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5.7 Some Applications of
Newton’s Law, cont
Multiple Objects


When two or more objects are
connected or in contact, Newton’s
laws may be applied to the system
as a whole and/or to each individual
object
Whichever you use to solve the
problem, the other approach can be
used as a check
Example 5.16 Multiple
Objects

First treat the system as a
whole:
F
x

 msystemax
Apply Newton’s Laws to
the individual blocks
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Solve for unknown(s)
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Check: |P21| = |P12|
Example 5.17 Two Boxes
Connected by a Cord
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Boxes A & B are
connected by a cord
(mass neglected). Boxes
are resting on a
frictionless table.
FP = 40.0 N
Find:

Acceleration (a) of

Tension (FT) in the
each box
cord connecting the
boxes
Example 5.17 Two Boxes
Connected by a Cord, final
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There is only horizontal motion
With: aA = aB = a
Apply Newton’s Laws for box A:
ΣFx = FP – FT = mAa (1)
 Apply Newton’s Laws for box B:
ΣFx = FT = mBa
(2)
Substituting (2) into (1):
FP – mBa = mAa  FP = (mA + mB)a 
a = FP/(mA + mB) = 1.82m/s2
Substituting a into (2) 
FT = mBa = (12.0kg)(1.82m/s2) = 21.8N

Example 5.18 The Atwood’s
Machine
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Forces acting on the objects:
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Tension (same for both objects,
one string)
Gravitational force
Each object has the same
acceleration since they are
connected
Draw the free-body diagrams
Apply Newton’s Laws
Solve for the unknown(s)
Example 5.18 The Atwood’s
Machine, 2
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The Atwood’s Machine:
Find: a and T
Apply Newton’s 2nd Law to each
Mass.
ΣFy = T – m1g = m1 a
(1)
ΣFy = T – m2g = – m2 a
(2)
Then:
T = m1g + m1 a
(3)
T= m g – m a
(4)
Example 5.18 The Atwood’s
Machine, 3

The Atwood’s Machine:
Equating: (3) = (4) and Solving
for a
m1g + m1 a = m2g – m2 a 
m1 a + m2 a = m2g – m1g 
a (m1 + m2) = (m2 – m1)g 

m2  m1 
a
g
m1  m2 
(5)
Example 5.18 The Atwood’s
Machine, final

The Atwood’s Machine:
Substituting (5) into (3) or (4):
T = m1g + m1 a
(3) 
  m  m1  
T  m1 g  m1  2
g 
  m1  m2  
m1m1 g  m1m2 g  m1m2 g  m1m1 g
m1m2 g  m1m1 g

T  m1 g 
 T
m1  m2
m1  m2
 m1m2 
T  2
g
 m1  m2 
Active Figure 5.14
Example 5.19 Two Objects and
Incline Plane
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Find: a and T
One cord: so tension
is the same for both
objects
Connected: so
acceleration is the
same for both objects
Apply Newton’s Laws
Solve for the
unknown(s)
Example 5.19 Two Objects and
Incline Plane, 2
xy plane:
ΣFx = 0 & ΣFy = m1 a 
T – m1g = m1 a 
T = m1g + m1 a (1)
x’y’ plane:
ΣFx = m2 a & ΣFy = 0 
m2gsinθ – T = m2 a (2)
n – m2gcosθ = 0
(3)

Example 5.19 Two Objects and
Incline Plane, Final
Substituting (1) in (2) gives:
m2gsinθ – (m1g + m1 a) = m2 a 
m2gsinθ – m1g – m1 a = m2 a 
a (m1 + m2) = m2gsinθ – m1g 
 Substituting a in (1) we’ll get:
T = m1g + m1 a (1)

