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CP Physics Chapter 4
Newton’s Laws
Force
• Force (F) is a push or a pull
• Measured in Newtons (N) for SI, pounds
(lb) in US. (4.45 N = 1 lb)
•It has magnitude & direction, so it’s a
vector
Force in general
• Name some forces
• How strong are they?
• Contact vs at-adistance (or field)
Force in motion
• Forces act on things, or “bodies”
• Forces can require physical contact or be “at a
distance”
• Forces are drawn as vectors and added as vectors
• To draw forces (vectors) acting on a body, isolate
(or “free”) the body from everything—only the
forces acting on the body are drawn. Called a Free
Body Diagram (FBD)—see samples
• Force, acceleration, and mass (F, a, m) can be used
to explain motion with Newton’s Laws of Motion
Newton’s First Law of Motion
• Law of Inertia
• Net Force (F)
a vector sum
• Constant
velocity means
F = 0, or
..\My Videos\inertia\train-hurricane.mp4
equilibrium, ..\My Videos\inertia\planeland.mp4
..\..\Videos\inertia\FirstSnowcarcrashfun.mp4
or a = 0
Bed of nails
1st Law: Inertia
• Old: A body at rest will stay at rest,
A body in motion will stay in motion,
Until acted on by an outside force
• New:
• A body at rest will stay at rest,
A body with velocity will maintain its
velocity,
Until acted on by a Net Force
Example Net Force
You place your book on a bathroom scale
and you find that it weighs 50 N.
A) What force does it exert on the scale?
B) You now push down on the book with
a force of 10 N, what is the force
applied to the book by the Earth?
What force is exerted on the scale?
C) You now lift the book with a force of
20 N, what is the force exerted on the
scale?
packet examples
Newton’s Second Law of Motion
• Net Force = mass  acceleration
or
F = ma
Basic Example F = ma
What is the net force required to
accelerate a 1.5 kg box at 2.0
m/sec2?
Basic Example 2
What is the net force exerted on a
1500 kg car if it is accelerated
from 5 m/s to 10 m/s in 3 sec?
nd
2
Law: Acceleration
• F = ma
• The most used equation in Physics
• Both sides of the equation can be a
problem to solve
• The F can be the difficult part
because F is a vector sum
• See the more advanced problems after
3rd law
A use of
nd
2
Law:
What is. . .
• Mass, m (kg)
• Weight, Fg (N), or force of gravity
They are NOT the same thing!
nd
• From 2 Law, when a is = g
Fg = m·g
calculates weight
Examples
How much does a 3 kg object
weigh?
What is the mass of a 900 N
teacher?
Newton’s Third Law of Motion
“Every action (force) has an equal &
opposite reaction (force)”
• Forces are always in pairs
• When 2 objects interact, the forces they exert are an
action-reaction pair
• Action-reaction forces are on different objects and
are equal but opposite in direction
Newton’s Third Law
• Action-Reaction Pairs
• The hammer exerts a force
on the nail to the right.
• The nail exerts an equal
but opposite force on the
hammer to the left.
Action “hammer hits nail”
Reaction “nail hits hammer”
Newton’s Third Law
• Action-Reaction Pairs
• The rocket exerts a downward
force on the exhaust gases.
• The gases exert an equal but
opposite upward force on the
rocket.
Ffuel
FRocket
One more on Newton’s Third Law
• Action-Reaction Pairs
• The earth pulls down on you
(weight)…
• You pull up on the earth with an
equal and opposite force (same
as your weight)
Common forces in Problems
• Weight (Fg), the force of gravity (= mg)
• Support Force; called Normal Force where
“Normal” is perpendicular. This force is
perpendicular to the supporting surface (FN)
• Tension (FT), when a rope, cable, spring,
etc is pulled or stretched apart
• Friction (Ff), but not yet!
Normal Force, FN
-Supporting force
-Perpendicular to supporting surface
-NOT always the same as weight
FN
Problem solving-• Identify Forces, Knows, and Unknowns
• Sketch the problem
– Choose your coordinate system (x & y)
– Isolate the object
– Draw external forces (free body diagram)
• Apply Newton’s Laws in x & y (Fx=max &
Fy=may ). Know if ax or ay = 0 !!
• Solve for unknowns
Example
A 52 N sled is pulled across a level
cement sidewalk. A horizontal
force of 36 N is exerted.
A) What is the acceleration of the
sled?
B) What is the acceleration of the
sled if the sled’s mass is 30 kg?
