Objectives
• Investigate the relationship between mass,
force, and acceleration.
• State Newton’s second law and give examples
to illustrate the law.
• Draw an accurate free body diagram
locating each of the forces acting on an
object or a system of objects.
• Use free body diagrams and Newton's
laws of motion to solve word problems.
Warm-up
•
•
•
•
Find the weight in Newtons for:
10 kg = _____ N
100 kg = _____ N
1 kg = _____ N
Engagement Activity
• Toliet Paper roll
Everything in the world does
one of two things:
Or……
Demo
• http://www.youtube.com/watch?v=iwP4he
WDhvw
• Demo with carts (varying force, varying
mass)
Objective
• To design two experiments to show the
effect mass has on acceleration keeping
force constant and the effect force has on
acceleration keeping mass constant.
Exploring Newton’s Second
Law
• Exploring Forces (discuss how mass is
constant in a system)
• Design two separate data collection
strategies to determine how two factors
affect the acceleration of the system: the
net force on the system and the total mass
of the system. (design for homework
Period 3)
Newton's 2nd law lab due
November 10.
Lab Report includes:
2 separate experiments in which one
experiment changes force keeping mass
constant and finding acceleration and the other
one changes mass keeping force constant and
finding acceleration.
-Problem or question
-Hypothesis for each
-Methods for each
- Data Table for each done in Excel if possible
- Graph for each done in Excel
Activity
• Computer Activity on Exploring Forces
Objectives
• Investigate the relationship between mass,
force, and acceleration.
• State Newton’s second law and give examples
to illustrate the law.
• Draw an accurate free body diagram
locating each of the forces acting on an
object or a system of objects.
• Use free body diagrams and Newton's
laws of motion to solve word problems.
Newton’s Second Law of Motion
The relation between acceleration and
force. Acceleration is proportional to force
and inversely proportional to mass.
“sigma” symbol means summation
If there is only one force present,
then you can leave it out.
Force is vector
Acceleration is a vector
Newton’s Second law
• Fnet = ma
(N) = (kg) (m/s2)
Force problems always must use
these units!
• If you prefer triangles
Fnet
a
• “net” means “the total sum”
or you can use the symbol Σ
m
Newton’s Second Law
• Force = Mass x Acceleration
• Force is measured in Newtons
ACCELERATION of GRAVITY(Earth) = 9.8 m/s2
• Weight (force) = mass x gravity (Earth)
Moon’s gravity is 1/6 of the Earth’s
If you weigh 420 Newtons on earth,
what will you weigh on the Moon?
70 Newtons
If your mass is 41.5Kg on Earth
what is your mass on the Moon?
Newton’s Second Law
• WEIGHT is a measure of the
gravity on the
force of ________
mass of an object
Newtons
• measured in __________
Newton’s Second Law
One rock weighs 5 Newtons.
The other rock weighs 0.5
Newtons. How much more
force will be required to
accelerate the first rock
at the same rate as the
second rock?
Ten times as much
Net Force & the 2nd Law
For a while, we’ll only deal with forces that
are horizontal or vertical.
When forces act in the same line, we can
just add or subtract their magnitudes to
find the net force.
32 N
15 N
2 kg
10 N
Fnet = 27 N to the right
a = 13.5 m/s2
Units
Fnet = m a
1N
= 1 kg
2
m/s
The SI unit of force is the Newton.
A Newton is about a quarter pound.
1 lb = 4.45 N
Graph of F vs. a
In the lab various known forces are
applied—one at a time, to the same mass—
and the corresponding accelerations are
measured. The data are plotted. Since F
and a are directly proportional, the
relationship is linear.
F
a
Slope
Since slope = rise / run = F / a, the slope
is equal to the mass. Or, think of y = mx
+ b, like in algebra class. y corresponds
to force, m to mass, x to acceleration,
and b (the
y-intercept) is zero.
F
F
a
a
W = mg
• Weight = mass  acceleration due to gravity.
• This follows directly from F = m a.
• Weight is the force of gravity on a body.
• Near the surface of the Earth,
g = 9.8 m/s2.
Video
• http://science360.gov/obj/video/58e62534e38d-430b-bfb1-c505e628a2d4/sciencenfl-football-newtons-second-law-motion
A scalar is simply a number, a magnitude alone.
