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Honors Physics Chapter 4
Forces
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Honors Physics
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Lecture
Q&A
Four Fundamental (Basic) Forces in
Nature
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Gravitational Force
Electromagnetic Force
Strong (Nuclear) Force
Weak (Nuclear) Force
Gravitational Force
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Gravity (Gravitational force Earth pulling on objects
around it), weight: W, Fg
On surface of Earth, gravity is always straight
downward, with a magnitude of
W  mg

Gravitational force always exist, but value of g is
different at different location.
–
–
–
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On surface of earth, g = 9.8 m/s2.
Higher elevation, smaller g.
Higher latitude, larger g.
Electromagnetic Force

All other forces we encounter in daily life are
electromagnetic force in nature.
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Electric force: F
Magnetic force: F
Tension (pull, string): T
Push: F
Normal (support): FN or N (not to be confused with
Newton, the unit of force)
Friction: ƒ (ƒs or ƒk)
Air resistance, air drag: Fd
Strong (Nuclear) Force
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Holds protons and neutrons within the nucleus
Exist in very short distance only (fm,  10-15 m)
Stable nucleus
Weak (Nuclear) Force
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Also holds protons and neutrons within the
nucleus
Also very short range (fm,  10-15 m)
Not strong enough. Protons and neutrons can
sometimes escape  Nuclear reaction
Unstable nucleus
Weakest
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Strong
Weak
Electromagnetic
Gravitational
Relative Strength of Forces
Strongest
Force at Distance
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Contact force:
 Force giver and receiver must be in contact
 Tension, push, normal force, friction, …

Field force / Distant force:
 Force giver and receiver do not have to be in contact
 Gravity, electric force, magnetic force, …
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Unification of Forces
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Electromagnetic and weak forces have been
unified into one force: Electroweak force
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It is believed that all four forces are different
aspects of a single force.
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Grand Unification Theories (GUTs) and
Supersymmetric theories
Not yet sucessful
Newton’s First Law of Motion
Law of Inertia

An object with no net force acting on
it remains at rest or moves with
constant velocity in a straight line.
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No net force is needed to keep a
motion/velocity.
A net force is needed to change a
motion/velocity.
Net force = 0  Object at Equilibrium.
A special case of Second Law of
Motion.
Newton’s Second Law of Motion

The acceleration of a body is directly proportional
to the net force on it and inversely proportional to
its mass.
Fnet
a
 Fnet  ma
m

Unit of Force:
m
F = ma  [F] = [m] [a] = kg 2 = Newton = N
s
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Example:
Together a motorbike and rider have a mass of 275 kg. The
motorbike is slowed down with an acceleration of –4.50 m/s2.
What is the net force on the motorbike? Describe the direction
of the force and the meaning of the negative sign.
m  275kg, a  4.50m / s 2 , Fnet  ?
Fnet  ma
m

 275kg   4.50 2 
s 

m
 1237.5kg  2
s
 1.24  103 N
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The negative sign indicates that the force is opposite
to the direction of motion.
Practice:
A car, mass 1225 kg, traveling at 105 km/h, slows to a stop
in 53 m. What is the size and direction of the force that
acted on the car? What provided the force?
km  1000m   1h 
m
m  1225kg , vi  105

29.2
, v f  0,

  3600s 
s
h  1 km  

2
d  53m, F  ?
m


v f 2  vi 2  2ad  a 
v f 2  vi 2
0   29.2 
m
s


 8.04 2
2  53m
s
2d
m

F  ma  1225kg   8.04 2   9849 N  9800 N
s 

“-”: direction of force is opposite to velocity
Road surface is providing the force as friction.
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Newton’s Third Law of Motion

When one object exerts a force on a second
object, the second object also exerts a force on the
first object that is equal in magnitude but opposite
in direction.
–
–
–
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Action-reaction forces always exist together.
Action and reaction forces are of the same kind of
fundamental force--both are electromagnetic, or both
are gravitational.
Action and reaction forces act on two different objects
and therefore do not cancel out each other when only
one object is considered.
Action and Reaction Forces
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In general, the force
acting on the object under
consideration is the action
force, and the other is the
reaction.
Force between box and
ground:
–
–
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Action
Reaction
Action force: ground supporting the box, Fgb
Reaction force: box pushing on the ground, Fbg
How To Find Reaction Force
1.
2.
Identify the force giver and receiver of the
force.
Reversing the order of force giver and
receiver, you will get the reaction force.
Example:
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Action force: a  b
Reaction force: b  a
Example on Action-Reaction
box
table
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Force Diagram,
Example (2)
Free-Body Diagram
N: normal force, force table supporting the box
forces on box
Wb: weight of box, force Earth pulling on box
Are N and W a pair of action and reaction force? No
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 2 different objects
 same fundamental force
Example (3)
Ngt: normal force of ground supporting table
forces on table
N
N: Force of box pushing on Wt: weight of table, gravitational force
Earth pulling on table
table
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Example (4)
Wb: reaction force to
weight of box, force box
pulling on Earth
Wb
Wt
Wt: reaction force to weight of
table, gravitational force table
pulling on Earth
forces on earth
Ntg: force of table pushing on ground (Earth),
reaction force of Ngt.
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Example (5)
N
Ngt
Wb
Wb
Wt
N
Wt
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Action and reaction forces
always exist in pairs and
act on different objects.
Ntg
What is mass?
Mass is an intrinsic property of an object.
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It does not change.
It does not depend on the object’s position, velocity,
acceleration, temperature, or …
Two methods to measure mass: two kinds of mass
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Inertial mass: m = F / a
Gravitational mass: comparing gravitational force on the
object whose mass to be measured to gravitational force on
an object of known mass
Tension
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Tension is along the line.
Tension points away from the object.
T
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Normal Force (N or FN)
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Normal force points from surface to object.
Normal force is always perpendicular to the surface in contact.
(Normal = perpendicular)
If there is no tendency for the object to go into the surface, there is
no normal force.
No contact, no normal force.
Fapp
N
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N
N
Weight and Apparent Weight
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Weight is force of gravity of Earth
pulling on the object.
N
W  mg
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Apparent weight is the
reading on the apparatus.
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Apparent weight is the normal force
the scale exerting on the object
Apparent weight is the tension the
spring exerting on the object
Apparent weight does not have to be the
same as the weight.
T
Free-Body Diagram (Force Diagram)
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Draw all the forces acting on the object of consideration.
Ignore all forces the object acting on other objects (reaction
forces)
Draw all forces starting from center of the object for simplicity.
(frictional force normally is drawn at surface of contact.)
T
N
f
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W
Example:
A person of 50 kg stands on a scale on the floor of an elevator that is
accelerating upward at 2.0 m/s2.
a) What is the weight of the person?
b) What is the apparent weight of the person? (What is the reading on
the scale?)
Given: m  50kg
a )W  mg  50kg  9.8
m
 490 N
2
s
+
N=?
b) Define upward as the positive direction.
a  2.0m / s 2 “-” because W is in the negative direction.
Diagram:
2nd Law:
Fnet  N  W
Fnet  ma

