Download 4.11 Equilibrium Application of Newton`s Laws of Motion

Chapter 4
Forces and Newton’s
Laws of Motion
4.1 The Concepts of Force and Mass
A force is a push or a pull.
Contact forces arise from physical
contact .
Action-at-a-distance forces do not
require contact and include gravity
and electrical forces.
4.1 The Concepts of Force and Mass
Arrows are used to represent forces. The length of the arrow
is proportional to the magnitude of the force.
15 N
5N
4.1 The Concepts of Force and Mass
Mass is a measure of the amount
of “stuff” contained in an object.
4.2 Newton’s First Law of Motion
Newton’s First Law
An object continues in a state of rest
or in a state of motion at a constant
speed along a straight line, unless
compelled to change that state by a
net force.
The net force is the vector sum of all
of the forces acting on an object.
4.2 Newton’s First Law of Motion
The net force on an object is the vector sum of
all forces acting on that object.
The SI unit of force is the Newton (N).
Individual Forces
4N
10 N
Net Force
6N
4.2 Newton’s First Law of Motion
Individual Forces
Net Force
5N
64
3N
4N
4.2 Newton’s First Law of Motion
Inertia is the natural tendency of an
object to remain at rest in motion at
a constant speed along a straight line.
The mass of an object is a quantitative
measure of inertia.
SI Unit of Mass: kilogram (kg)
4.2 Newton’s First Law of Motion
An inertial reference frame is one in
which Newton’s law of inertia is valid.
All accelerating reference frames are
noninertial.
4.3 Newton’s Second Law of Motion
Mathematically, the net force is
written as

F

where the Greek letter sigma
denotes the vector sum.
4.3 Newton’s Second Law of Motion
Newton’s Second Law
When a net external force acts on an object
of mass m, the acceleration that results is
directly proportional to the net force and has
a magnitude that is inversely proportional to
the mass. The direction of the acceleration is
the same as the direction of the net force.

a


F
m



F  ma
4.3 Newton’s Second Law of Motion
SI Unit for Force
 m  kg  m
kg   2   2
s
s 
This combination of units is called a newton (N).
4.3 Newton’s Second Law of Motion
4.3 Newton’s Second Law of Motion
A free-body-diagram is a diagram that
represents the object and the forces that
act on it.
4.3 Newton’s Second Law of Motion
The net force in this case is:
275 N + 395 N – 560 N = +110 N
and is directed along the + x axis of the coordinate system.
4.3 Newton’s Second Law of Motion
If the mass of the car is 1850 kg then, by
Newton’s second law, the acceleration is
F  110 N

a

 0.059 m s
m
1850 kg
2
4.4 The Vector Nature of Newton’s Second Law
The direction of force and acceleration vectors
can be taken into account by using x and y
components.



F  ma
is equivalent to
F
y
 ma y
F
x
 max
4.4 The Vector Nature of Newton’s Second Law
4.4 The Vector Nature of Newton’s Second Law
The net force on the raft can be calculated
in the following way:
Force

P

A
x component
y component
+17 N
0N
+(15 N) cos67
+(15 N) sin67
+23 N
+14 N
4.4 The Vector Nature of Newton’s Second Law
ax
F


ay
F


x
m
y
m
 23 N
2

 0.018 m s
1300 kg
 14 N
2

 0.011 m s
1300 kg
4.5 Newton’s Third Law of Motion
Newton’s Third Law of Motion
Whenever one body exerts a force on a
second body, the second body exerts an
oppositely directed force of equal
magnitude on the first body.
4.5 Newton’s Third Law of Motion
Suppose that the magnitude of the force is 36 N. If the mass
of the spacecraft is 11,000 kg and the mass of the astronaut
is 92 kg, what are the accelerations?
4.5 Newton’s Third Law of Motion
 
F  P.


F  P.

