Clicker Question
Room Frequency BA
A glider (Yoda) is gliding along an air track at constant speed.
There is no friction and no air resistance.
What can you say about the net force (total force) on the glider?
A) The net force is zero.
B) The net force is non-zero and is in the direction of motion.
C) The net force is non-zero and is in the direction opposite the
motion.
D) The net force is non-zero and is perpendicular to the motion.
Velocity is constant, thus acceleration is zero.
F = ma = 0. So, net force is zero.
Announcements
• CAPA Set #5 due Friday at 10 pm
• This week in Section 
Assignment #3: Forces, Newton’s Laws, Free-Body Diagrams
Print out and bring the assignment to your section…
• Reading in Chapter 4.1-4.6
(Forces and Force diagrams)
• Exam #1 grades posted on CULearn
Solutions for version #1 posted on CULearn
More details at “Exam Information” link
Newton’s Three Laws
Newton’s Laws (as stated) hold in an
“Inertial Reference Frame” (non-rotating / non-accelerating):
I. “Law of Inertia”: Every object continues in its state
of rest, or uniform motion (constant velocity in a
straight line), as long as no net force acts on it.
II. Acceleration is proportional to force and inversely
proportional to mass. Famous F net  ma equation.
III.Whenever one object exerts a force on a second
object, the second exerts an equal force in the
opposite direction on the first.
F AB  F BA
Newton’s Second Law
Acceleration is proportional to force and
inversely proportional to mass.
F net  ma
Knowing the forces acting on an object of mass (m), we
can calculate the accelerations.
Then we can use these accelerations to calculate the
kinematics (motion).
Note that the direction of the acceleration is the same as
the direction of the net force vector.
The unit of force in the SI
system is the Newton (N).
Note that the pound is a unit of
force, not of mass, and can
therefore be equated to
Newtons but not to kilograms.
1 Newton ~ 0.22 pounds.
Force is a vector, so Newton’s
2nd Law is a vector equation:
F  ma

Fx  max
Fy  may
All Kinds of Forces
Pulling / Pushing
Friction
Gravity
Air Resistance / Drag
Springs
Lots of others too (electric, magnetic, etc.)
Clicker Question
Room Frequency BA
Forces add like vectors, not like scalars.
Consider two forces, F1 and F 2, both of magnitude 1 N.
One points to the right and other straight up.
What’s the magnitude and direction of the net force?
A)
B)
C)
D)
E)
2 N, straight up.
√2 N, straight up.
0 N.
√2 N, up and right.
-√2 N, left.
F2
Fnet  F1  F 2
√2 N
1N
1N
F1
Newton’s Third Law
Whenever one object exerts a force on a
second object, the second object exerts an
equal and opposite direction force
on the first.
F AB  F BA
Room Frequency BA
Clicker Question
A moving van collides with a sports car in a
high-speed head-on collision. Crash!
M
F truck
F car
m
During the impact, the truck exerts a force with magnitude Ftruck on
the car and the car exerts a force with magnitude Fcar on the truck.
Which of the following statements about these forces is true:
A) The magnitude of the force exerted by the truck on the car is
the same size as the magnitude of the force exerted by the car on
the truck: Ftruck = Fcar
B) Ftruck > Fcar
C) Ftruck < Fcar
Guaranteed by Newton’s Third Law
Room Frequency BA
Clicker Question
A moving van collides with a sports car in a
high-speed head-on collision. Crash!
M
F truck
F car
m
During the collision, the imposed forces cause the
truck and the car to undergo accelerations with
magnitudes atruck and acar. What is the relationship
between atruck and acar?
A) atruck > acar
B) acar > atruck
Newton’s Second Law: F = ma
C) atruck = acar
D) Indeterminate from information given.
Free-Body Diagram
Draw all forces for each object (truck and car).
Consider only horizontal direction.
m
F truck
CAR
F car
Ftruck = m acar
Force of truck acting on the car
Fcar = M atruck
M
TRUCK
Force of car acting on the truck


| Ftruck || Fcar | By Newton’s Third Law (and with opposite directions)
ma car  Matruck
acar
M

atruck
m
Demonstration with Skateboard
Consider an Everyday Occurrence
Assume you are in a tie with your sumo-wrestler friend.
Draw the free-body diagrams (i.e. forces) acting on you.
F (tension in rope)
F (pulling on the rope)?
F (friction with floor)
Consider an Everyday Occurrence
Draw the free-body diagrams (i.e. forces) acting on you.
F (tension in rope)
F (Normal)
F (friction with floor)
F (gravity)
Any other forces acting on you?
Room Frequency BA
Clicker Question
What is the direction of the friction force acting on the
sumo wrestler’s feet? A) Left
B) Down
C) Right
D) Up
F (tension in rope)
F (Normal)
F (friction with floor)
F (gravity)
We often draw free-body diagrams schematically

 on you)
F ( tension pulling
F ( tension pulling on Sumo)
m
M
N = “normal force”
N
F friction
F friction
Fgr  mg
Fgr  M g
= force of gravity
Be very careful to indicate on which object the force is acting
(not who does the acting).
Point-Like Objects
We often treat objects as point-like with all forces acting on that point.
Reasonable for rigid (non-deformable) objects and ignoring rotations.
Since you are not moving….
No acceleration in x or y
directions
N
F friction
m

F ( tension)
F  mg
y
x
Label Coordinate Axes
Net force in x must be zero.


F (friction)  F (tension)  0
Net force in y must be zero.


F (gravity)  F ( Normal)  0
Slip-n-Slide
Part I – going down the slide
Physics approximation….
1) You are a small rectangular box (that we will
treat as a point-like object anyway).
2) The slope of the slide is constant (angle = q)
How large a net force will you feel down the slope?
What is the acceleration down the slope?
How fast will you be going after traveling down the
slope for some time t?
Step 1: Draw a free-body diagram
Note that “Normal Force” is
always Perpendicular
(i.e. normal) to the surface.
Fg = mg
In this case, it is not in the
opposite direction to the
gravitational force.