Example
A) A 24 N fish hangs from
a scale in an elevator. The
elevator accelerates
upward at 3 m/sec2. What
does the scale read?
B) Again; this time the car
accelerates downward at
1.8 m/sec2.
So, when is the elevator cable most likely to break?
Example
An 800 N student stands in an
elevator. The elevator accelerates
upward at 2.5 m/sec2. What is the
tension in the cables if the elevator
car’s mass is 1000 kg?
When the car accelerates downward
at 2 m/sec2?
Example
What is the
acceleration of an
Atwood machine if
the two masses
hanging on it is 3 kg
and 4 kg?
What is the tension in
the cable?
CP Physics Chapter 4
Newton’s Laws with friction
Study two types of
Friction
•Static (s) stopped
•Kinetic (k) moving
Drag is fluid friction , which works differently
What Affects the Force
of Friction (Ff ) on a sliding object?
• Speed? (assuming v > 0)
• Area in contact?
• Weight of object?
• Materials in contact (surface
roughness)?
What if . . . ?
Fpush < Ff, max Fpush = Ff, max Fpush > Ff, max
• Ff is NOT an “applied” force
 (mu) is the coefficient of friction
General: Ff = FN
Fs  sFN
so Fs,max  sFN
Fk = kFN
FYI:
 s > k
 is a property of the materials,
which depends on their surface
roughness and contact force, FN
s=static k=kinetic
Example #1
What is the acceleration of a 30 kg
crate if the coefficient of friction is
0.034 and the object has a 15 N
horizontal pulling force applied to
it?
Try this problem again but the crate
weighs 30 N.
Example #2A
A 52 N sled is pulled across a
cement sidewalk at constant speed.
A horizontal force of 36 N is
exerted. What is the coefficient of
friction between the metal runners
and the cement?
Example #2B
Now the sled runs on packed snow
where the coefficient of friction
is 0.12. If a 650 N person sits
on the sled:
A)What is the applied force needed to move it
at constant speed?
B) How much force is needed to accelerate the
sled at 2 m/s2?
Example #3A
What coefficient of friction is needed to keep
the system at equilibrium?
m1=4 kg
=?
m2=3 kg
Example #3B
What is the acceleration of the system shown
below? What is the tension in the cable?
m1=4 kg
=0.2
m2=3 kg
1) What is the minimum mass of the fish
needed to just cause the Spam to move?
mspam = 20 kg
s = 0.5
mfish = ?
2) What is  if the fish was 12.5 kg
causing acceleration of 1.2 m/s2?
Example #4
A penguin slides across the ice
15 m to a stop. The coefficient of
friction between the penguin and
the ice is 0.01. What was the
acceleration and how fast was the
penguin going initially?
CP Physics Chapter 4
Newton’s Laws if any Angles
SOH CAH TOA !
Vector Addition
• Calculate x & y components of all vectors.
Add all components for x, and all for y
direction to find the total in the x (Fx) and
y (Fy).
• Knowing Fx and Fy , calculate Resultant
Force with Pythagorean Thm:
F  R  Fx  F
2
2
y
Example
• A 80 kg skier starts from rest and accelerates
down a hill which is angled 35 above
horizontal. The coefficient of friction
between the skis and snow is 0.10.
• Draw the forces acting on the skier
• What direction is the acceleration in?
Problem solving-• Identify Forces, Knows, and Unknowns
• Sketch the problem
–
–
–
–
Choose your coordinate system
Isolate the object(s)
RESOLVE any Forces at angles into x & y
Draw external forces
• Apply Newton’s Laws in x & y (Fx=max &
Fy=may )
• Solve for unknowns
Example
A sled of mass 30 kg is pulled with
a force of 50 N at a 25 degree
angle. If the coefficient of friction
between the runners and the snow
is 0.1, what is the acceleration of
the sled?
Example
A 50 kg boy is inside a 20 kg crate that is pushed with a
200 N force at 20 degrees below the horizontal.
a) What coefficient of friction would keep the crate from
accelerating?
b) What coefficient of friction would keep the crate
accelerating at 1.5 m/sec2?
F = 200 N
Review
You are standing on a scale while
riding an elevator and your mass is
35 kg. Determine how much the
Normal force is when:
• The elevator is coming down
at a constant speed of 2.3 m/s
• The elevator is accelerating
down at 1.35 m/s2
• The elevator is accelerating up
at 2.25 m/s2