On the astronaut 
On the spacecraft


P
 36 N
2
as 

 0.0033 m s
ms 11,000 kg


 P  36 N
2
aA 

 0.39 m s
mA
92 kg
4.6 Types of Forces: An Overview
In nature there are two general types of forces,
fundamental and nonfundamental.
Fundamental Forces
1. Gravitational force
2. Strong Nuclear force
3. Electroweak force
4.6 Types of Forces: An Overview
Examples of nonfundamental forces:
friction
tension in a rope
normal or support forces
4.7 The Gravitational Force
Newton’s Law of Universal Gravitation
Every particle in the universe exerts an attractive force on every
other particle.
A particle is a piece of matter, small enough in size to be
regarded as a mathematical point.
The force that each exerts on the other is directed along the line
joining the particles.
4.7 The Gravitational Force
For two particles that have masses m1 and m2 and are
separated by a distance r, the force has a magnitude
given by
m1m2
F G 2
r
G  6.673 1011 N  m 2 kg 2
4.7 The Gravitational Force
m1m2
F G 2
r

 6.67 10
8
11
 1.4 10 N
N  m kg
2
2

12 kg 25 kg 

2
1.2 m
4.7 The Gravitational Force
4.7 The Gravitational Force
Definition of Weight
The weight of an object on or above the earth is the
gravitational force that the earth exerts on the object.
The weight always acts downwards, toward the center
of the earth.
On or above another astronomical body, the weight is the
gravitational force exerted on the object by that body.
SI Unit of Weight: newton (N)
4.7 The Gravitational Force
Relation Between Mass and Weight
W G
M Em
r
W  mg
ME
g G 2
r
2
4.7 The Gravitational Force
On the earth’s surface:
ME
g G 2
RE

 6.67 10
 9.80 m s
11
2
N  m kg
2
 5.98 10 kg 
6.38 10 m
24
2
6
2
4.8 The Normal Force
Definition of the Normal Force
The normal force is one component of the force that a surface
exerts on an object with which it is in contact – namely, the
component that is perpendicular
to the surface.
4.8 The Normal Force
FN  11 N  15 N  0
FN  26 N
FN  11 N  15 N  0
FN  4 N
4.8 The Normal Force
Apparent Weight
The apparent weight of an object is the reading of the scale.
It is equal to the normal force the man exerts on the scale.
4.8 The Normal Force
F
y
  FN  mg  ma
FN  mg  ma
apparent
weight
true
weight
4.9 Static and Kinetic Frictional Forces
When an object is in contact with a surface there is a force
acting on that object. The component of this force that is
parallel to the surface is called the
frictional force.
4.9 Static and Kinetic Frictional Forces
When the two surfaces are
not sliding across one another
the friction is called
static friction.
4.9 Static and Kinetic Frictional Forces
The magnitude of the static frictional force can have any value
from zero up to a maximum value.
fs  f
f
MAX
s
0  s  1
MAX
s
  s FN
is called the coefficient of static friction.
4.9 Static and Kinetic Frictional Forces
Note that the magnitude of the frictional force does
not depend on the contact area of the surfaces.
4.9 Static and Kinetic Frictional Forces
Static friction opposes the impending relative motion between
two objects.
Kinetic friction opposes the relative sliding motion motions that
actually does occur.
f k   k FN
0  k  1
is called the coefficient of kinetic friction.
4.9 Static and Kinetic Frictional Forces
4.9 Static and Kinetic Frictional Forces
The sled comes to a halt because the kinetic frictional force
opposes its motion and causes the sled to slow down.
4.9 Static and Kinetic Frictional Forces
Suppose the coefficient of kinetic friction is 0.05 and the total
mass is 40kg. What is the kinetic frictional force?
f k  k FN  k mg 


0.0540kg  9.80 m s  20kg
2
4.10 The Tension Force
Cables and ropes transmit
forces through tension.
4.10 The Tension Force
A massless rope will transmit
tension undiminished from one
end to the other.
If the rope passes around a
massless, frictionless pulley, the
tension will be transmitted to
the other end of the rope
undiminished.
4.11 Equilibrium Application of Newton’s Laws of Motion
Definition of Equilibrium
An object is in equilibrium when it has zero acceleration.

