Transcript Force and 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
Homework for Chapter 4
• Read Chapter 4.1-4.3
• HW 4.1-4.3 pg 130-132 #1-36
• Omit 5,6,13,32,33
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
4.1 The Concepts of Force and Net Force
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
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
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
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).
Homework 4A
Covers Sections 4.1-4.4: Newton’s Laws
• HW 4A: pp. 131-132: 13, 22, 34, 35, 43, 44.
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.
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
Homework 4.B
(Section 4.5)
• HW 4B : pp. 132-134: 46-48, 51-53, 56-59, 64.
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
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
4.5 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?
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.
Homework 4C
Covers Section 4.6: Friction
• HW 4C: pp. 135-136: 76-78, 80, 83, 85, 88.
Formulas for Chapter 4