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Newton’s Laws of Motion
• We begin our discussion of dynamics (study of how
forces produce motion in objects)
• We’ll use kinematic quantities (displacement,
velocity, acceleration) along with force & mass to
analyze principles of dynamics
– What is the relationship between motion and the forces
that cause the motion?
• Newton’s Laws of Motion (3 of them) describe the
behavior of dynamical motion
– Form the foundation of classical mechanics (or
Newtonian mechanics) – the physics of “everyday life”
– They are deterministic in nature
Newton’s Laws of Motion
• Concept of force: Push or pull experienced as a result
of interaction between objects or between object and
its environment
– Contact forces (involves direct contact between objects)
– Long-range forces (forces that act at a distance, e.g. gravity)
– Weight (force of gravitational attraction that Earth exerts on
an object)
• Force is a vector quantity (has magnitude & direction)
• SI unit of force: the Newton (N)
– 1 N = 1 kgm/s2
• Diagrams of forces acting on bodies: free-body
diagrams
 Draw all the external forces
acting on the hot dog cart
Newton’s Laws of Motion
• Free-body diagram for the hot dog cart (neglecting
friction):
“Normal” force ( to
surface of contact) FN
Pulling
 force

F p1
Weight of cart FW
• Effect of all 3 forces acting on the cart same as
effect of a single force equal to vector sum of
individual forces

– Total (net)force = vector
sum

 of individual forces = Fnet
– So Fnet  Fp1  FW  FN   F
– Since cart does not move up or down, sum of vertical
forces must be zero (same effect as no vertical forces):


Fnet  Fp1 (in this case)
Newton’snd Laws of Motion
• Suppose we add a 2
pulling force:

Fp 2

F p1

FN

FW
– Easier to add forces if we use the components of each force
– Any force can be replaced by its component vectors, acting
at the same point
y
• Set up coordinate system
Fp2y
& determine vector
components:
q
= hot dog cart

Fp 2
Fp2x
q

F p1

FN

FW
x
Newton’s Laws of Motion
• The net force acting on the hot dog cart can be
determined from the components of each individual
force:
F  F
F  F
net x

net y
x

y
Fnet x  Fp1  Fp 2 cosq
Fnet y  Fp 2 sin q  FN  FW

2
2
Fnet  Fnet x   Fnet y 
• How do forces affect motion?
• First consider what happens when net force on a
body is zero:
– Box at rest on floor:
– Box will remain at rest

FN

FW

F  0
Newton’s Laws of Motion
– Box sliding along freshly waxed floor:
v = constant  a = 0

F  0

FN
v = const.

FW
– v will remain constant (a = 0) if there is no friction
between box and floor (approximately true for a slick
floor)
– No force is needed to keep box sliding once it has started
moving (it would slow down and stop only if friction,
another force, were present)
• Newton’s 1st Law of Motion: A body with zero net
force acting on it moves with constant velocity
(which may be zero) and zero acceleration
– It’s the net force that matters in Newton’s 1st Law
Newton’s Laws of Motion
• Inertia is a property that indicates the tendency of a
body to keep moving once it’s set in motion, or the
tendency of a body at rest to remain at rest
– For example, your inertia is what causes you to feel like
you are being “pushed” against the side of your car when
you exit quickly from the highway onto an exit ramp
• When the net force on a body is zero, we say that
the body is in equilibrium:

F  0
 Fx  0
 Fy  0
(Newton’s 1st Law)
• Newton’s 1st Law is valid only in an inertial
reference frame, i.e. not a reference frame that is
accelerating with respect to the earth
– For example, an accelerating car does not form an
inertial reference frame
Newton’s Laws of Motion
• Now consider what happens when net force on a
body is not zero:
– From experiments, we learn that the presence of a net
force acting on a body causes the body to accelerate
• What is the relationship between net force and
acceleration?
• Let’s examine 3 different pulling forces on the cart:



2a
a
2 Fp1

F p1
1
a
2
1 
F p1
2
Newton’s Laws of Motion
– Magnitude of acceleration is directly proportional to the
magnitude of the net force acting on the body
– Constant of proportionality is the mass m of the body
• Newton’s
2nd


