Download 4.3 Newton`s Second Law of Motion

AP Physics
Chapter 4
Force and Motion
Chapter 4: Force and Motion
4.1
4.2
4.3
4.4
4.5
The Concepts of Force and Net Force
Inertia and Newton’s First Law of Motion
Newton’s Second Law of Motion
Newton’s Third Law of Motion
Free-body Diagrams and Translational
Equilibrium
4.6 Friction
4.1 The Concepts of Force and Net Force
dynamics – what causes motion and
changes in motion.
• Isaac Newton summarized the “why”
of motion with three laws.
force – something capable of changing an object’s state of
motion (its velocity)
• a vector quantity
• SI unit is the newton
• A newton of force causes a 1 kg mass to accelerate
1 m/s2.
• 1 N = 1kg · m
s2
net force – the vector sum, or resultant, of all
the forces acting on an object.
• can be written as ∑ F
• balanced forces create zero net force
• unbalanced forces create a net force
• a net force produces an acceleration
4.1 The Concepts of Force and Net Force
This figure illustrates what happens in the
presence of zero and nonzero net force.
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4.1 The Concepts of Force and Net Force
4.1 The Concepts of Force and Net Force
4.2 Inertia and Newton’s First Law of
Motion
According to Aristotle, the natural state of objects
was to be at rest, and if you got them moving,
eventually they would come to rest again.
Galileo did experiments rolling balls down and up
inclined planes, and realized that, in the absence
of some kind of force, an object would keep
moving forever once it got started.
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4.2 Inertia and Newton’s First Law of
Motion
Galileo called this inertia:
Inertia is the natural tendency of an object to maintain
a state of rest or to remain in uniform motion in a
straight line (constant velocity).
Later, Newton realized that mass is a measure of
inertia.
Inertia is Italian for lazy. Inertia is the lazy force.
© 2010 Pearson Education, Inc.
4.2 Inertia and Newton’s First Law of
Motion
Newton’s first law is called the law of inertia:
In the absence of an unbalanced applied force (Fnet
= 0), a body at rest remains at rest, and a body
already in motion remains in motion with a constant
velocity (constant speed and direction).
The bottom line: There is NO ACCELERATION in this case AND the object
must be at EQILIBRIUM ( All the forces cancel out).
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acc  0   F  0
4.2 Inertia and Newton’s First Law of Motion
4.2 Inertia and Newton’s First Law of Motion
4.2 Inertia and Newton’s First Law of Motion
inertia – the natural tendency of an object to maintain a state of rest or to remain in
uniform motion in a straight line (constant velocity).
inertial frame of reference – a non-accelerating frame of reference.
- a frame of reference where Newton’s 1st Law holds.
ex: juggling on a bus travelling at constant velocity.
non-inertial frame of reference – an accelerating frame of reference where fictitious
forces arise.
ex: having to hang on to a merry-go-round, or be thrown off!
mass – a measure of inertia
** If the net force acting on an object is zero, then its acceleration is zero.***
4.2 Inertia and Newton’s First Law of Motion
4.2 Inertia and Newton’s First Law of Motion
4.3 Newton’s Second Law of Motion
4.3 Newton’s Second Law of Motion
4.3 Newton’s Second Law of Motion
Experiments show that the acceleration of an
object is proportional to the force exerted on it and
inversely proportional to its mass.
Double the force -> double the acceleration.
Double the mass -> half the acceleration.
And so on…
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4.3 Newton’s Second Law of Motion
The units of force are called Newtons after Isaac
Newton.
1 N = 1 kg . m/s2.
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4.3 Newton’s Second Law of Motion
An object’s weight is the force exerted on it by
gravity.
Here, g is the acceleration of gravity:
g = 9.81 m/s2
Weight therefore has the same units as force
in the SI system—Newtons.
© 2010 Pearson Education, Inc.
4.3 Newton’s Second Law of Motion
An object’s weight is the force exerted on it by
gravity.
