GF =FFg

22. Physics 111: Lecture 4, Pg 22
Gravity...Gravity...
Near the Earth’s surface... 





== 2
E
E
2
E
E
g
R
M
Gm
R
mM
GF
So |Fg| = mg = ma
a = g
All objects accelerate with
acceleration g, regardless of
their mass!
2
2
E
E
s/m81.9
R
M
Gg ==Where:

=g
Leaky Cup

23. Physics 111: Lecture 4, Pg 23
Example gravity problem:Example gravity problem:
What is the force of gravity exerted by the earth on a typical
physics student?
Typical student mass m = 55kg
g = 9.8 m/s2
.
Fg = mg = (55 kg)x(9.8 m/s2
)
Fg = 539 N
FFg
The force that gravity exerts on any object is
called its Weight
W = 539 N

24. Physics 111: Lecture 4, Pg 24
Lecture 4,Lecture 4, Act 2Act 2
Force and accelerationForce and acceleration
Suppose you are standing on a bathroom scale in 141 Loomis
and it says that your weight is W. What will the same scale say
your weight is on the surface of the mysterious Planet X ?
You are told that RX ~ 20 REarth and MX ~ 300 MEarth.
(a)(a) 00.75.75 W
(b)(b) 1.5 W
(c)(c) 2.25 W
E
X

25. Physics 111: Lecture 4, Pg 25
Lecture 4,Lecture 4, Act 2Act 2
SolutionSolution
The gravitational force on a person
of mass m by another object (for instance
a planet) having mass M is given by: F G
Mm
R
= 2
W
W
F
F
X
E
X
E
=Ratio of weights = ratio of forces:
=
G
M m
R
G
M m
R
X
X
E
E
2
2
= ⋅






M
M
R
R
X
E
E
X
2
W
W
X
E
= ⋅




 =300
1
20
75
2
.

26. Physics 111: Lecture 4, Pg 26
Newton’s Third Law:Newton’s Third Law:
Forces occur in pairs: FFA ,B = - FFB ,A.
For every “action” there is an equal and opposite “reaction”.
We have already seen this in the case of gravity:
F F12
1 2
12
2 21= =G
m m
R
R12
m1
m2
FF12 FF21
Newton’s
Sailboard

27. Physics 111: Lecture 4, Pg 27
Newton's Third Law...Newton's Third Law...
FFA ,B = - FFB ,A. is true for contact forces as well:
FFm,w FFw,m
FFm,f
FFf,m
2 Skateboards

28. Physics 111: Lecture 4, Pg 28
Example of Bad ThinkingExample of Bad Thinking
Since FFm,b = -FFb,m, why isn’t FFnet = 0 and aa = 0 ?
a ??a ??
FFm,b FFb,m
ice

29. Physics 111: Lecture 4, Pg 29
Example of Good ThinkingExample of Good Thinking
Consider only the boxonly the box as the system!
FFon box = maabox = FFb,m
Free Body Diagram (next time).
aaboxbox
FFm,b FFb,m
ice

30. Physics 111: Lecture 4, Pg 30
Lecture 4,Lecture 4, Act 3Act 3
Newton’s 3rd LawNewton’s 3rd Law
Two blocks are stacked on the ground. How many action-reaction
pairs of forces are present in this system?
(a) 2
(b) 3
(c) 4
a
b

31. Physics 111: Lecture 4, Pg 31
Lecture 4,Lecture 4, Act 3Act 3
Solution:Solution:
(c) 4
FFE,aE,a
FFa,Ea,E
a
b
FFE,bE,b
a
b
FFb,Eb,E
FFb,ab,a
FFa,ba,b
a
b
FFg,bg,b
FFb,gb,g
a
b

32. Physics 111: Lecture 4, Pg 32
Recap of today’s lectureRecap of today’s lecture
Newton’s 3 Laws:
Law 1: An object subject to no external forces is at rest or
moves with a constant velocity if viewed from an inertial
reference frame.
Law 2: For any object, FFNET = Σ FF = maa
Law 3: Forces occur in pairs: FFA ,B = - FFB ,A.
Extinguisher
Cart