Lecture 8

Goals:
 Differentiate between Newton’s 1st, 2nd and 3rd Laws
 Use Newton’s 3rd Law in problem solving
Assignment: HW4, (Chapters 6 & 7, due 2/18, Wednesday)
Finish Chapter 7
1st Exam Wed., Feb. 18th from 7:15-8:45 PM Chapters 1-7
in room 2103 Chamberlin Hall
Physics 207: Lecture 8, Pg 1
Inclined plane with “Normal” and Frictional Forces
“Normal” means perpendicular
1. Static Equilibrium Case
2. Dynamic Equilibrium (see 1)
Normal
Force
3. Dynamic case with non-zero acceleration
SF=0
Friction
f Force
Fx= 0 = mg sin q – f
mg sin q
Fy= 0 = mg cos q – N
with mg sin q = f ≤ mS N
if mg sin q > mS N, must slide
Critical angle ms = tan qc
q
mg cos q
q
q
Block weight is mg
Physics 207: Lecture 8, Pg 2
y
x
Inclined plane with “Normal” and Frictional Forces
1. Static Equilibrium Case
“Normal” means perpendicular
Normal
Force
2. Dynamic Equilibrium
Friction opposite velocity
v
(down the incline)
SF=0
fK
Friction
Force
mg sin q
Fx= 0 = mg sin q – fk
Fy= 0 = mg cos q – N
fk = mk N = mk mg cos q
Fx= 0 = mg sin q – mk mg cos q
mk = tan q (only one angle)
q
mg cos q
q
q
mg
Physics 207: Lecture 8, Pg 3
y
x
Inclined plane with “Normal” and Frictional Forces
3. Dynamic case with non-zero acceleration
Normal
Result depends on direction of velocity
Force
Friction Force
Sliding Down
v
mg sin q
q
Fx= max = mg sin q ± fk
q
fk
Sliding
Up
Weight of block is mg
Fy= 0 = mg cos q – N
fk = mk N = mk mg cos q
Fx= max = mg sin q ± mk mg cos q
ax = g sin q ± mk g cos q
Physics 207: Lecture 8, Pg 4
Friction in a viscous medium
Drag Force Quantified


With a cross sectional area, A (in m2), coefficient of drag of
1.0 (most objects),  sea-level density of air, and velocity, v
(m/s), the drag force is:
D = ½ C  A v2  c A v2
in Newtons
c = ¼ kg/m3
In falling, when D = mg, then at terminal velocity
Example: Bicycling at 10 m/s (22 m.p.h.), with projected area
of 0.5 m2 exerts ~30 Newtons
 Minimizing drag is often important
Physics 207: Lecture 8, Pg 7
“Free” Fall


Terminal velocity reached when Fdrag = Fgrav (= mg)
For 75 kg person with a frontal area of 0.5 m2,
vterm  50 m/s, or 110 mph
which is reached in about 5 seconds, over 125 m of fall
Physics 207: Lecture 8, Pg 8
Newton’s 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, FNET = S F = ma
Law 3:
Forces occur in pairs: FA , B = - FB , A
(For every action there is an equal and opposite reaction.)
Read: Force of B on A
Physics 207: Lecture 8, Pg 10
Newton’s Third Law:
If object 1 exerts a force on object 2 (F2,1 ) then object 2
exerts an equal and opposite force on object 1 (F1,2)
F1,2 = -F2,1
For every “action” there is an equal and opposite “reaction”
IMPORTANT:
Newton’s 3rd law concerns force pairs which
act on two different objects (not on the same object) !
Physics 207: Lecture 8, Pg 11
Gravity
Newton also recognized that gravity is an
attractive, long-range force between any two
objects.
When two objects with masses m1 and m2 are
separated by distance r, each object “pulls” on the
other with a force given by Newton’s law of gravity,
as follows:
Physics 207: Lecture 8, Pg 12
Cavendish’s Experiment
F = m1 g = G m1 m2 / r2
g = G m2 / r2
If we know big G, little
g and r then will can
find m2 the mass of
the Earth!!!
Physics 207: Lecture 8, Pg 13
Example (non-contact)
Consider the forces on an object undergoing projectile motion
FB,E = - mB g
FB,E = - mB g
FE,B = mB g
FE,B = mB g
EARTH
Question: By how much does g change at an altitude of
40 miles? (Radius of the Earth ~4000 mi)
Physics 207: Lecture 8, Pg 14
Example
Consider the following two cases (a falling ball and ball
on table),
Compare and contrast Free Body Diagram
and
Action-Reaction Force Pair sketch
Physics 207: Lecture 8, Pg 15
Example
The Free Body Diagram
mg
FB,T= N
mg
Ball Falls
For Static Situation
N = mg
Physics 207: Lecture 8, Pg 16
Normal Forces
Certain forces act to keep an object in place.
These have what ever force needed to balance all others
(until a breaking point).
FB,T
FT,B
Main goal at this point : Identify force pairs and apply
Newton’s third law
Physics 207: Lecture 8, Pg 17
Example
First: Free-body diagram
Second: Action/reaction pair forces
FB,E = -mg
FB,T= N
FT,B= -N
FE,B = mg
FB,E = -mg
FE,B = mg
Physics 207: Lecture 8, Pg 18
The flying bird in the cage

You have a bird in a cage that is resting on your upward
turned palm. The cage is completely sealed to the
outside (at least while we run the experiment!). The bird
is initially sitting at rest on the perch. It decides it needs
a bit of exercise and starts to fly.
Question: How does the weight of the cage plus bird
vary when the bird is flying up, when the bird is flying
sideways, when the bird is flying down?