Here, g is the acceleration of gravity:
g = 9.81 m/s2
Weight therefore has the same units as force
in the SI system—Newtons.
© 2010 Pearson Education, Inc.
4.3 Newton’s Second Law of Motion
Newton’s second law may be applied to a
system as a whole, or to any part of a system. It
is important to be clear about what system or
part you are considering!
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4.3 Newton’s Second Law of Motion
Newton’s second law applies separately to each
component (x’s separate from the y’s) of the force.
Must find the Cartesian components.
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4.3 Newton’s Second Law of Motion
4.3 Newton’s Second Law of Motion
4.3 Newton’s Second Law of Motion
4.3 Newton’s Second Law of Motion
4.3 Newton’s Second Law of Motion
4.3 Newton’s Second Law of Motion
4.3 Newton’s Second Law of Motion
4.3 Newton’s Second Law of Motion
Example #1
Onyxx, the ballet dancer, has mass of 45.0kg
a) What is Onyxx’s weight on Earth?
b) What is Onyxx’s mass on Jupiter?
c) What is Onyxx’s weight on Jupiter?
Note: The acceleration due to gravity on Jupiter
is 25 m/s2
Example # 1
Onyxx, the ballet
dancer, has mass of
45.0kg
a) What is Onyxx’s
weight on Earth?
Given:
m = 45kg
g = 10 m/s2
Unknown:
w
w = mg = (45kg)(10m/s2)
w = 450 N
Example #1
b) What is Onyxx’s mass
on Jupiter?
c) What is Onyxx’s
weight on Jupiter?
b) The mass stays the
same.
c) Given:
m = 45kg
g = 25 m/s2
w= mg = (45kg)9 25
m/s2)
Example # 2
• Carleon, the 72kg star quarterback of Enloe HS
football team, collides with , Solomon, a
stationary left tackle, and is brought to a stop
with an acceleration of 20 m/s2
• a) What force does Solomon exert on Carleon?
• b) What force does Carleon exert on
Solomon?
Example #2
• Carleon, the 72kg star
quarterback of Enloe HS
football team, collides
with , Solomon, a
stationary left tackle,and
is brought to a stop with
an acceleration of 20 m/s2
• a) What force does
Solomon exert on
Carleon?
• b) What force does
Carleon exert on
Solomon?
Given:
m = 72kg
a = 20 m/s2
Unknown
F = m a = (72kg)(-20 m/s2)
= -1440 N
b) For every action there is
an equal and opposite
reaction ;therefore ,
Carleon exerts a force of
1440N on Solomon.
Example #3
A 20 g sparrow flying toward a bird
feeder mistakes the pane of glass in a
window for an opening and slams
into it with a force of 2 N. What is
the bird’s acceleration?
Example # 3
• A 20 g sparrow flying
toward a bird feeder
mistakes the pane of
glass in a window for an
opening and slams into
it with a force of 2 N.
What is the bird’s
acceleration?
Given:
m = 20 g = .02kg
F=2N
Unknown
a
F = m a or a = F / m
a = 2N / .02kg
= 100 m/s2
Example #4
A 30 g arrow is shot by William Tell through
an 8 cm thick apple sitting on top of his
son’s head. If the arrow enters the apple
at 30 m/s and emerges at 25 m/s in the
same direction, with what force has the
apple resisted the arrow?
Example # 4
A 30 g arrow is shot by William
Tell through an 8 cm thick
apple sitting on top of his
son’s head. If the arrow
enters the apple at 30 m/s
and emerges at 25 m/s in
the same direction, with
what force has the apple
resisted the arrow?
Given:
m= 30 g =.03 kg
d = 8 cm =.08 m
vo = 30 m/s
vf = 25 m/s
Example #5
Sandeep is sledding down an ice-covered hill
inclined at an angle of 15° with the horizontal.
If Sandeep and the sled have a combined mass
of 54kg, what is the force pulling them down
the hill?
