Chapter 7. NEWTON’S SECOND LAW
A PowerPoint Presentation by
Paul E. Tippens, Professor of Physics
Southern Polytechnic State University
©
2007
The space shuttle
Endeavor lifts off for
an 11-day mission in
space. All of Newton’s
laws of motion - the
law of inertia, actionreaction, and the
acceleration produced
by a resultant force are exhibited during
this lift-off. Credit:
NASA Marshall Space
Flight Center (NASA-
NASA
MSFC).
Objectives: After completing this
module, you should be able to:
• Write Newton’s second law using appropriate
units for mass, force, and acceleration.
• Demonstrate your understanding of the
distinction between mass and weight.
• Draw free-body diagrams for objects at rest
and in motion.
• Apply Newton’s second law to problems
involving one or more bodies in constant
acceleration.
Newton’s First Law Reviewed
Newton’s First Law: An object at rest or an
object in motion at constant speed will remain
at rest or at constant speed in the absence of a
resultant force.
A glass is placed on a tablecloth and the
tablecloth is jerked quickly to the right. The
glass tends to remain at rest while the
tablecloth is removed.
Newton’s First Law (Cont.)
Newton’s First Law: An object at rest or an
object in motion at constant speed will remain
at rest or at constant speed in the absence of a
resultant force.
Assume glass and tablecloth move together at
constant speed. If the tablecloth stops
suddenly, the glass tends to maintain its
constant speed.
Understanding the First Law:
Discuss what the driver
experiences when a car
accelerates from rest and
then applies the brakes.
(a) The driver is forced to move forward. An
object at rest tends to remain at rest.
(b) Driver must resist the forward motion as
brakes are applied. A moving object tends
to remain in motion.
Newton’s Second Law:
• Second Law: Whenever a resultant force
acts on an object, it produces an
acceleration: an acceleration that is
directly proportional to the force and
inversely proportional to the mass.
F
a
m
Acceleration and Force With
Zero Friction Forces
Pushing the cart with twice the force
produces twice the acceleration. Three
times the force triples the acceleration.
Acceleration and Mass
Again With Zero Friction
F
F
a/2
a
Pushing two carts with same force F produces
one-half the acceleration. The acceleration
varies inversely with the amount of material
(the mass).
Measuring Mass and Force
The SI unit of force is the newton (N) and
the unit for mass is the kilogram (kg).
.
Force and Acceleration
F
F
F
4N
8N
12N
a = 2 m/s2
a = 4 m/s2
a = 6 m/s2
Acceleration a is directly proportional to force F
and is in the direction of the force. Friction
forces are ignored in this experiment.
Force and Acceleration
DF
F
DF
Da
= Constant
8N
Da
4 m/s
kg
=
2
2
a
Inertia or mass of 2 kilograms
Mass = 2 kg
MASS: A Measure of Inertia
6N
3
kg
12 N
18 N
a = 2 m/s2
3 kg
a = 4 m/s2
3 kg
a = 6 m/s2
One kg is that mass on which a constant force
of 1 N will produce an acceleration of 1 m/s2.
Friction forces are ignored in this experiment.
Metric Systems of Units
SI system: Accept kg as unit of mass, m as
unit of length, and s as unit of time. Derive
new unit of force, the newton (N).
F (N) = m (kg) a (m/s2)
Newton: The Unit of Force
One newton is that resultant force which imparts
an acceleration of 1 m/s2 to a mass of 1 kg.
F (N) = m (kg) a (m/s2)
What resultant force will give a 3 kg mass an
acceleration of 4 m/s2? Remember F = m a
3 kg
F=?
a = 4 m/s2
F  (3 kg)(4 m/s )
2
F = 12 N
Comparing the
Newton to the Pound
1 N = 0.225 lb
1 lb = 4.45 N
1 lb
4.45 N
A 160-lb person weighs about 712 N
A 10-N hammer weighs about 2.25 lb
Example 1: What resultant force F is required
to give a 6 kg block an acceleration of 2 m/s2?
a = 2 m/s2
6 kg
F=?
F = ma = (6 kg)(2 m/s2)
F = 12 N
Remember consistent units for force, mass,
and acceleration in all problems.
Example 2: A 40-N resultant force causes a
block to accelerate at 5 m/s2. What is the
mass?
2
a = 5 m/s
F = 40 N
m=?
F  ma
or
F
m
a
m = 8 kg
Example 3. A net force of 4.2 x 104 N acts on
a 3.2 x 104 kg airplane during takeoff. What
is the force on the plane’s 75-kg pilot?
First we find the
acceleration a of
plane.
F = 4.2 x 104 N
F 4.2 x 104 N
a 
m 3.2 x 104 kg
+
F = ma
m = 3.2 x 104 kg
a = 1.31 m/s2
To find F on 78-kg pilot, assume same acceleration:
F = ma = (75 kg)(1.31 m/s2);
F = 98.4 N
A Word About Consistent Units
Now that we have derived units of
newtons, we can no longer use units that
are inconsistent with those definitions.