So, what is holding the airplane up in the sky?
Physics 207: Lecture 8, Pg 19
Exercise
Newton’s Third Law
A fly is deformed by hitting the windshield of a speeding bus.

v
The force exerted by the bus on the fly is,
A.
B.
C.
greater than
equal to
less than
that exerted by the fly on the bus.
Physics 207: Lecture 8, Pg 20
Exercise 2
Newton’s Third Law
Same scenario but now we examine the accelerations
A fly is deformed by hitting the windshield of a speeding bus.

v
The magnitude of the acceleration, due to this collision, of
the bus is
A. greater than
B. equal to
C. less than
that of the fly.
Physics 207: Lecture 8, Pg 21
Exercise 2
Newton’s Third Law
Solution
By Newton’s third law these two forces form an interaction
pair which are equal (but in opposing directions).

Thus the forces are the same
However, by Newton’s second law Fnet = ma or a = Fnet/m.
So Fb, f = -Ff, b = F0
but |abus | = |F0 / mbus | << | afly | = | F0/mfly |
Answer for acceleration is (C)
Physics 207: Lecture 8, Pg 22
Exercise 3
Newton’s 3rd Law


Two blocks are being pushed by a finger on a horizontal
frictionless floor.
How many action-reaction force pairs are present in this
exercise?
a
A.
B.
C.
D.
b
2
4
6
Something else
Physics 207: Lecture 8, Pg 23
Exercise 3
Solution:
Fa,f
Ff,a
a
Fb,a
FE,a
Fa,b
bF
Fg,a
Fg,b
Fa,g
Fb,g
Fa,E
E,b
Fb,E
6
Physics 207: Lecture 8, Pg 24
Example: Friction and Motion

A box of mass m1 = 1 kg is being pulled by a horizontal
string having tension T = 40 N. It slides with friction
(mk= 0.5) on top of a second box having mass m2 = 2 kg,
which in turn slides on a smooth (frictionless) surface.
 What is the acceleration of the second box ?
Central question: What is force on mass 2?
(A) a = 0 N
(B) a = 5 N
(C) a = 20 N (D) can’t tell
v
T
a=?
m1
m2
slides with friction (mk=0.5 )
slides without friction
Physics 207: Lecture 8, Pg 25
Example
Solution

First draw FBD of the top box:
v
N1
T
fk = mKN1 = mKm1g
m1
m1g
Physics 207: Lecture 8, Pg 26
Example
Solution

Newtons 3rd law says the force box 2 exerts on box 1 is
equal and opposite to the force box 1 exerts on box 2.

As we just saw, this force is due to friction:
Reaction
f2,1 = -f1,2
Action
m1
f1,2 = mKm1g = 5 N
m2
(A) a = 0 N
(B) a = 5 N
(C) a = 20 N (D) can’t tell
Physics 207: Lecture 8, Pg 27
Example
Solution

Now consider the FBD of box 2:
N2
f2,1 = mkm1g
m2
m1g
m2g
Physics 207: Lecture 8, Pg 28
Example
Solution

Finally, solve Fx = ma in the horizontal direction:
mK m1g = m2a
m1m k g 5 N
a

m2
2 kg
= 2.5 m/s2
f2,1 = mKm1g
m2
Physics 207: Lecture 8, Pg 29
Home Exercise
Friction and Motion, Replay

A box of mass m1 = 1 kg, initially at rest, is now pulled by a
horizontal string having tension T = 10 N. This box (1) is on
top of a second box of mass m2 = 2 kg. The static and kinetic
coefficients of friction between the 2 boxes are ms=1.5 and
mk= 0.5. The second box can slide freely (frictionless) on an
smooth surface.
Compare the acceleration of box 1 to the acceleration of box 2 ?
a1
T
m1
a2
m2
friction coefficients
ms=1.5 and mk=0.5
slides without friction
Physics 207: Lecture 8, Pg 30

Home Exercise
Friction and Motion, Replay in the static case
A box of mass m1 = 1 kg, initially at rest, is now pulled by a
horizontal string having tension T = 10 N. This box (1) is on
top of a second box of mass m2 = 2 kg. The static and kinetic
coefficients of friction between the 2 boxes are ms=1.5 and
mk= 0.5. The second box can slide freely on an smooth
surface (frictionless).
If no slippage then the maximum frictional force between 1 & 2 is
(A) 20 N
(B) 15 N (C) 5 N
a1
T
m1
a2
m2
(D) depends on T
friction coefficients
ms=1.5 and mk=0.5
slides without friction
Physics 207: Lecture 8, Pg 31
Home Exercise
Friction and Motion
fS  mS N = mS m1 g = 1.5 x 1 kg x 10 m/s2
which is 15 N (so m2 can’t break free)
N
fS
T
m1 g
fs = 10 N and the acceleration of box 1 is
Acceleration of box 2 equals that of box 1, with |a| = |T| / (m1+m2)
and the frictional force f is m2a
(Notice that if T were in excess of 15 N then it would break free)
a1
T
m1
a2
m2
friction coefficients
ms=1.5 and mk=0.5
slides without friction
Physics 207: Lecture 8, Pg 32
Lecture 8 Recap
Assignment: HW4, (Chapters 6 & 7, due 2/18, Wednesday)
Finish Chapter 7
1st Exam Wed., Feb. 18th from 7:15-8:45 PM Chapters 1-7
in room 2103 Chamberlin Hall
Physics 207: Lecture 8, Pg 33