Example # 5
Sandeep is sledding down
an ice-covered hill
inclined at an angle of
15° with the horizontal.
If Sandeep and the sled
have a combined mass
of 54kg, what is the
force pulling them
down the hill?
4.3 Newton’s Second Law of Motion
4.3 Newton’s Second Law of Motion
4.3 Newton’s Second Law of Motion
4.3 Newton’s Second Law of Motion
4.4 Newton’s Third Law of Motion
For every force (action), there is an equal and opposite
force (reaction).
Note that the action and reaction forces act on
different objects.
This image shows how a
block exerts a
downward force on a
table; the table exerts an
equal and opposite force
on the block, called the
normal force N.
© 2010 Pearson Education, Inc.
4.4 Newton’s Third Law of Motion
Action: HAMMER HITS NAIL
Reaction: NAIL HITS HAMMER
Action: Earth pulls on YOU
Reaction: YOU pull on the earth
All you do is SWITCH the wording!
© 2010 Pearson Education, Inc.
4.4 Newton’s Third Law of Motion
This figure shows the force during a
collision between a truck and a train. You
can clearly see the forces are EQUAL
and OPPOSITE. To help you understand
the law better, look at this situation from
the point of view of Newton’s Second
Law.
FTruck  FTrain
mTruck ATruck  M TrainaTrain
There is a balance between the mass and acceleration. One object usually
has a LARGE MASS and a SMALL ACCELERATION, while the other has a
SMALL MASS (comparatively) and a LARGE ACCELERATION.
© 2010 Pearson Education, Inc.
4.4 Newton’s Third Law of Motion
4.4 Newton’s Third Law of Motion
4.4 Newton’s Third Law of Motion
4.4 Newton’s Third Law of Motion
4.4 Newton’s Third Law of Motion
4.4 Newton’s Third Law of Motion
Check for Understanding
Fill in the blanks using the word bank.
velocity
acceleration
mass
non-inertial
dynamics
kinematics
weight
inertial
“the same”
different
1. _________________is used to analyze motion, but ________________
explains what causes motion and changes in motion.
2. A(n)_____________ frame of reference is non-accelerating.
3. In a(n) ________________ frame of reference, fictitious forces arise.
4. The gravitational force on an object near Earth’s surface is called ________.
5. With no forces acting upon it, an object moves with constant ____________.
6. In an action/reaction pair, both forces must act on __________ object(s).
Check for Understanding
Fill in the blanks using the word bank.
velocity
acceleration
mass
non-inertial
dynamics
kinematics
weight
inertial
“the same”
different
1. Kinematics is used to analyze motion, but dynamics explains what causes
motion and changes in motion.
2. A(n) inertial frame of reference is non-accelerating.
3. In a(n) non-inertial frame of reference, fictitious forces arise.
4. The gravitational force on an object near Earth’s surface is called weight.
5. With no forces acting upon it, an object moves with constant velocity.
6. In an action/reaction pair, both forces must act on different object(s).
Lesson 4.5: Free Body
Diagrams and
Translational
Equilibrium
Fground on car
Fforward
Fresistance
Fgravity
4.5 Free-body Diagrams and Translational Equilibrium
Problems we will solve will generally have constant forces and therefore have
constant accelerations. This allows us to use the kinematics equations and
Newton’s Laws to analyze motion.
Free-body Diagrams
1. Sketch the problem. Draw all force vectors.
2. Pick the body to be analyzed using a free-body diagram. Draw a dot at
the center of this body. Draw the origin of your x-y axes at this point.
Draw one of the axes along the direction of the body’s acceleration.
3. Draw and label all force vectors acting on the body with their tails on the
dot. If the body is accelerating, draw an acceleration vector.
4. Resolve any forces not directed along an x or y axis into their
components.
5. Use Newton’s second law to write equations for the x and y directions.
Solve!
4.5 Free-body Diagrams and Translational Equilibrium
Solution:
Isolate the body which we wish to analyze and draw a free-body diagram.
