Download AP Physics 1 * Unit 2

AP Physics 1 – Unit 2
DYNAMICS OF FORCE AND MOTION
Newton’s Laws
Learning Objectives:
BIG IDEA 1: Objects and systems have properties such as mass and charge. Systems
may have internal structure.1.C.1.1: I can design an experiment for collecting data
to determine the relationship between the net force exerted on an object its
inertial mass and its acceleration.
[SP 4.2]
1.C.3.1: I can design a plan for collecting data to measure gravitational mass and to
measure inertial mass and to distinguish between the two experiments. [SP 4.2]
BIG IDEA 2: Fields existing in space can be used to explain interactions.
2.B.1.1: I can calculate the gravitational force on an object with mass m in a
gravitational field of strength g in the context of the effects of a net force on objects
and systems. [SP 2.2, 7.2]
Learning Objectives
BIG IDEA 3: The interactions of an object with other objects can be described by forces.
3.A.2.1: I can represent forces in diagrams or mathematically using appropriately labeled vectors with
magnitude, direction, and units during the analysis of a situation.
[SP 1.1]
3.A.3.1: I can analyze a scenario and make claims (develop arguments, justify assertions) about the forces
exerted on an object by other objects for different types of forces or components of forces. [SP 6.4, 7.2]
3.A.3.2: I can challenge a claim that an object can exert a force on itself. [SP 6.1]
3.A.3.3: I can describe a force as an interaction between two objects and identify both objects for any force. [SP
1.4]
3.A.4.1: I can construct explanations of physical situations involving the interaction of bodies using Newton’s
third law and the representation of action-reaction pairs
of forces. [SP 1.4, 6.2]
3.A.4.2: I can use Newton’s third law to make claims and predictions about the action-reaction pairs of forces
when two objects interact. [SP 6.4, 7.2]
3.B.1.1: I can predict the motion of an object subject to forces exerted by several objects using an application of
Newton’s second law in a variety of physical situations with acceleration in one dimension. [SP 6.4, 7.2]
Learning Objectives
3.B.1.2: I can design a plan to collect and analyze data for motion (static, constant, or accelerating) from
force measurements and carry out an analysis to determine the relationship between the net force and
the vector sum of the individual forces. [SP 4.2, 5.1]
3.B.1.3: I can reexpress a free-body diagram representation into a mathematical representation and solve
the mathematical representation for the acceleration of the object. [SP 1.5, 2.2]
3.B.2.1: I can create and use free-body diagrams to analyze physical situations to solve problems with
motion qualitatively and quantitatively. [SP 1.1, 1.4, 2.2]
3.C.4.1: I can make claims about various contact forces between objects based on the microscopic cause
of those forces. [SP 6.1]
3.C.4.2: I can explain contact forces (tension, friction, normal, spring) as arising from interatomic electric
forces and that they therefore have certain directions. [SP 6.2]
Learning Objectives
BIG IDEA 4: Interactions between systems can result in changes in those systems.
4.A.1.1: I can use representations of the center of mass of an isolated two-object system to analyze the
motion of the system qualitatively and semi-quantitatively. [SP 1.2, 1.4, 2.3, 6.4]
4.A.2.1: I can make predictions about the motion of a system based on the fact that acceleration is equal to
the change in velocity per unit time, and velocity is equal to the change in position per unit time. [SP 6.4]
4.A.2.2: I can evaluate using given data whether all the forces on a system or whether all the parts of a
system have been identified. [SP 5.3]
4.A.2.3: I can create mathematical models and analyze graphical relationships for acceleration, velocity, and
position of the center of mass of a system and use them to calculate properties of the motion of the center
of mass of a system. [SP 1.4, 2.2]
4.A.3.1: I can apply Newton’s second law to systems to calculate the change in the center-of-mass velocity
when an external force is exerted on the system. [SP 2.2]
4.A.3.2: I can use visual or mathematical representations of the forces between objects in a system to
predict whether or not there will be a change in the center-of-mass velocity of that system. [SP 1.4]
Force
•Unit is the NEWTON(N)
•Is by definition a push or a pull
•Can exist during physical contact(Tension,
Friction, Applied Force)
•Can exist with NO physical contact, called
FIELD FORCES ( gravitational, electric, etc)
st
1
Law of Motion - Inertia
INERTIA – a quantity of matter, also called MASS. Italian
for “LAZY”. Unit for MASS = kilogram.
Weight or Force due to Gravity is how your MASS is
effected by gravity.
W  mg
NOTE: MASS and WEIGHT are NOT the same thing. MASS never changes
When an object moves to a different planet.
What is the weight of an 85.3-kg person on earth? On Mars=3.2 m/s/s)?
W  mg  W  (85.3)(9.8)  835.94 N
WMARS  (85.3)(3.2)  272.96 N
st
1
Law of Motion - Inertia
An object in motion remains in motion in a straight line and at a constant speed
OR an object at rest remains at rest, UNLESS acted upon by an EXTERNAL
(unbalanced) Force.
There are TWO conditions here and one constraint.
Condition #1 – The object CAN move but must be at a CONSTANT SPEED
Condition #2 – The object is at REST
Constraint – As long as the forces are BALANCED!!!!! And if all the forces
are balanced the SUM of all the forces is ZERO.
The bottom line: There is NO ACCELERATION in this case AND the object
must be at EQILIBRIUM ( All the forces cancel out).
acc  0   F  0
Free-Body Diagrams
A pictorial representation of forces complete with labels.
FN
T
Ff
T
W1,Fg1
or m1g
m2g
•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.
Free-Body Diagrams
Ff
FN
mg
st
Newton’s 1
Law and Equilibrium
Since the Fnet = 0, a system moving at a constant speed or at rest MUST be at
EQUILIBRIUM.
TIPS for solving problems
•Draw a FBD
•Resolve anything into COMPONENTS
•Write equations of equilibrium
•Solve for unknowns
Example:
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
Example:
Suppose the same box is now pulled at an angle of 30 degrees above the horizontal.
a)Calculate the Force of Friction
Fax  Fa cos   50 cos 30  43.3N
b)Calculate the Force Normal
F f  Fax  43.3N
FN
Ff
Fa
Fay
30
Fax
mg
FN  mg!
FN  Fay  mg
FN  mg  Fay  (10)(9.8)  50 sin 30
FN  73N
What if it isn’t at equilibrium?
If an object is NOT at rest or moving at a constant speed, that means the
FORCES are UNBALANCED. One force(s) in a certain direction over power the
others.
THE OBJECT WILL THEN ACCELERATE.
nd
Newton’s 2
Law
The acceleration of an object is directly proportional to the NET FORCE and
inversely proportional to the mass.
a  FNET
1
a
m
FNET
a
 FNET  ma
m
FNET   F
Tips:
•Draw an FBD
•Resolve vectors into components
•Write equations of motion by adding and
subtracting vectors to find the NET FORCE.
Always write larger force – smaller force.
•Solve for any unknowns
nd
Newton’s 2
Law
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.
FN
Ff
Fa
In which direction,
is this object
accelerating?
The X direction!
mg
So N.S.L. is worked
out using the forces
in the “x” direction
only
FNet  ma
Fa  F f  ma
50  12  10a
a  3.8 m / s 2
Example:
A mass, m1 = 3.00kg, 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  m1a
T
m2 g  m1a  m2 a
m2 g  m2 a  m1a
m2 g  a (m2  m1 )
a
m2 g
(11)(9.8)