 N  W  ma
m
N  W  ma  490 N  50kg  2.0 2  590 N
s
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W=mg
Example: 106-32
You are helping to repair a roof by loading equipment into a bucket that
workers hoist to the rooftop. If the rope is guaranteed not to break as long
as the tension does not exceed 450 N and you fill the bucket until it has a
mass of 42 kg, what is the greatest acceleration that the workers can give
the bucket as they pull it to the roof?
Define upward as the positive direction.
T  450 N ,m  42kg , a  ?
Diagram:
2nd Law:
a

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Fnet  T  W
Fnet  ma

+
T
 T  W  ma
T  mg
T

g
m
m
450 N
m
m
 9.8 2  0.914 2
42kg
s
s
W=mg
Practice 113-68
A 873-kg dragster, starting from rest, attains a speed of 26.3 m/s (58.9 mph) in 0.59 s.
a) Find the average acceleration of the dragster during this time interval.
B) What is the magnitude of the average net force on the dragster during this time?
C) Assume that the driver has a mass of 68 kg. What horizontal force does the seat
exert on the driver?
m  873kg , vi  0, v f  26.3
a)a  ?
v f  vi
v


a
t
t
m
, t  0.59s
s
m
0
m
s
 44.6 2
0.59s
s
+
v
26.3
N
b) Fnet  ?
Fnet  ma  873kg  44.6
Fseat
m
 38900 N
2
s
c) Fseat  ?
30
Fseat
?
m
68
kg

44.6
 3030 N  680Lb
 Fnet  ma 
2
s
W
Example: 115-92
Two blocks, one of mass 5.0 kg and the other of mass 3.0
kg, are tied together with a massless rope as in Figure 424. This rope is strung over a massless, resistance-free
pulley. The blocks are released from rest. Find a) the
tension in the rope, and b) the acceleration of the blocks.
Let downward = + for ma = 5 kg, and upward = + for mb = 3 kg.
Then two masses will have the same acceleration, a = ?. And the
tensions will be the same, T = ?
ma:
Fnet  Wa  T  ma a  T  Wa  ma a
mb:
Fnet  T  Wb  mb a
+
T
T
3
Wb
5
+
Wa
 Wa  ma a   Wb  mb a
 Wa  Wb  ma a  mb a  ma g  mb g   ma  mb  a   ma  mb  g   ma  mb  a
a
 ma  mb  g 
ma  mb
 5kg  3kg   9.8
5kg  3kg
m
s 2  2.45 m
s2
m
m

 T  Wa  ma a  ma g  ma a  ma  g  a   5kg   9.8 2  2.45 2   36.8N
s
s 

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Another Approach: 115-92
Two blocks, one of mass 5.0 kg and the other of mass 3.0
kg, are tied together with a massless rope as in Figure 424. This rope is strung over a massless, resistance-free
pulley. The blocks are released from rest. Find a) the
tension in the rope, and b) the acceleration of the blocks.
+
T
3
5
Take the two masses as a single system. Let clockwise as the
positive direction of the motions.
Then the net force accelerating the system is:
Fnet  Wa  Wb  ma g  mb g   ma  mb  g
Wb
And this net force is acceleration a total mass of m  ma  mb
So the acceleration of the system is
a
Fnet  ma  mb  g


m
ma  mb
 5kg  3kg   9.8
5kg  3kg
m
s 2  2.45 m
s2
Then we apply Newton’s second law on one of the mass to find the tension.
Let downward = + for 5kg. Then
Fnet  W  T  ma
32
m
m

 T  W  ma  mg  ma  m  g  a   5kg   9.8 2  2.45 2   36.8 N
s
s 

+
Wa
Practice:
Two blocks, 3.0kg and 5.0kg, are connected
by a light string and pulled with a force of
16.0N along a frictionless surface. Find (a)
the acceleration of the blocks, and (b) the
tension between the blocks.
+
N1
N2
3.0kg
Let right = +, m1 = 5.0 kg, m2 = 3.0 kg, F = 16.0 N
T
T
W2
5.0kg
W1
Consider only forces in the horizontal direction.
m1: Fnet  F  T  m1a 
m2: Fnet  T  m2 a
  F  m2 a  m1a

 F  m2a  m1a   m1  m2  a
a
F
16.0 N
m

 2.0 2
m1  m2 3.0kg  5.0kg
s
 T  m2 a  3.0kg  2.0
33
m
 6.0 N
2
s
F
16.0N