T = tension holding up scale
(this is what the scale will read)
a = 20 m/s2
F.B.D. of spring scale
weight = mg
Fnet = T – mg
and,
So,
Fnet = ma
ma = T – mg
Solve for T:
T = ma + mg
= m (a + g)
= 100 kg (20 m/s2 + 10 m/s2)
T = 3000 N, which is what the scale will read. The answer is d.
4.5 More on Newton’s Laws: Free-Body
Diagrams and Translational Equilibrium
A free-body diagram
(FBD) draws the
forces on an object
as though they all act
at a given point. You
should always draw
such a diagram
whenever you are
solving Force
problems.
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Example 4.5:
4.5 Free-body Diagrams and Translational Equilibrium
Pulley Problems
• a string is considered massless for our purposes
• the tension is the same everywhere in a string
• a single, fixed pulley simply changes the direction of a force
Atwood’s Machine
• The acceleration of an object depends on the net applied and the object’s
mass.
• In an Atwood’s Machine, the difference in weight between two hanging
masses determines the net force acting on both masses.
• This net force accelerates both hanging masses; the heavier mass is
accelerated downward, and the lighter mass is accelerated upward.
Example 60:
An Atwood machine has suspended masses of 0.25 kg and 0.20 kg. Under ideal
conditions, what will be the acceleration of the smaller mass?
4.5 Free-body Diagrams and Translational Equilibrium
Inclined Plane Problems
Suppose we have a mass being accelerated by a rope up a frictionless
inclined plane.
1. Make a sketch. Identify the forces acting on the
mass. Here we have identified tension, weight,
and the normal force.
2. Free-body diagram the mass. Draw a dot to
represent the center of mass. Establish the xaxis along the plane, where positive is the
direction of acceleration. The y-axis is
perpendicular to the x-axis.
3. Draw force vectors in the proper directions
pointing away from the dot. Draw an acceleration
vector in the direction of the net force.
4. Resolve any forces that are not
directed along the x or y axes into x
or y components. Typically you will
need to do this for the weight vector.
Use the free-body diagram to analyze
the forces in terms of Newton’s
second law of motion.
Example 4.6:
Example 65:
In the ideal setup shown, m1 = 3.0 kg, m2 = 2.5 kg.
a) What is the acceleration of the masses?
b) What is the tension in the string?
4.5 Free-body Diagrams and Translational Equilibrium
translational equilibrium – the object is at rest or constant velocity
• the sum of the forces, Fnet = 0
•a=0
static translational equilibrium – the object is at rest; v = 0 m/s.
Hints for Static Equilibrium Problems
• Remember, tension in a string is the same everywhere in the string.
• Resolve force vectors into their components.
• Sum x and y components separately.
• Fnet = 0 for the x and y components
4.5 Free-body Diagrams and Translational Equilibrium
4.5 Free-body Diagrams and Translational Equilibrium
Check for Understanding: Practice Problem
A 2-kg block and a 5-kg block are shown. The surface is frictionless. Find the
tension in the rope connecting the two blocks.
5 kg
2 kg
Solution: Draw free-body diagrams of both masses.
FN
M = 5 kg
Mg
Fnet x = T = Ma
x+
a
T
T
m = 2 kg
a
mg
Fnet y = - T + mg = ma
We have two equations and two unknowns, so we can solve it…
y+
Check for Understanding: Practice Problem
Equation 1:
Equation 2:
T = Ma
- T + mg = ma
Substitute Equation 1 into Equation 2:
- Ma + mg = ma
Solve for a: a
mg = ma + Ma
mg = a (m + M)
a = mg
= (2 kg)(9.8 m/s2) = 2.8 m/s2
(m + M)
7 kg
So, T = Ma = (5 kg)(2.8 m/s2) = 14 N
Check for Understanding: Practice Problem 1
Check for Understanding: Practice Problem 1
Lesson 4.6: Friction
Warmup: Newton’s Influence
Isaac Newton (1642 – 1727) had a tremendous influence on
our understanding of the rules by which nature behaves. He
was the first to mathematically describe the universal nature
of gravity and the effect of forces on motion, and to realize
that white light is composed of separate colors. Along the
way, he developed calculus to describe his findings
mathematically.
