```Chapter 5
• Applying
Newton’s Laws
Equilibrium
An object is in equilibrium when
the net force acting on it is zero.
In component form, this is
The net force on each
man in the tower is
zero.
Slide 5-13
Equilibrium
A hanging street sign with more than one force acting on it
1
2
3
1 + 2 + 3 =0
1 + 2 + 3 =0
Equilibrium
What are the components of the forces?
1
1
1
3
2
2
0
1 + 2 + 3 = 0
3
0
1 + 2 + 3 =0
Slide 5-14
Example Problem
A 100 kg block with a weight of 980 N hangs on a rope. Find
the tension in the rope if
A. the block is stationary.
B. it’s moving upward at a steady speed of 5 m/s.
C. it’s accelerating upward at 5 m/s2.
Slide 5-15
Example Problem
A 100 kg block with a weight of 980 N hangs on a rope. Find
the tension in the rope if
A.
the block is stationary.
= 0
=  +  = 0

100kg
+  = 0
m
= 100kg ∙ 9.8 2 = 980N
s

Slide 5-15
A 100 kg block with a weight of 980 N hangs on a rope. Find the tension in the
rope if
A.
B.
C.
the block is stationary.
it’s moving upward at a steady speed of 5 m/s.
it’s accelerating upward at 5 m/s2.
= 0
=  +  = 0

100kg
+  = 0
m
= 100kg ∙ 9.8 2 = 980N
s

Slide 5-15
A 100 kg block with a weight of 980 N hangs on a rope. Find the tension in the
rope if
A.
B.
C.
the block is stationary.
it’s moving upward at a steady speed of 5 m/s.
it’s accelerating upward at 5 m/s2.
=

+  =
100kg
=  −
=   −
m
m
= 100kg 5 2 − −9.8 2
s
s
= 1480N

Slide 5-15
Example Problem
A wooden box, with a mass of 22 kg, is pulled at a constant
speed with a rope that makes an angle of 25° with the wooden
floor. If the coefficient of kinetic friction is  = .5, what is the
tension in the rope?
Slide 5-16
A wooden box, with a mass of 22 kg, is pulled at a constant
speed with a rope that makes an angle of 25° with the wooden
floor. If the coefficient of kinetic friction is  = .5, what is the
tension in the rope?
=  +  +  = 0
=  +  = 0
= − −
= −

25°

Slide 5-16
A wooden box, with a mass of 22 kg, is pulled at a constant speed with a rope that makes
an angle of 25° with the wooden floor. If the coefficient of kinetic friction is  = .5, what is
the tension in the rope?
= −
= − −
cos  = −
sin  = − −
cos
sin  = − +

cos
= −

25°

Slide 5-16
A wooden box, with a mass of 22 kg, is pulled at a constant speed with a rope that makes
an angle of 25° with the wooden floor. If the coefficient of kinetic friction is  = .5, what is
the tension in the rope?
cos
sin  = − +

cos
sin  −
= −

cos
sin  −
= −

−
=
cos
sin  −

Example Problem
A ball weighing 50 N is pulled back by a rope to an angle of 20°.
What is the tension in the pulling rope?
Slide 5-19
Example Problem
A ball weighing 50 N is pulled back by a rope to an angle of 20°.
What is the tension in the pulling rope?

Slide 5-19
A ball weighing 50 N is pulled back by a rope to an angle of 20°.
What is the tension in the pulling rope?
=  +  = 0
= −

°
cos 20 =

=  +  = 0
= − = 50N

Slide 5-19
Using Newton’s Second Law
Slide 5-20
Dynamics with the 2nd law

s

Σ = s +  =

Σ = s +  =
Example Problem
A sled with a mass of 20 kg slides along frictionless ice at 4.5 m/s.
It then crosses a rough patch of snow which exerts a friction force
of -12 N. How far does it slide on the snow before coming to rest?
Slide 5-21
A sled with a mass of 20 kg slides along frictionless ice at 4.5 m/s.
It then crosses a rough patch of snow which exerts a friction force
of -12 N. How far does it slide on the snow before coming to rest?