Law:
 F  ma
– In (2 – D) component form:
F
x
• Remember that:
 max
F
y
or

a

F
m
 may
– Newton’s 2nd Law is a vector equation
– Newton’s 2nd Law refers to external forces (ma is not a
force, so don’t include it on free-body diagrams!)
– Newton’s 2nd Law valid only in inertial reference frames,
like the 1st Law
– A nonzero net force is a cause, acceleration is the effect
CQ1: A 50-kg skydiver and a 100-kg
skydiver open their parachutes and reach a
constant velocity. The net force on the
larger skydiver is:
A) equal to the net force on the smaller
skydiver.
B) twice as great as the net force on the smaller
skydiver.
C) four times as great as the net force on the
smaller skydiver.
D) half as great as the net force on the smaller
skydiver.
CQ2: If F is the force of air resistance on an
object with mass m moving at a constant
velocity, which of the following best
describes the acceleration of the object
when the force of air resistance is reduced
by a factor of 4?
A)
B)
C)
D)
F/m
½ F/m
¼ F/m
¾ F/m
CQ3: Interactive Example Problem:
Predict the Satellite’s Motion
Which animation correctly shows the motion of
the satellite after the thruster force is applied?
A) Animation 1
B) Animation 2
C) Animation 3
D) Animation 4
PHYSLET #8.2.2, Prentice Hall (2001)
Weight
• We’ve already seen (from the hot dog cart example)
that there is a force called weight that is exerted on
bodies due to the gravitational pull of the earth
• What is the magnitude of this force?
– From Newton’s 2nd Law:

 F  ma
– Gravitational forces accelerate bodies with constant
magnitude a = g = 9.8 m/s2, so W  m g (with a downward
direction)
• Weight acts on a body all of the time
• Magnitude of g can change, depending on the
location (gmoon = 1.62 m/s2, for example)
Newton’s Third Law
• Newton’s 3rd Law: When a force from object
A is exerted on object B, B will exert a force
on A that is equal in magnitude but opposite
in direction to the force that A exerts on B:


FA on B   FB on A
– If I push on a wall, the wall pushes back on me
with a force that is equal to mine in magnitude
but opposite in direction
– Forces thus always come in pairs
– Force pairs resulting from Newton’s 3rd Law are
called action – reaction pairs and they never act
on the same body
CQ4: A book is at rest on a horizontal table.
What is the reaction force (as dictated by
Newton’s 3rd law) to the weight of the book?
A) The force that the table exerts upward
on the book.
B) The force that the book exerts
downward on the table.
C) The force of gravity on the book.
D) The force of gravity that the book exerts
on Earth.
Problem-Solving Strategy for Newton’s Laws
1) Draw cartoon of physical situation & define your
coordinate system
2) Draw free-body diagram of the object of interest
a) Draw force vectors for each external force acting on the
object
b) NEVER include ma in a free-body diagram
3) Apply Newton’s Laws of motion as appropriate
4) Repeat steps 1 – 3 for multiple objects if
necessary
5) Check your results – do they make sense?
Example Problem #4.31
A setup similar to the one
shown at right is often used
in hospitals to support and
apply a traction force to an
injured leg. (a) Determine
the force of tension in the
rope supporting the leg. (b)
What is the traction force
exerted on the leg? Assume
the traction force is
horizontal.
Solution (details given in class):
(a) 78.4 N
(b) 105 N
Example Problem #4.30
An object of mass 2.0 kg starts from rest and slides down an
inclined plane 80 cm long in 0.50 s. What net force is acting on
the object along the incline?
Solution (details given in class):
13 N
Geometry:
q
q
q
90°  q q
Example Problem #4.36
Find the acceleration of each block and the tension in the
cable for the following frictionless system:
5.00 kg
Coupled system: mass m2 moves
m1
same distance in same time as
mass
m