A force is usually shown as a vector, which includes both
magnitude and a direction.
Force (or free-body) diagrams show the relative magnitude
and direction of all forces acting upon an object. The object
must be isolated and “free” of its surroundings.
This is a free-body diagram of the Statue
of Liberty. She is represented by a simple
box. The forces acting on her are labeled
with a magnitude and the arrow shows
direction. Notice the surrounding objects
are stripped away and the forces acting on
the object are shown.
495,704 lb
495,704 lb
W here represents the force of the weight
of the statue.
W =495,704 lb
N is the normal force, which represents
the force Liberty Island is pushing back
up on the statue.
The island has a great resistance to
compression. The ground is exerting a
force upward on the statue perpendicular,
or normal, to the surface.
N = 495,704 lb
(Positive y-direction)
+y
Think of the diagram on an XY
plane.
W =495,704 lb
If “up” is assumed to be the positive
direction, then N is positive and W is
negative.
N = 495,704 lb
+x
(Positive x-direction)
The first line of this calculation reads,
“The sum of the Forces in the positive y direction is
W + N” (  is the Greek symbol for “sum” )
+
(Positive y-direction)
+y
W =495,704 lb
Fy = W + N
Fy = (-495704 lb) + (+495704 lb )
Fy = 0
The sum of the forces in the y is zero.
N = 495,704 lb
+x
(Positive x-direction)
The forces acting on the object cancel each other out.
•We know F = m * a, where “a” is acceleration.
•If a = 0, then F = m * 0 = 0.
•When  F = 0, the object is not accelerating.
•We we can then say that the forces acting on the
object cancel each other out and it is in a state of
static equilibrium.
Create a free body diagram (FBD) for each of the following
situations. Draw a FBD of the gorilla:
N
W
Sitting Gorilla
Free Body Diagram of the Sitting
Gorilla (The box represents the
gorilla, W = weight of the gorilla
(Earth’s mass on gorilla), N =
Normal force (ground on gorilla)
Create a free body diagram (FBD) for each of the following
situations. Draw a FBD of the gorilla:
W
N
This is also an acceptable
diagram.
Sitting Gorilla
Draw a FBD of the wooden swing:
T1
T2
W
Parrot on wooden
swing hung by ropes
Free Body Diagram of the wooden
swing (The box represents the wooden
swing, W = weight of the swing (Earth’s
mass on swing) and the parrot, T
represents the ropes that are in tension
supporting the weight)
Draw a FBD of bucket the bungee jumper leaped from:
Bungee jumping
from crane
Draw a FBD of bucket the bungee jumper leaped from:
T
W
Bungee jumping
from crane
Free Body Diagram of the bucket (T
represents the tensile force of the cable
the bucket is suspended from, and W is
the weight of the diver and the bucket or
Earth’s mass on diver and bucket)
Draw a FBD of the ring at point C:
A
B
C
D
Traffic Light
supported by cables
Draw a FBD of the ring at point C:
A
B
C
TCA
TCB
D
TCD
Traffic Light
supported by cables
Free Body Diagram of the ring at
point C (T represents the force of the
cables that are in tension acting on
the ring)
Draw a FBD of the traffic light:
A
B
C
D
Traffic Light
supported by cables
Draw a FBD of the traffic light:
A
B
TCD
C
D
W
Traffic Light
supported by cables
Free Body Diagram of the traffic light
(TCD represents the force of the
cables acting on the light and W is
the weight acting on the light or
Earth’s mass on traffic light)
Draw a FBD of the pin at point A:
A
C
B
E
D
Pin-Connected Pratt Through
Truss Bridge
Draw a FBD of the pin at point A:
TAB
A
TAC T TAE
AD
B
Free Body Diagram of pin A
C
E
D
Pin-Connected Pratt Through
Truss Bridge
(If you consider the third dimension, then
there is an additional force acting on point A
into the paper: The force of the beam that
connects the front of the bridge to the back of
the bridge.)
Class work Chapter 4
• Questions on page 89 #s 1-5
• P.93 6-8
• Free body diagram worksheet
Homework Chapter 4
• Finish classwork.
• Section Review page 95.
9-14
Closure
• If a loaded truck that can accelerate at 1
m/s2 loses its load and has three-fourths of
the original mass, what acceleration can it
attain from the same driving force.
• Kahoot
42