In 1730, the poet Alexander Pope wrote the following epitaph for Isaac
Newton:
Nature and Nature’s laws lay hid in night;
God said, Let Newton be! and all was light.
Write a paragraph explaining why you think Pope
gave Newton this tribute.
Physics Daily Warmup, # 40
4.6 Friction
The force of friction always
opposes the direction of motion
(or of the direction the motion
would be in the absence of
friction).
Depending on the
circumstances, friction may be
desirable or undesirable.
Image life without friction…
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4.6 Friction
Types of friction:
Static friction: when the frictional force is large
enough to prevent motion. Static – not moving.
Kinetic friction: when two surfaces are sliding
along each other. Kinetic – motion.
Rolling friction: when an object is rolling without
slipping. A ball rolling down a hill. Without rolling
friction the ball would only translate down the hill
– slide.
© 2010 Pearson Education, Inc.
4.6 Friction
friction – resistance to motion that occurs whenever two materials, or media, are in
contact with each other.
• air resistance is a form of friction
• Sometimes we want to increase friction. (ex: cinder an icy road)
• Other times we want to reduce friction. (ex: change your oil)
• For ordinary solids, friction is caused mostly by local adhesion between the
high spots, or asperities, of contacting surfaces (particularly metals).
• When the contacting surfaces move against one another, the asperities of
the harder material “plow” through the softer material.
The types of friction are:
• static friction – the force of friction that prevents sliding. Static friction is
often responsible for movement. example: walking on ice vs. the floor. Static friction
acts in the direction of movement.
• kinetic (sliding) friction – there is relative motion at the surfaces in contact.
Kinetic friction acts opposite the direction of movement.
• rolling friction – one surface rotates as it moves over another surface but
does not slip or slide at the point or area of contact
4.5 Friction
4.5 Friction
4.6 Friction
This figure illustrates what happens as the applied
force increases: first, the static frictional force
increases; then the kinetic frictional force takes over
as the object begins to move.
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4.6 Friction
The coefficients of friction depend on both materials
involved.
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4.6 Friction
This form for the frictional force is an approximation;
the actual phenomenon is very complicated. The
coefficient of friction may vary somewhat with
speed; there may be some dependence on the
surface area of the objects.
Also, remember that these equations are for the
magnitude of the frictional force—it is always
perpendicular to the normal force. (Why?)
Note: Friction ONLY depends on the MATERIALS
sliding against each other, NOT on surface area.
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4.6 Friction
Air resistance is another form of friction. It depends
on an object’s shape and size, as well as its speed.
VERY complex relationship = calculus.
For an object in free fall, as the force of air
resistance increases with speed, it eventually
equals the downward force of gravity. At that point,
there is no net force on the object and it falls with a
constant velocity called the terminal velocity.
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4.6 Friction
This figure shows the velocity as a function of time for a
falling object with air resistance.
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Free-Body Diagrams
A pictorial representation of forces complete with labels.
FN
Ff
T
T
W1,Fg1
or m1g
m2g
© 2010 Pearson Education, Inc.
•Weight(mg) – Always
drawn from the center,
straight down
•Force Normal(FN) – A
surface force always drawn
perpendicular to a surface.
•Tension(T or FT) – force in
ropes and always drawn
AWAY from object.
•Friction(Ff)- Always drawn
opposing the motion.
FBD - Inclines



Ff

FN
mg cos 

mg
mg sin 


Tips for inclines:
•Rotate Axis
•Break weight into components
Free-Body Diagrams
TIPS for solving problems
•Draw a FBD – ALWAYS!
•Resolve anything into COMPONENTS
•Write equations of equilibrium (NFL, NSL)
•Solve for unknowns
© 2010 Pearson Education, Inc.