  m sin   m1  
T  m1 g  m1  2
g 
  m1  m2 

m2 sin   m1 

a
g
 m1  m2 
m1m1 g  m2 m1 g  m1m2 g sin   m1m1 g
T

m1  m2
m2 m1 g  m1m2 g sin  m1m2 g  sin   1
T


m1  m2
m1  m2
Problem-Solving Hints
Newton’s Laws
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Conceptualize the problem – draw a
diagram
Categorize the problem
Equilibrium (SF = 0) or
 Newton’s Second Law (SF = m a)
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Analyze
Draw free-body diagrams for each object
 Include only forces acting on the object
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Problem-Solving Hints
Newton’s Laws, cont
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Analyze, cont.
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Establish coordinate system
Be sure units are consistent
Apply the appropriate equation(s) in component form
Solve for the unknowns.
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This always requires Kinder Garden Algebra
(KGA). Like solving two linear equations with two
unknowns
Finalize
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Check your results for consistency with your free- body
diagram
Check extreme values
5.8 Forces of Friction
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When an object is in motion on a
surface or through a viscous medium,
there will be a resistance to the
motion
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This is due to the interactions between
the object and its environment
This resistance is called the Force of
Friction
Forces of Friction, 2
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Friction exists between any 2 sliding surfaces.
Two types of friction:
 Static (no motion) friction
 Kinetic (motion) friction
The size of the friction force depends on:
 The microscopic details of 2 sliding surfaces.
 The materials they are made of
 Are the surfaces smooth or rough?
 Are they wet or dry?
Forces of Friction, 3
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Friction is proportional to the normal force
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ƒs  µs n (5.8) and
ƒk = µk n (5.9)
These equations relate the magnitudes of the
forces, THEY ARE NOT vector equations
The force of static friction (maximum) is
generally greater than the force of kinetic
friction
ƒs > ƒk
The coefficients of friction (µk,s) depends
on the surfaces in contact
Forces of Friction, final
The direction of the frictional
force is opposite the direction of
motion and parallel to the
surfaces in contact
 The coefficients of friction (µk,s)
are nearly independent of the
area of contact
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Static Friction
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Static friction acts to
keep the object from
moving: ƒs = F
If F increases, so does ƒs
If F decreases, so does ƒs
ƒs  µs n where the
equality holds when the
surfaces are on the
verge of slipping

Called impending motion
Static Friction, cont
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Experiments determine the relation
used to compute friction forces.
The friction force ƒs exists ║ to the
surfaces, even if there is no motion.
Consider the applied force F 
∑F = ma = 0 & also v = 0
There must be a friction force
ƒs to oppose F 
F – ƒs = 0  ƒs = F
Kinetic Friction
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The force of kinetic friction (ƒk )
acts when the object is in motion
Friction force ƒk is proportional to
the magnitude of the normal force
n between 2 sliding surfaces.
ƒk  n  ƒk  k n (magnitudes)
 k  Coefficient of kinetic friction
 k : depends on the surfaces &
their conditions
 k : is dimensionless & < 1
Static & Kinetic Friction
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Experiments find that the
Maximum Static
Friction Force ƒs,max is
proportional to the
magnitude (size) of the
normal force n between
the 2 surfaces.
DIRECTIONS:
ƒs,max  n
Then: ƒs,max = sn
(magnitudes)
Static & Kinetic Friction
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s  Coefficient of static friction
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s : depends on the surfaces & their conditions
s : is dimensionless & < 1
Always:
 ƒs,max > ƒk  s n > k n (Cancel n)  s > k
 ƒs  ƒs,max = µsn  ƒs  µs n
Some Coefficients of
Friction
Active Figure 5.16
Friction in Newton’s
Laws Problems
Friction is a force, so it simply is
included in the Net Force (SF ) in
Newton’s Laws
 The rules of friction allow you to
determine the direction and
magnitude of the force of friction

Example 5.20 Pulling Against
Friction
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Assume: mg = 98.0N  n = 98.0 N,
s = 0.40, k = 0.30 
ƒs,max = sn = 0.40(98N) = 39N
Find Force of Friction if the force
applied FA is:
A. FA = 0  ƒs = FA = 0 
ƒs = 0
Box does not move!!
n
ƒs,k
B. FA = 10N  FA < ƒs,max or (10N < 39N)
ƒs – FA = 0  ƒs = FA = 10N
The box still does not move!!
Example 5.20 Pulling
Against Friction, 2
C. FA = 38N < ƒs,max  ƒs – FA = 0 
ƒs = FA = 38N
This force is still not quite large
enough to move the box!!!
n
ƒs,k
D. FA = 40N > ƒs,max  kinetic friction.
This one will start moving the box!!!
ƒk  k n = 0.30(98N) = 29N.
The net force on the box is:
∑F = max  40N – 29N = max  11N = max
 ax = 11 kg.m/s2/10kg = 1.10 m/s2
ƒs,k
Example 5.20 Pulling Against
Friction, final
ƒs,max = 39N
ƒk= 29N
Example 5.21 To Push or Pull a
Sled
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Similar to Quiz 5.14
Will you exerts less force
if you push or pull the
girl?
θ is the same in both
cases
Newton’s 2nd Law:
∑F = ma
Pushing
Pulling
Example 5.21 To Push or Pull a
Sled, 2
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x direction: ∑Fx = max 
Fx – ƒs,max = max
n
Pushing
y direction: ∑Fy = 0 
n – mg – F y = 0 
n = m g + Fy 
Fx
ƒs,max
Fy
ƒs,max = μsn 
ƒs,max = μs (mg + Fy )
Pushing
Example 5.21 To Push or Pull a
Sled, final