 7.7 m / s 2
m1  m2
14
Example:
FNet  ma
m2 g  T  m2 a
T  m1a
FNET
FNet  ma 
m
a
Rise
Slope 
Run
T  (3)(7.7)  23.1 N
rd
Newton’s 3
Law
“For every action there is an EQUAL and OPPOSITE reaction.
◦ This law focuses on action/reaction pairs (forces)
◦ They NEVER cancel out
All you do is SWITCH the wording!
•PERSON on WALL
•WALL on PERSON
rd
Newton’s 3
Law
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.
Examples:
Action: HAMMER HITS NAIL
Reaction: NAIL HITS HAMMER
Action: Earth pulls on YOU
Reaction: YOU pull on the earth
Friction
KINETIC AND STATIC
Learning Objectives:
Types of Friction
Static – Friction that keeps an object at rest and prevents it from moving
Kinetic – Friction that acts during motion
Force of Friction
F f  FN
The Force of Friction is
directly related to the Force   constant of proportion ality
Normal.
  coefficien t of friction
◦ Mostly due to the fact that BOTH
are surface forces
Fsf   s FN
The coefficient of
Fkf   k FN
friction is a unitless
constant that is
specific to the
material type and
usually less than one.
Note: Friction ONLY depends on the MATERIALS sliding against each
other, NOT on surface area.
Example:
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.
Fa
a) What is the coefficient of kinetic
friction between the crate and the
floor?
F f   k FN
FN
Fay
F f  Fax  Fa cos   600(cos 20)  563.82 N
FN  Fay  mg  Fa sin   1500
20
FN  600(sin 20)  1500  1705.21N
Fax
563.82   k 1705.21
 k  0.331
Ff
mg
Example:
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)
FN
20
Fax
FNet  ma
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
Fa
𝐹𝑓
mg
Fay
Inclines

