PHY300

1. Giambattista Physics
Chapter 4
©McGraw-Hill Education. All rights reserved. Authorized only for instructor use in the classroom. No reproduction or further distribution permitted without the prior written consent of McGraw-Hill Education.

2. ©McGraw-Hill Education
Chapter 4: Forces and Newton’s Laws of Motion
4.1 Interactions and Forces
4.2 Inertia and Equilibrium:
Newton’s First Law of Motion
4.3 Net Force, Mass, and
Acceleration: Newton’s Second
Law of Motion
4.4 Interaction Pairs: Newton’s
Third Law of Motion
4.5 Gravitational Forces
4.6 Contact Forces
4.7 Tension
4.8 Applying Newton’s Laws
4.9 Reference Frames
4.10 Apparent Weight
4.11 Air Resistance
4.12 Fundamental Forces

3. ©McGraw-Hill Education
4.1 Interactions and Forces
Isaac Newton was the first to discover that the laws that govern
motions on the Earth also applied to celestial bodies.
Over the next few chapters we will study how bodies interact
with one another.

4. ©McGraw-Hill Education
What is a Force?
Simply, a force is a “push” or “pull” on an object.

5. ©McGraw-Hill Education
Measuring Forces
How can a force be measured? One way is with a spring scale.
By hanging masses on a
spring we find that the
spring stretch  applied
force.
The units of force are Newtons (N).

6. ©McGraw-Hill Education
Net Force
The net force is the vector sum of all the forces acting on a body.






 3
2
1
net F
F
F
F
F

7. ©McGraw-Hill Education
Free-Body Diagrams
Should be drawn for problems when forces are involved.
Should be large so that they are readable.
Draw an idealization of the body in question (a dot, a box,…).
You will need one free-body diagram for each body in the
problem that will provide useful information for you to solve the
given problem.
Indicate only the forces acting on the body. Label the forces
appropriately. Do not include the forces that this body exerts on
any other body.

8. ©McGraw-Hill Education
Free-Body Diagrams Continued
A coordinate system is a must.
Do not include fictitious forces. Remember that ma is itself not a
force!
You may indicate the direction of the body’s acceleration or
direction of motion if you wish, but it should be done well off to
the side of the free-body diagram.

9. ©McGraw-Hill Education
4.2 Inertia and Equilibrium:
Newton’s First Law of Motion
Newton’s 1st Law (The Law of Inertia):
If no force acts on an object, then its speed and direction of
motion do not change.
Inertia is a measure of an object’s resistance to changes in its
motion.

10. ©McGraw-Hill Education
If No Force Acts...
If the object is at rest, it remains at rest (speed = 0).
If the object is in motion, it continues to move in a straight line
with the same speed (constant velocity).
No force is required to keep a body in straight-line motion when
effects such as friction are negligible.
An object is in translational equilibrium if the net force on it is
zero.

11. ©McGraw-Hill Education
4.3 Net Force, Mass, and Acceleration:
Newton’s Second Law of Motion
Newton’s 2nd Law:
The acceleration of a body is directly proportional to the net
force acting on the body and inversely proportional to the body’s
mass.
Mathematically: a
F
F
a m
m

 net
net
or

12. ©McGraw-Hill Education
Force, Mass, Equilibrium
An object’s mass is a measure of its inertia. The more mass, the
more force is required to obtain a given acceleration.
Recall: the net force is just the vector sum of all of the forces
acting on the body, often written as F.
If a = 0, then F = 0 (and vice-versa). The object is in translational
equilibrium. It can have speed = 0, or speed  0, but constant
magnitude and direction (i.e., constant velocity).

13. ©McGraw-Hill Education
4.4 Interaction Pairs:
Newton’s Third Law of Motion
Newton’s 3rd Law:
When 2 bodies interact, the forces on the bodies from each
other are always equal in magnitude and opposite in direction.
Or, forces come in pairs.
Mathematically: .
12
21 F
F 


14. ©McGraw-Hill Education
Interaction Pairs, Example
Consider a box resting on a table.
F1
(a) F1 is the force of the
Earth on the box, what is
the interaction partner of
this force?

15. ©McGraw-Hill Education
Interaction Pairs, Example Continued
Consider a box resting on a table.
F2
(b) F2 is the force of the box
on the table, what is the
interaction partner of this
force?

16. ©McGraw-Hill Education
External Forces
Any force on a system from a body outside of the system is an
external force.
F
Example: Pulling a box
across the floor. The pulling
force is external to the box.