A 10-kg box is being pulled across the table to the
right at a constant speed with a force of 50N.
a)Calculate the Force of Friction
b)Calculate the Force Normal
FN
Ff
mg
Fa
Fa  F f  50 N
mg  Fn  (10)(9.8)  98N
Suppose the same box is now pulled at an angle of 30
degrees above the horizontal (10-kg box, 50 N force).
a) Calculate the Force of Friction
Fax  Fa cos   50 cos 30  43.3N
b) Calculate the Force Normal
F f  Fax  43.3N
Fa
FN
Ff
30
mg
Fax
Fay
FN  mg!
FN  Fay  mg
FN  mg  Fay  (10)(9.8)  50 sin 30
FN  73N
A 10-kg box is being pulled across the table to the
right by a rope with an applied force of 50N.
Calculate the acceleration of the box if a 12 N
frictional force acts upon it.
FNet  ma
FN
Ff
mg
Fa
In which direction, is
this object accelerating?
Fa  F f  ma
The X direction!
50  12  10a
So N.S.L. is worked out
using the forces in the
“x” direction only
a  3.8 m / s 2
A mass, m1 = 3.00 kg, is resting on a frictionless horizontal table
is connected to a cable that passes over a pulley and then is
fastened to a hanging mass, m2 = 11.0 kg as shown below. Find
the acceleration of each mass and the tension in the cable.
FNet  ma
FN
m2 g  T  m2 a
T
T
m1g
T  m1a
m2 g  m1a  m2 a
m2 g  m2 a  m1a
m2 g  a (m2  m1 )
m 2g
a
m2 g
(11)(9.8)

 7.7 m / s 2
m1  m2
14
FNet  ma
m2 g  T  m2 a
T  m1a
FNET
FNet  ma 
m
a
Rise
Slope 
Run
T  (3)(7.7)  23.1 N
Example 5
A 1500 N crate is being pushed
across a level floor at a constant
speed by a force F of 600 N at an
angle of 20° below the horizontal
as shown in the figure.
F f   k FN
Fa
Fay
20
a) What is the coefficient of kinetic
friction between the crate and the
floor?
F f  Fax  Fa cos   600(cos 20)  563.82 N
FN
FN  Fay  mg  Fa sin   1500
FN  600(sin 20)  1500  1705.21N
Fax
563.82   k 1705.21
 k  0.331
Ff
mg
Example 6
Fa
If the 600 N force is instead pulling the block at
an angle of 20° above the horizontal as
shown in the figure, what will be the
acceleration of the crate. Assume that the
coefficient of friction is the same as found in
(a)
FNet  ma
FN
Fax
Ff
Fax  F f  ma
Fa cos   FN  ma
Fa cos    (mg  Fa sin  )  ma
600 cos 20  0.331(1500  600 sin 20)  153.1a
563.8  428.57  153.1a
a  0.883 m / s 2
20
mg
Fay
Example 7
Masses m1 = 4.00 kg and m2 = 9.00 kg are connected by a light string that passes over
a frictionless pulley. As shown in the diagram, m1 is held at rest on the floor and m2 rests
on a fixed incline of angle 40 degrees. The masses are released from rest, and m2 slides
1.00 m down the incline in 4 seconds. Determine (a) The acceleration of each mass (b)
The coefficient of kinetic friction and (c) the tension in the string.