Pulling
y direction: ∑Fy = 0 
n + Fy – mg = 0 
n = mg – Fy 
ƒs,max = μsn 

ƒs,max = μs (mg – Fy )
n
Fy
ƒs,max
Fx
NOTE:
ƒs,max (Pushing) > ƒs,max (Pulling)
Friction Force would be less
if you pull than push!!!
Pulling
Conceptual Example 5.22 Why Does the
Sled Move? (Example 5.11 Text Book)
 To determine if the horse (sled) moves: consider only the
horizontal forces exerted ON the horse (sled) , then
apply 2nd Newton’s Law: ΣF = m a.
 Horse:
T : tension exerted by the sled.
fhorse : reaction exerted by the Earth.
 Sled:
T : tension exerted by the horse.
fsled : friction between sled and snow.
Conceptual Example 5.22 Why Does the
Sled Move? final
 Horse: If fhorse > T , the horse accelerates to the right.
 Sled:
If T > fsled , the sled accelerates to the right.
 The forces that accelerates the system (horse-sled) is
the net force fhorse  fsled
 If fhorse = fsled the system will move with constant
velocity.
Example 5.23 Sliding Hockey
Puck
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Example 5.13 (Text Book)
Draw the free-body
diagram, including the force
of kinetic friction
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Opposes the motion
Is parallel to the surfaces in
contact
Continue with the solution as
with any Newton’s Law
problem
Example 5.23 Sliding Hockey
Puck, 2
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Given: vxi = 20.0 m/s vxf = 0,
xi = 0, xf = 115 m
Find μk?
y direction: (ay = 0)
∑Fy = 0 
n – mg = 0  n = mg (1)
x direction: ∑Fx = max 
– μkn = max (2)
Substituting (1) in (2) :
– μk(mg) = max 
ax = – μk g
Example 5.23 Sliding Hockey
Puck, final
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ax = – μk g
To the left (slowing down) &
independent of the mass!!
Replacing ax in the Equation:
vf2 = vi2 + 2ax(xf – xi) 
0 = (20.0m/s)2 + 2(– μk g)(115m) 
μk 2(9.80m/s2)(115m) = 400(m2/s2) 
μk = 400(m2/s2) / (2254m2/s2) 
μk = 0.177
Example 5.24 Two Objects
Connected with Friction
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Example 5.13 (Text Book)
Known: ƒk  kn
Find: a
Mass 1: (Block)
y direction: ∑Fy = 0, ay = 0 
n + Fsinθ – m1g = 0 
n = m1g – Fsinθ
(1)
x direction: ∑Fx = m1a 
Fcosθ – T – ƒk = m1a 
Fcosθ – T – kn = m1a 
T = Fcosθ – kn – m1a (2)
Example 5.24 Two Objects
Connected with Friction, 2
Mass 2: (Ball)
 y direction: ∑Fy = m2a 
T – m2g = m2a 
T = m2g + m2a (3)
 x direction: ∑Fx = 0, ax = 0
n = m1g – Fsinθ
(1)
T = Fcosθ – kn – m1a (2)
Substitute (1) into (2):
T = Fcosθ – k(m1g – Fsinθ ) – m1a (4)
 Equate: (3) = (4) and solve for a:

Example 5.24 Two Objects
Connected with Friction, final
m2 g  m2 a  F cos   k (m1 g  F sin  )  m1a 
m1a  m2 a  F cos   k m1 g   k F sin   m2 g 
m1  m2 a  F cos  k sin    k m1  m2 g 
F cosθ  μ k sinθ   μ k m1  m2 g
a
m1  m2 
Inclined Plane Problems
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

Tilted coordinate system:
Convenient, but not necessary.
K-Trigonometry:
Fgx= Fgsinθ = mgsinθ
Fgy= Fgcosθ = – mgcosθ
Understand:
 ∑F = m a ,
ƒk  k n
ax ≠ 0 ay = 0
 y direction: ∑Fy = 0 
n – mgcosθ = 0 
n = mgcosθ
(1)
 x direction: ∑Fx = max 
mgsinθ – ƒ = max (2)
ax
Is the normal force n equal
& opposite to the weight Fg ?
NO!!!!
Experimental Determination
of µs and µk


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The block is sliding down the
plane, so friction acts up the
plane
This setup can be used to
experimentally determine the
coefficient of friction µs,k
µs,k = tan s,k


For µs use the angle where the
block just slips
For µk use the angle where the
block slides down at a constant
speed
Active Figure 5.19
Example 5.25 The Skier




Assuming: FG = mg , ay = 0
ƒk  kn, k = 0.10
Find: ax
Components:
FGx= FGsin30o = mgsin30o
FGy= FGcos30o = – mgcos30o
Newton’s 2nd Law
 y direction: ∑Fy = 0 
n – mgcos30o = 0 
n = mgcos30o (1)
 x direction: ∑Fx = max 
mgsin30o – ƒk = max (2)
ax
Example 5.25 The Skier
n = mgcos30o
(1)
mgsin30o – ƒk = max (2)
 Replacing ƒk  kn in (2)
mgsin30o – kn = max (3)
 Substituting (1) into (3)
mgsin30o – kmgcos30o = max
ax = gsin30o – μkgcos 30o
ax = g(0.5) – 0.10g(0.87) 
ax = 0.41g 
ax = 4.00m/s2
ax
Material for the Midterm

Examples to Read!!!
Example 5.2
 Example 5.3
 Example 5.12


(Page 120)
(Page 122)
(Page 134)
Homework to be solved in Class!!!
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