17. ©McGraw-Hill Education
Internal Forces
Force between bodies of a system are internal forces.
Fext
Example: Pulling 2 boxes across the floor where the two
boxes are attached to each other by a rope. The pulling
force is external, but the forces on each box from the
tension in the rope are internal.

18. ©McGraw-Hill Education
4.5 Gravitational Forces
Gravity is the force between two masses. Gravity is a long-range
or field force. No contact is needed between the bodies. The
force of gravity is always attractive!
2
2
1
r
M
GM
F 
r is the distance between the two masses
M1 and M2 and G = 6.671011 Nm2/kg2.
M2
r
M1
F21 F12
.
12
21 F
F 


19. ©McGraw-Hill Education
Gravitational Forces Continued 1
Let M1 = mass of the Earth. 2
2
M
r
GM
F E







Here F = the force the Earth exerts on mass M2. This is the force
known as weight, w.
.
2
2
2
gM
M
r
GM
w
E
E










km
6400
kg
10
98
.
5
E
24
E



r
M
2
2
m/s
8
.
9
N/kg
8
.
9
where 


E
E
r
GM
g Near the surface
of the Earth

20. ©McGraw-Hill Education
Gravitational Forces Continued 2
Note that
m
F
g  is the gravitational force per unit
mass. This is called the gravitational
field strength. It is also referred to as
the acceleration due to gravity.
What is the direction of g?
What is the direction of w?

21. ©McGraw-Hill Education
Gravitational Forces Example
Example: What is the weight of a 100 kg astronaut on the
surface of the Earth (force of the Earth on the astronaut)? How
about in low Earth orbit? This is an orbit about 300 km above
the surface of the Earth.
On Earth: N
980

 mg
w
In low Earth orbit:
 
N
890
)
( 2












h
R
GM
m
r
mg
w
E
E
o
Their weight is reduced by about 10%.
The astronaut is NOT weightless!

22. ©McGraw-Hill Education
4.6 Contact Forces
Contact forces: these forces arise because of an interaction
between the atoms in the surfaces in contact.

23. ©McGraw-Hill Education
Normal Force
Normal force: this force acts in the direction perpendicular to
the contact surface.
N
w
Example: normal force of the
ground on the box acts upward
(perpendicular to the horizontal
surface).
Example: normal force of the
inclined ramp on the box acts
at an angle (perpendicular to
the surface of the ramp).
N
w

24. ©McGraw-Hill Education
Normal Force Example
Consider a box on a table.
FBD for
the box
N
w
x
y
Apply
Newton’s
2nd law mg
w
N
w
N
Fy






that
So
0
This just says the magnitude of the
normal force equals the magnitude of the
weight; they are not Newton’s third law
interaction partners.

25. ©McGraw-Hill Education
Friction
Friction: a contact force parallel to the contact surfaces.
Static friction acts to prevent objects from sliding.
Kinetic friction acts to make sliding objects slow down.

26. ©McGraw-Hill Education
Static Friction
The force of static friction is modeled as s s
f N


where s is the coefficient of static friction and N is the normal
force.

27. ©McGraw-Hill Education
Kinetic Friction
The force of kinetic friction is modeled as k k
f N


where k is the coefficient of kinetic friction and N is the normal
force.

28. ©McGraw-Hill Education
Contact Forces Example 1
Example (text problem 4.102): A box full of books rests on a
wooden floor. The normal force the floor exerts on the box is
250 N.
(a) You push horizontally on the box with a force of 120 N, but it
refuses to budge. What can you say about the coefficient of
friction between the box and the floor?
FBD for
box N
w
x
y
F
fs
Apply
Newton’s
2nd Law 0
)
2
(
0
)
1
(








s
x
y
f
F
F
w
N
F

29. ©McGraw-Hill Education
Contact Forces Example 1 Continued 1
From equation (2): 48
.
0
s 




N
F
N
f
F s
s 

This is the minimum value of s, so s > 0.48.
(b) If you must push horizontally on the box with 150 N force to
start it sliding, what is the coefficient of static friction?
Again from equation (2): 60
.
0
s 




N
F
N
f
F s
s 


30. ©McGraw-Hill Education
Contact Forces Example 1 Continued 2
(c) Once the box is sliding, you only have to push with a force of
120 N to keep it sliding. What is the coefficient of kinetic
friction?
FBD for
box
N
w
x
y
F
fk
Apply
Newton’s
2nd Law 0
)
2
(
0
)
1
(








k
x
y
f
F
F
w
N
F
From equation (2):
48
.
0
N
250
N
120
k
k
k





N
F
N
f
F



31. ©McGraw-Hill Education
Contact Forces Continued
Consider a box of mass m that is at rest on an incline.
Its FBD is:
FBR
w x
y
There is one long-range force
acting on the box: gravity.
There is one contact force
acting on the box from the
ramp. FBR = force on box
from ramp.
If the net force acting on the box is zero, then the contact force
from the ramp must have the same magnitude as the weight
force, but be in the opposite direction.