Slide 5-21
A sled with a mass of 20 kg slides along frictionless ice at 4.5 m/s.
It then crosses a rough patch of snow which exerts a friction force
of -12 N. How far does it slide on the snow before coming to rest?

Slide 5-21
A sled with a mass of 20 kg slides along frictionless ice at 4.5 m/s.
It then crosses a rough patch of snow which exerts a friction force
of -12 N. How far does it slide on the snow before coming to rest?
2

=
2
−
2

=
+ 2∆

=2
∆

−2
∆ =
2
∆

=

Slide 5-21
Mass and Weight
• Mass and weight are not the same
=
m
The moon’s gravity
• The moon has about 1/6 of the
gravity of earth
m =
1
6
m
Weightlessness
• Falling doesn’t mean you have no
weight
• But this is the term we use anyway

Mass and Weight
–w = may = m(–g)
w = mg
Slide 5-23
Which of the following statements
about mass and weight is correct?
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A. Your mass is a measure of the
force gravity exerts on you
B. Your mass is same everywhere
in the universe
C. Your weight is the same
everywhere in the universe
D. Your weight is a measure of
accelerated
30
Apparent Weight
Slide 5-24
Example Problem
A 50 kg student gets in a 1000 kg elevator at rest. As the elevator
begins to move, she has an apparent weight of 600 N for the first 3
s. How far has the elevator moved, and in which direction, at the
end of 3 s?
Slide 5-25
A 50 kg student gets in an elevator at rest. As the elevator begins to
move, she has an apparent weight of 600 N for the first 3 s. How far
has the elevator moved, and in which direction, at the end of 3 s?
=  +  =
= 600N

+
=

=
1 2
1 2
∆ =   +  → ∆ =
2
2
0
1  +
∆ =
2

2

Slide 5-25
Normal Forces
N
m

N =
Normal Forces
+
N

+
−  cos

−  sin
Normal Forces
=
N
+
+
Σ = − sin  =
Σ = − cos  + N =
− cos  + N = 0
N =  cos

−  cos

−  sin
Static Friction
fs max = µsn
Slide 5-30
Kinetic Friction
fk = µkn
Slide 5-31
Coefficients of static friction are typically greater
than coefficients of kinetic friction.
50%
50%
Fa
lse
Tr
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e
A. True
B. False
Include static friction
f =  N
N

+
+
−  cos

−  sin
Kinetic and Rolling friction
f =  N
f =  N

f
f
Working with Friction Forces
Slide 5-32
Drag
An object moving in a gas or liquid experiences a drag force
Drag coefficient. Depends on
details of the object’s shape.
“Streamlining” reduces drag by
making CD smaller. For a typical
object, CD 0.5.
Density of gas or liquid. Air has
a density of 1.29 kg/m3.
A is the object’s cross section area
when facing into the wind.
Drag depends on the square of the speed.
This is a really important factor that limits the
top speed of cars and bicycles. Going twice
as fast requires 4 times as much force and,
as we’ll see later, 8 times as much power.
Slide 5-34
Drag

=
1
2
2
Direction opposite of motion
Drag
=
1
2
4

is the fluid density
A is the cross-sectional area
is the drag coefficient(most objects
have  = 1/2)
Terminal Speed
Sky divers go no faster
than this
average objects
At this speed…
Σ = 0

Example Problem
A car traveling at 20 m/s stops in a distance of 50 m. Assume that
the deceleration is constant. The coefficients of friction between a
passenger and the seat are μs = 0.5 and μk = 0.3. Could a 70 kg
passenger slide off the seat if not wearing a seat belt?
Slide 5-33
Example Problem
A car traveling at 20 m/s stops in a distance of 50 m. Assume that
the deceleration is constant. The coefficients of friction between a
passenger and the seat are μs = 0.5 and μk = 0.3. Could a 70 kg
passenger slide off the seat if not wearing a seat belt?
m
= 20
s
∆ = 50m
Slide 5-33
Example Problem
A car traveling at 20 m/s stops in a distance of 50 m. Assume that
the deceleration is constant. The coefficients of friction between a
passenger and the seat are μs = 0.5 and μk = 0.3. Could a 70 kg
passenger slide off the seat if not wearing a seat belt?
m
= 20
s
∆ = 50m
0
2 = 2 + 2∆
=
2
−
2∆
=   =
≷
Slide 5-33
Cross-Section Area
Slide 5-35
Terminal Speed
A falling object speeds up
until reaching terminal speed,
then falls at that speed
without further change.