Ff
FN




mg
mg sin 

Tips
•Rotate Axis
•Break weight into components
•Write equations of motion or
equilibrium
•Solve
Example:
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.
T
FN
Ff
m2gcos40
m2g
m1
40
m2gsin40
m1g
T  m1 g  m1a  T  m1a  m1 g
m2 g sin   ( F f  T )  m2 a
40
T
FNET  ma
FNET  ma
Example:
T  m1 g  m1a  T  m1a  m1 g
m2 g sin   ( F f  T )  m2 a
x  voxt  1 at 2
2
1  0  1 a ( 4) 2
2
a  0.125 m / s 2
T  4(.125)  4(9.8)  39.7 N
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 
k 
m2 g sin   m1a  m1 g  m2 a
m2 g cos 
k 
56.7  0.5  39.2  1.125
 0.235
67.57
Newton’s Law of
Gravitation
Learning Objectives:
Gravity
What causes YOU to be pulled down? THE EARTH….or more
specifically…the EARTH’S MASS. Anything that has MASS has a
gravitational pull towards it.
Fg Mm
What the proportionality above is
saying is that for there to be a FORCE
DUE TO GRAVITY on something there
must be at least 2 masses involved,
where one is larger than the other.
Gravity
As you move AWAY from the earth, your DISTANCE
increases and your FORCE DUE TO GRAVITY
decrease. This is a special INVERSE relationship
called an Inverse-Square.
1
Fg  2
r
The “r” stands for SEPARATION DISTANCE and is the
distance between the CENTERS OF MASS of the 2
objects. We us the symbol “r” as it symbolizes the
radius. Gravitation is closely related to circular
motion as you will discover later.
Putting it all together
m1m2
r2
G  constant of proportion ality
Fg 
G  Universal Gravitatio nal Constant
G  6.67 x10
Fg  G
 27
Nm 2
m1m2
r2
Fg  mg  Use this when you are on the earth
Fg  G
m1m2
 Use this when you are LEAVING th e earth
r2
kg 2
Example:
Let’s set the 2 equations equal to each other since they BOTH represent your
weight or force due to gravity
Fg  mg  Use this when you are on the earth
Fg  G
m1m2
 Use this when you are LEAVING th e earth
2
r
Mm
mg  G 2
r
M
g G 2
r
M  Mass of the Earth  5.97 x10 24  kg
r  radius of the Earth  6.37 x10 6  m
SOLVE FOR g!
(6.67 x1027 )(5.97 x1024 )
2
g

9
.
81
m
/
s
(6.37 x106 ) 2