FNET  ma
T
FN
m2 g sin   ( F f  T )  m2 a
Ff
m2gcos40
40
T
m2g
m1
40
m2gsin40
m1g
T  m1 g  m1a  T  m1a  m1 g
Example 7
FNET  ma
x  voxt  1 at 2
2
1  0  1 a ( 4) 2
2
a  0.125 m / s 2
T  m1 g  m1a  T  m1a  m1 g
m2 g sin   ( F f  T )  m2 a
T  4(.125)  4(9.8)  39.7 N
Just Algebra Sub :
m2 g sin   F f  T  m2 a
m2 g sin   F f  (m1a  m1 g )  m2 a
m2 g sin    k FN  m1a  m1 g  m2 a
m2 g sin    k m2 g cos   m1a  m1 g  m2 a
m2 g sin   m1a  m1 g  m2 a   k m2 g cos 
m2 g sin   m1a  m1 g  m2 a
k 
m2 g cos 
k 
56.7  0.5  39.2  1.125
 0.235
67.57
4.5 Friction
Example: Find the magnitude of P, the force exerted on this box, necessary to
keep the box moving at a constant velocity.
37°
Ff
P
µ = .36
15 kg
Solution:
Draw a free-body diagram. Hint: Never draw a force vector pointing into an
object, even when something is pushing.
FN
FN
Ff
P
mg
Py
mg
Break P into x and y components.
Px
4.5 Friction
The problem states the box will be moving at constant velocity.
This means Fnet = 0. µ = .36, m= 15 kg, Ө = 37°. Also Ff = µ FN
Equations for the vertical direction:
FN – mg – Py = 0 ; Py = P sin Ө
Equations for the horizontal direction:
Ff – Px = 0 ; Px = P cos Ө
Reduce the number of variables by rewriting the equation for the horizontal
forces as
µ FN - P cos 37° = 0
Use the vertical equation to substitute for FN : FN = mg + P sin 37°
µ (mg + P sin 37°) - P cos 37° = 0
µ mg + µ P sin 37° - P cos 37° = 0
P (µ sin 37° - cos 37°) = - µ mg
P=
- µ mg_______
µ sin 37° - cos 37°
P = 91 N
4.5 Friction
 = 0.25

A 150 N box sits motionless on an inclined plane as shown above. What is
the maximum angle of the incline before the block starts to slide?
Review of Chapter 4
A force is capable of changing an object’s state of
motion; that state of motion will change if and only
if there is a net force on the object.
Newton’s first law: In the absence of unbalanced
external forces, an object’s velocity will not
change.
Newton’s second law:
© 2010 Pearson Education, Inc.
Review of Chapter 4
Weight is the force exerted on an object by
gravity:
Newton’s second law holds separately for
each component of the force and acceleration.
© 2010 Pearson Education, Inc.
Review of Chapter 4
Newton’s third law: For every force, there is an
equal and opposite force (reaction force) acting on
the other object.
NFL – Inertia
NSL – F=ma
NTL – Action/reaction
Translational equilibrium: an object having no net
force on it in any direction. Its velocity is constant
or zero.
© 2010 Pearson Education, Inc.
Review of Chapter 4
Friction is the resistance to motion that occurs
when different surfaces are in contact.
Static friction:
Kinetic friction:
An object falling in air experiences air resistance;
this resistance increases until the object reaches
its terminal velocity.
© 2010 Pearson Education, Inc.
Check for Understanding
Fill in the blanks using the word bank.
cos Ө
sin Ө
kinetic
static
dynamic
coefficient
“opposite to”
“along with”
1. _________________friction occurs when two objects are in contact but not
moving.
2. _____________ friction occurs when two objects are rubbing against each
other.
3. Mu is the ________________ of friction.
4. The force of friction is ________ the direction of motion.
5. For an object on an inclined plane, the parallel component of weight is
mg __________.
6. For an object on an inclined plane, the perpendicular component of weight is
mg __________.
Check for Understanding
Fill in the blanks using the word bank.
cos Ө
sin Ө
kinetic
static
dynamic
coefficient
“opposite to”
“along with”
1. Static friction occurs when two objects are in contact but not moving.
2. Kinetic friction occurs when two objects are rubbing against each other.
3. Mu is the coefficient of friction.
4. The force of friction is opposite to the direction of motion.
5. For an object on an inclined plane, the parallel component of weight is
mg sin Ө.
6. For an object on an inclined plane, the perpendicular component of weight is
mg cos Ө.
Formulas for Chapter 4