32. ©McGraw-Hill Education
Contact Forces Continued 2
The force FBR can be resolved into components that are
perpendicular and parallel to the ramp.
FBR
w
N
fs
x
y The perpendicular component is
what we call the normal force.
The parallel component is the
static friction force.
In the FBD, the force FBR is dashed to indicate that it has been
resolved into components.

33. ©McGraw-Hill Education
Contact Forces Example 2
Example: Let the box on the ramp have a mass 2.5 kg. If the
angle between the incline and the horizontal is 25, what are
the magnitudes of the weight force, normal force, and static
friction force acting on the box?
FBR
w
N
fs
x
y

Apply
Newton’s
2nd Law 0
sin
0
cos








s
x
y
f
w
F
w
N
F


The forces are:
  
 
  N
4
.
10
25
sin
N
5
.
24
sin
N
2
.
22
25
cos
N
5
.
24
cos
N
5
.
24
N/kg
8
.
9
kg
5
.
2













w
f
w
N
mg
w
s

34. ©McGraw-Hill Education
4.7 Tension
This is the force transmitted through a “rope” from one end to
the other.
An ideal cord has zero mass, does not stretch, and the tension is
the same throughout the cord.

35. ©McGraw-Hill Education
Tension Example
Example (text problem 4.88): A pulley is hung from the ceiling by a rope. A
block of mass M is suspended by another rope that passes over the pulley
and is attached to the wall. The rope fastened to the wall makes a right
angle with the wall. Neglect the masses of the rope and the pulley. Find the
tension in the rope from which the pulley hangs and the angle .
FDB for the
mass M
w
T
x
y
Apply Newton’s 2nd
Law to the mass M.
Mg
w
T
w
T
Fy





 0

36. ©McGraw-Hill Education
Tension Example Continued
FBD for the pulley:
x
y
T
F

T
Apply Newton’s 2nd Law:
0
sin
0
cos








T
F
F
T
F
F
y
x



 sin
cos F
F
T 

This statement can only be
true if  = 45 and
Mg
T
F 2
2 

Note: F is the tension in the rope that connects the pulley to the
ceiling. It also represents the force of the support on the rope.

37. ©McGraw-Hill Education
4.8 Applying Newton’s Laws
a
F m



38. ©McGraw-Hill Education
Applying Newton’s Laws Example 1
Example: Two blocks on a frictionless surface are connected by a
rope. An external force pulls one of the blocks horizontally. Find
the tension in the cord connecting the two blocks, assuming the
external force of magnitude 10.0 N is applied to the right on
block 1. The masses are m1 = 3.00 kg and m2 = 1.00 kg.
F
block 2 block 1
Assume that the rope stays taut so that both blocks have the
same acceleration.

39. ©McGraw-Hill Education
Applying Newton’s Laws Example 1 Continued 1
FBD for block 2 (on the left):
x
T
w2
N2
y
FBD for block 1 (on the right):
T F
w1
N1
x
y
Apply Newton’s 2nd Law to each block:
0
2
2
2







w
N
F
a
m
T
F
y
x
0
1
1
1








w
N
F
a
m
T
F
F
y
x

40. ©McGraw-Hill Education
Applying Newton’s Laws Example 1 Continued 2
1
2 )
(1)
(2
F T m a
T m a
 

These two equations contain the
unknowns: a and T.
To solve for T, a must be eliminated. Solve for a in equation (2)
and substitute into equation (1):
N
5
.
2
kg
1
kg
3
1
N
10
1
1
2
1
2
1
2
1
2
1
1




















































m
m
F
T
T
m
m
T
m
T
m
F
m
T
m
a
m
T
F

41. ©McGraw-Hill Education
Applying Newton’s Laws Example 2
Example: A box slides across a rough surface. If the coefficient
of kinetic friction is 0.3, what is the acceleration of the box?
FBD for
box:
fk
w
N
x
y
Apply Newton’s 2nd Law:
0
k








w
N
F
ma
f
F
y
x

42. ©McGraw-Hill Education
Applying Newton’s Laws Example 2 Continued
k
(1)
(2) 0
f ma
N w N w mg
 