Acceptable measures of LENGTH are:
SI units: meter (m)
Unacceptable units are: centimeters (cm);
millimeters (mm); kilometers (km);
yards (yd); inches (in.); miles (mi)
Consistent Units (Continued . . .)
Acceptable measures of MASS are:
SI units: kilogram (kg)
Unacceptable units are: grams (gm);
milligrams (mg); newtons (N);
pounds (lb); ounces (oz)
The last three unacceptable units are
actually units of force instead of mass.
Consistent Units (Cont.)
When we say that the acceptable units for
force and mass are the newton and the
kilogram, we are referring to their use in
physical formulas. ( Such as F = m a)
The centimeter, the millimeter, the
milligram, the mile, and the inch may be
useful occasionally in describing quantities.
But they should not be used in formulas.
Problem Solving Strategy
(For the Simpler Problems.)
• Read problem; draw and label sketch.
• List all given quantities and state what is to
be found.
• Make sure all given units are consistent with
Newton’s second law of motion (F = m a).
• Determine two of the three parameters in
Newton’s law, then solve for the unknown.
Example 4. A 54-gm tennis ball is in contact
with the racket for a distance of 40 cm as it
leaves with a velocity of 48 m/s. What is the
average force on the ball?
First, draw sketch and list
given quantities:
Given: vo = 0; vf = 48 m/s
x = 40 cm; m = 54 gm
a=?
Consistent units require converting grams
to kilograms and centimeters to meters:
Given: vo = 0; vf = 48 m/s x = 0.40 m;
Cont. . .
m = 0.0540 kg; a = ?
Example 4 (Cont). A 54-gm tennis ball is in
contact with the racket for a distance of 40
cm as it leaves with a velocity of 48 m/s.
What is the average force on the ball?
Knowing that F = m a, we need
first to find acceleration a:
0
2ax  v  v ; a 
2
f
(48 m/s)2
a
;
2(0.40 m)
2
0
v
2
f
2x
a  2880 m/s 2
F= (0.054 kg)(2880 m/s2);
F = ma
F = 156 N
Weight and Mass
• Weight is the force due to gravity. It is
directed downward and it varies from
location to location.
• Mass is a universal constant which is a
measure of the inertia of a body.
F = m a so that: W = mg and m =
W
g
Weight and Mass: Examples
• What is the weight of a 10-kg block?
10 kg
m
9.8 m/s2 W
W = 98 N
• What is the mass of a 200 N block?
W = mg
9.8 ft/s2
200N
W/g = m = (200 N)/(9.8 m/s2)
m =20.4 kg
Mass is Constant; W Varies.
16N
1.6 m/s2
Moon
98 N
9.8 m/s2
Earth
m=
W
g
= 10 kg
Description of Objects
• Objects described by mass or weight:
W (N) = m (kg) x 9.8 m/s2
• Conversions made by Newton’s 2nd Law:
W = mg
m=
W
g
Always Remember!!
In Physics, the use of Newton’s second law
and many other applications makes it
absolutely necessary to distinguish between
mass and weight. Use the correct units!
Metric SI units: Mass is in kg; weight is in N.
Always give preference to the SI units.
Example 5. A resultant force of 40 N
gives a block an acceleration of 8 m/s2.
What is the weight of the block near
the surface of the Earth?
a
8 m/s2
W=?
F = 40 N
To find weight, we must first
find the mass of the block:
F
F  ma; m 
a
40 N
m
 5 kg
2
8 m/s
Now find weight of a
5-kg mass on earth.
W = mg
= (5 kg)(9.8 m/s2)
W = 49.0 N
Newton’s Third Law (Reviewed):
• Third Law: For every action force, there
must be an equal and opposite reaction
force. Forces occur in pairs.
Action
Reaction
Reaction
Action
Acting and Reacting Forces
• Use the words by and on to study
action/reaction forces below as they
relate to the hand and the bar:
Action
The action force is exerted by
bar
the _____
hands on the _____.
The reaction force is exerted
bar on the _____.
hands
by the _____
Reaction
Example 6: A 60-kg athlete exerts a force on
a 10-kg skateboard. If she receives an
acceleration of 4 m/s2, what is the
acceleration of the skateboard?
Force on runner = -(Force on board)
mr ar = -mb ab
Force on
Board
(60 kg)(4 m/s2) = -(10 kg) ab
Force on
Runner
(60 kg)(4 m/s)
2
a
 24 m/s
-(10 kg)
a = - 24 m/s2
Review of Free-body Diagrams:
• Read problem; draw and label sketch.
• Construct force diagram for each object,
vectors at origin of x,y axes.
• Dot in rectangles and label x and y components opposite and adjacent to angles.
• Label all components; choose positive
direction.
Example of Free-body Diagram
A
300
600
B
Ay
30
A
0
Ax
B
600
By
Bx
4 kg
1. Draw and label sketch.
W = mg
2. Draw and label vector force diagram.
3. Dot in rectangles and label x and y components opposite and adjacent to angles.
Applying Newton’s Second Law
• Read, draw, and label problem.