v
=
v

a
=
a
=
a
1
1
2
1
2
+x
m2
+y
N
T
10.0 kg
Free–body diagrams:
m1
m1g
T
m2
m2g
Apply Newton’s 2nd Law to block m1:
S Fx = m1ax  T = m1ax = m1a (1)
S Fy = 0  m1g – N = 0  m1g = N
(a = 0 in y–direction)
Example Problem (continued)
Apply Newton’s 2nd Law to block m2:
S Fy = m2ay = m2a
 m2g – T = m2a (2)
Combining equations (1) and (2):
 m2g – m1a = m2a  a = [m2 / (m1 + m2)]g = 6.53 m/s2
Using equation (1) and plugging in for a:
 T = m1a = 32.7 N
(Note that this analysis will be useful for Experiment 5 in lab!)
CQ5: Interactive Example Problem:
Rocket Blasts Off
What is the maximum height reached by the
rocket?
A) 0 m
B) 240 m
C) 960 m
D) 2880 m
E) 3840 m
ActivPhysics Problem #2.4, Pearson/Addison Wesley (1995–2007)
Friction
• Friction is a contact force between two surfaces
that always opposes motion
F
f


f is always  to N
N
W
• The kind of friction that acts when a body slides 
over a surface is called the kinetic friction force f k
(“kinetic” for motion)
– The magnitude of f k is proportional to the magnitude of
the normal force N:
f k  mk N
– Constant mk = coefficient of kinetic friction (depends on
the two surfaces in contact)
Friction
• When pushing a car, have you ever noticed that it’s
harder to start car moving than to keep it moving?
• The magnitude of the frictional force varies!

• There is a static frictional force f s , with variable
magnitude, that is almost always larger (at it’s
maximum value) than the kinetic frictional force
• For the case of pushing a piano across the floor:
N
(1)
(no pushing)
W
(2)
F
fs
N
(pushing but no sliding)
W
(3)
N
F
fs,max
(4)
Friction
(just about to slide)
W
N
F
fk
fs,max = fs max. value
(piano sliding)
W
• fs has a variable magnitude: f s  ms N
– Constant ms is the coefficient of static friction (ms > mk)
• In situation (1) above: f  0
• In situation (2) above: f  f s  ms N
• In situation (3) above: f  f s  ms N
• In situation (4) above: f  f k  mk N
Friction
• Variable magnitude of the
friction force as a function
of pushing force F
summarized in the graph
at right
• Friction responsible for motion of wheeled vehicles
(Note that f is in the same direction
as the motion of the car, but
opposite to the motion of the tires!)
CQ6: If the rear wheels of a truck drive the
truck forward, then the frictional force on the
rear tires due to the road is:
A) kinetic and in the backward direction.
B) kinetic and in the forward direction.
C) static and in the backward direction.
D) static and in the forward direction.
Example Problem #4.44
A crate of mass 45.0 kg is being transported on the flatbed of a
pickup truck. The coefficient of static friction between the crate
and the truck’s flatbed is 0.350, and the coefficient of kinetic
friction is 0.320. (a) The truck accelerates forward on level
ground. What is the maximum acceleration the truck can have
so that the crate does not slide relative to the truck’s flatbed?
(b) The truck barely exceeds this acceleration and then moves
with constant acceleration, with the crate sliding along its bed.
What is the acceleration of the crate relative to the ground?
Solution (details given in class):
(a) 3.43 m/s2
(b) 3.14 m/s2
Example Problem #4.69
Two boxes of fruit on a frictionless horizontal surface are
connected by a light string (see figure below), where m1 = 10 kg
and m2 = 20 kg. A force of 50 N is applied to the 20-kg box. (a)
Determine the acceleration of each box and the tension in the
string. (b) Repeat the problem for the case where the
coefficient of kinetic friction between each box and the surface
is 0.10.
N2
N1 T T
f1
m1g
f2
( f1 and f2 = 0 in part a)
m2gSolution (details given in class):
(a) a = 1.7 m/s2, T = 17 N
(b) a = 0.69 m/s2, T = 17 N
Example Problem #4.53
Find the acceleration reached by
each of the two objects shown in
the figure at right if the coefficient of
kinetic friction between the 7.00-kg
object and the plane is 0.250.
+y
T
N
f
+x
T
m1g
m2g
Solution (details given in class):
3.30 m/s2
Incline with Friction Interactive