If two objects have the same
size and shape, the more
massive object has a larger
terminal speed.
At terminal speed, the net
force is zero and the object
falls at constant speed with
zero acceleration.
Slide 5-36
Summer 2014 Beginning of lecture
Applying Newton’s Third Law: Interacting Objects
Slide 5-37
Example Problem
Block A has a mass of 1 kg; block B’s mass is 4 kg. They are
pushed with a force of magnitude 10 N across a frictionless
surface.
a. What is the acceleration of the blocks?
b. With what force does A push on B? B push on A?
Slide 5-39
Block A has a mass of 1 kg; block B’s mass is 4 kg. They are
pushed with a force of magnitude 10 N across a frictionless
surface.
a. What is the acceleration of the blocks?
b. With what force does A push on B? B push on A?
= 10N = 1kg + 4kg
10N
m
=
=2 2
5kg
s
10N
Slide 5-39
Block A has a mass of 1 kg; block B’s mass is 4 kg. They are
pushed with a force of magnitude 10 N across a frictionless
surface.
a. What is the acceleration of the blocks?
b. With what force does B push on A? A push on B?
Force from block A
on hand
Force from hand on
block A
Slide 5-39
Block A has a mass of 1 kg; block B’s mass is 4 kg. They are
pushed with a force of magnitude 10 N across a frictionless
surface.
a. What is the acceleration of the blocks?
b. With what force does A push on B? B push on A?
= ℎ +  = 1kg
m
2 2
s
10N +  = 2N
Force from hand on
block A
= −8N
Force from block B
on block A
Slide 5-39
Block A has a mass of 1 kg; block B’s mass is 4 kg. They are pushed with
a force of magnitude 10 N across a frictionless surface.
a. What is the acceleration of the blocks?
b. With what force does A push on B? B push on A?
=  =
From Newton’s 3rd law
= −
− =
− −8N =
8N =   = 4kg
m
2 2
s
Force from block B
on block A
Force from block A
on block B
Slide 5-39
Which pair of forces is a “third law pair”?
The string tension and the friction force acting on A
The normal force on A due to B and the weight of A
The normal force on A due to B and the weight of B
The friction force acting on A and the friction force acting on B
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Example Problem
What is the acceleration of block B?
Slide 5-42
Sum the forces on block B?
= 20N +  =
=  =   =
20N +
m
=
= 5.764 2

s
Slide 5-42
Ropes and Pulleys
All ropes are assumed massless with
uniform tension
All pulleys are assumed frictionless

Example Problem
Block A, with mass 4.0 kg, sits on a frictionless table. Block B, with
mass 2.0 kg, hangs from a rope connected through a pulley to
block A. What is the acceleration of block A?
Slide 5-44
Block A, with mass 4.0 kg, sits on a frictionless table. Block B, with
mass 2.0 kg, hangs from a rope connected through a pulley to
block A. What is the acceleration of block A?

Same string, same tension??

=   = 0 ∙

Slide 5-44
Block A, with mass 4.0 kg, sits on a frictionless table. Block B, with mass 2.0 kg,
hangs from a rope connected through a pulley to block A. What is the acceleration
of block A? SUM THE FORCES on each mass separately

=  =
=  +  =
Note that the motions of these two masses are coupled

Slide 5-44
Block A, with mass 4.0 kg, sits on a frictionless table. Block B, with
mass 2.0 kg, hangs from a rope connected through a pulley to
block A. What is the acceleration of block A?
=
+  =
−  +  =
= −
Newton’s Third
But…  =

So just call them both
−  +  =
Now solve for

2.0kg ∙  1
=
=
=
+
6.0kg
3
Slide 5-44
Summary
Slide 5-45
Summary
Slide 5-46
```