• Draw free-body diagram for each body.
• Choose x or y-axis along motion and choose
direction of motion as positive.
• Write Newton’s law for both axes:
SFx = m ax
SFy = m ay
• Solve for unknown quantities.
Example 7: A cart and driver have a mass of
120 kg. What force F is required to give an
acceleration of 6 m/s2 with no friction?
1. Read problem and draw a sketch.
Diagram for Cart:
n
W
F
+
x
2. Draw a vector force diagram and label forces.
3. Choose x-axis along motion and indicate the
right direction as positive (+).
Example 7 (Cont.) What force F is required
to give an acceleration of 6 m/s2?
4. Write Newton's Law equation for both axes.
m = 120 kg
Diagram for cart:
n
F
W
ay = 0
n-W=0
The normal force n
SFy = 0;
is equal to weight W
+
x
SFx = max; F = ma
F = (120 kg)(6 m/s2)
F = 720 N
Example 8: What is the tension T in the
rope below if the block accelerates upward
at 4 m/s2? (Draw sketch and free-body.)
T
a
10 kg
a = +4 m/s2
T
mg
+
SFx = m ax = 0 (No info)
SFy = m ay = m a
T - mg = m a
mg = (10 kg)(9.8 m/s) = 98 N
m a= (10 kg)(4 m/s) = 40 N
T - 98 N = 40 N
T = 138 N
Example 9: In the absence of friction, what
is the acceleration down the 300 incline?
n
n
+
mg sin 600
600
W
300
mg
SFx = m ax
mg cos 600 = m a
a = g cos 600
a = (9.8 m/s2) cos 600
a = 4.9 m/s2
Example 10. Two-Body Problem: Find tension in the
connecting rope if there is no friction on the surfaces.
12 N
2 kg
4 kg
Find acceleration of
system and tension
in connecting cord.
First apply F = ma to entire system (both masses).
n
SFx = (m2 + m4) a
12 N
12 N = (6 kg) a
(m2 + m4)g
a=
12 N
6 kg
a = 2 m/s2
Example 10 (Cont.) The two-body problem.
12 N
2 kg
4 kg
Now find tension T
in connecting cord.
Apply F = m a to the 2 kg mass where a = 2 m/s2.
n
T
SFx = m2 a
T = (2 kg)(2 m/s2)
m2 g
T=4N
Example 10 (Cont.) The two-body problem.
2 kg
4 kg
12 N Same answer for T
results from focusing
on 4-kg by itself.
Apply F = m a to the 4 kg mass where a = 2 m/s2.
T
n
12 N
m2 g
SFx = m4 a
12 N - T = (4 kg)(2 m/s2)
T=4N
Example 11 Find acceleration of system and
tension in cord for the arrangement shown.
First apply F = m a to entire
system along the line of motion.
2 kg
SFx = (m2 + m4) a
n
4 kg
T
m4g = (m2 + m4) a
+a
T
m2 g
m4 g
Note m2g is balanced by n.
a=
m4g
m2 + m4
=
(4 kg)(9.8 m/s2)
2 kg + 4 kg
a = 6.53 m/s2
Example 11 (Cont.) Now find the tension T given
that the acceleration is a = 6.53 m/s2.
To find T, apply F = m a to just
the 2 kg mass, ignoring 4 kg.
2 kg
n
4 kg
T
+a
T
m2 g
m4 g
F
x
 m2 a or T  m2a
T = (2 kg)(6.53 m/s2)
T = 13.1 N
Same answer if using 4 kg.
m4g - T = m4 a
T = m4(g - a) = 13.1 N
Example 11. Find the acceleration of the system
shown below. (The Atwood machine.)
First apply F = ma to entire system
along the line of motion.
SFx = (m2 + m5) a
m5 g  m2 g  (m2  m5 )a
m5 g  m2 g (5 kg  2 kg)(9.8 m/s 2 )
a

m2  m5
2 kg + 5 kg
a = 4.20 m/s2
2 kg
5 kg
+a
T
T
m2 g m5 g
Summary
Newton’s First Law: An object at rest or an object
in motion at constant speed will remain at rest or
at constant speed in the absence of a resultant
force.
Newton’s Second Law: A resultant force produces
an acceleration in the direction of the force that
is directly proportional to the force and inversely
proportional to the mass.
Newton’s Third Law: For every action force,
there must be an equal and opposite reaction
force. Forces occur in pairs.
Summary: Procedure
FR  ma;
FR
a
m
N = (kg)(m/s2)
• Read, draw and label problem.
• Draw free-body diagram for each body.
• Choose x or y-axis along motion and choose
direction of motion as positive.
• Write Newton’s law for both axes:
SFx = m ax
SFy = m ay
• Solve for unknown quantities.
CONCLUSION: Chapter 7
Newton’s Second Law of Motion