   
From equation (1): ma
mg
N
f 



 k
k
k 

Solving for a:    2
2
k m/s
94
.
2
m/s
8
.
9
3
.
0 




 g
a 

43. ©McGraw-Hill Education
Applying Newton’s Laws Example 3
In the previous example, a box sliding across a rough surface
was found to have an acceleration of 2.94 m/s2. If the initial
speed of the box is 10.0 m/s, how long does it take for the box
to come to rest?
Know: a = 2.94 m/s2, vix = 10.0 m/s, vfx = 0.0 m/s
Want: t.
sec
40
.
3
m/s
94
2
m/s
0
.
10
0
2












.
a
v
t
t
a
v
v
x
ix
x
ix
x

44. ©McGraw-Hill Education
4.9 Reference Frames
For Newton’s Second Law to be valid it must be applied in an
inertial reference frame. An inertial reference frame is one
where Newton’s First law is valid.

45. ©McGraw-Hill Education
4.10 Apparent Weight
Imagine a person standing
on a bathroom scale.
FBD for the
person:
w
N
x
y
Apply Newton’s 2nd Law:
y
y
y
ma
mg
N
ma
w
N
F







46. ©McGraw-Hill Education
Apparent Weight Continued
The normal force is the force the scale exerts on you. By
Newton’s 3rd Law this is also the force (magnitude only) you
exert on the scale. A scale will read the normal force.
 
y
a
g
m
N 
 is what the scale reads.
When ay = 0, N = mg. The scale reads your true weight.
When ay  0, N > mg or N < mg.

47. ©McGraw-Hill Education
Apparent Weight Example
Example (text problem 4.171): A woman of mass 51 kg is
standing in an elevator. If the elevator pushes up on her feet
with 408 newtons of force, what is the acceleration of the
elevator?
FBD for
woman:
w
N
x
y
Apply Newton’s 2nd Law:
y
y
y
ma
mg
N
ma
w
N
F







48. ©McGraw-Hill Education
Apparent Weight Example Continued
Given: N = 408 newtons, m = 51 kg, g = 9.8 m/s2
Unknown: ay
Solving Newton’s
2nd Law for ay:
2
m/s
8
.
1




m
mg
N
ay
The elevator could be (1) traveling upward with
decreasing speed, or (2) traveling downward with
increasing speed.

49. ©McGraw-Hill Education
4.11 Air Resistance
A stone is dropped from the edge of a cliff; if air resistance
cannot be ignored, the FBD for the stone is:
x
y
w
Fd
Apply Newton’s Second Law
ma
w
F
F d
y 



Where Fd is the magnitude of the drag
force on the stone. This force is
directed opposite the object’s velocity

50. ©McGraw-Hill Education
Air Resistance Continued
Assume that 2
bv
Fd 
b is a parameter that depends on
the size and shape of the object.
Since Fd  v2, can the object be in equilibrium?
b
mg
v
v
mg
bv
ma
w
F
F
t
d
y








when
yes,
0
2

51. ©McGraw-Hill Education
Air Resistance Example
Example: A paratrooper with a fully loaded pack has a mass of
120 kg. The force due to air resistance has a magnitude of Fd =
bv2 where b = 0.14 N s2/m2.
(a) If the paratrooper falls with a speed of 64 m/s, what is the
force of air resistance on the paratrooper?
   N
570
m/s
64
/m
s
N
14
.
0
2
2
2
2


 bv
Fd

52. ©McGraw-Hill Education
Air Resistance Example Continued
(b) What is the paratrooper’s acceleration?
FBD:
x
y
w
Fd
Apply Newton’s Second Law
and solve for a.
2
m/s
1
.
5








m
mg
F
a
ma
w
F
F
d
d
y
(c) What is the paratrooper’s terminal speed?
m/s
92
0
0
2









b
mg
v
mg
bv
ma
w
F
F
t
t
d
y

53. ©McGraw-Hill Education
4.12 Fundamental Forces
The four fundamental forces of nature are:
Gravity which is the force between two masses; it is the weakest
of the four.
Strong Force which helps to bind atomic nuclei together; it is the
strongest of the four.
Weak Force plays a role in some nuclear reactions.
Electromagnetic is the force that acts between charged particles.

54. ©McGraw-Hill Education
Contact Forces Example 2 Appendix
The free-body diagram for the box on the ramp shows the
normal force perpendicular to the surface in the +y-direction.
The weight is downward, and the static friction force is up
(parallel to) the incline. The +x-direction is down the incline. The
weight makes an angle of  relative to the normal vector (surface
perpendicular) direction.