Fx  0

Fy  0
4.11 Equilibrium Application of Newton’s Laws of Motion
Reasoning Strategy
• Select an object(s) to which the equations of equilibrium are
to be applied.
• Draw a free-body diagram for each object chosen above.
Include only forces acting on the object, not forces the object
exerts on its environment.
• Choose a set of x, y axes for each object and resolve all forces
in the free-body diagram into components that point along these
axes.
• Apply the equations and solve for the unknown quantities.
4.11 Equilibrium Application of Newton’s Laws of Motion
Figure 4.28a shows a traction device used with a
foot injury. The weight of the 2.2 kg object creates
a tension in the rope that passes around the
pulleys. Therefore, tension forces T1 and T2 are
applied to the pulley on the foot. (It may seem
surprising that the rope applies a force to either
side of the foot pulley. Ignoring the weight of the
foot, find the magnitude of F?
4.11 Equilibrium Application of Newton’s Laws of Motion
 T1 sin 35  T2 sin 35  0


 T1 cos 35  T2 cos 35  F  0


4.11 Equilibrium Application of Newton’s Laws of Motion
An automobile engine has a weight W, whose
magnitude is W = 3150 N. This engine is being
positioned above an engine compartment, as Figure
4.29a illustrates. To position the engine, a worker is
using a rope. Find the tension T1 in the supporting
cable and the tension T2 in the positioning rope.
4.11 Equilibrium Application of Newton’s Laws of Motion
4.11 Equilibrium Application of Newton’s Laws of Motion
Force

T1

T2

W
x component
y component
 T1 sin 10.0
 T1 cos10.0
 T2 sin 80.0
 T2 cos 80.0
0


W
W  3150 N
4.11 Equilibrium Application of Newton’s Laws of Motion

F
Fx   T1 sin 10.0  T2 sin 80.0  0
y
  T1 cos 10.0  T2 cos 80.0  W  0
The first equation gives


 sin 80.0 
T
T1  
  2
 sin 10.0 
Substitution into the second gives
 sin 80.0 




T cos10.0  T2 cos 80.0  W  0
  2
 sin 10.0 
4.11 Equilibrium Application of Newton’s Laws of Motion
T2 
W
 sin 80.0 



 cos10.0  cos 80.0
 
 sin 10.0 
T2  582 N
T1  3.30 10 N
3
4.12 Nonequilibrium Application of Newton’s Laws of Motion
When an object is accelerating, it is not in equilibrium.

Fx  max

Fy  may
4.12 Non-equilibrium Application of Newton’s Laws of Motion
• A supertanker of mass m = 1.50 × 108 kg is being towed by two
tugboats, as in Figure 4.31a. The tensions in the towing cables
apply the forces T1 and T2 at equal angles of 30.0° with respect to
the tanker's axis. In addition, the tanker's engines produce a
forward drive force D, whose magnitude is D = 75.0 × 103 N.
Moreover, the water applies an opposing force R, whose
magnitude is R = 40.0 × 103 N. The tanker moves forward with an
acceleration that points along the tanker's axis and has a
magnitude of 2.00 × 10-3 m/s2. Find the magnitudes of the
tensions T1 and T2 .
4.12 Nonequilibrium Application of Newton’s Laws of Motion
The acceleration is along the x axis so
ay  0
4.12 Nonequilibrium Application of Newton’s Laws of Motion
Force

T1

T2

D

R
x component
y component
 T1 cos 30.0
 T2 cos 30.0

 T1 sin 30.0

 T2 sin 30.0
D
0
R
0


4.12 Nonequilibrium Application of Newton’s Laws of Motion
F
y
  T1 sin 30.0  T2 sin 30.0  0

 T1  T2
F
x
 max
  T1 cos 30.0  T2 cos 30.0  D  R

4.12 Nonequilibrium Application of Newton’s Laws of Motion
T1  T2  T
max  R  D
5
T
 1.53 10 N

2 cos 30.0