Chapter 4 Forces and Newtons Laws of Motion. 4.1 The Concepts of Force and Mass A force is a push or...
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Transcript of Chapter 4 Forces and Newtons Laws of Motion. 4.1 The Concepts of Force and Mass A force is a push or...
![Page 1: Chapter 4 Forces and Newtons Laws of Motion. 4.1 The Concepts of Force and Mass A force is a push or a pull. Arrows are used to represent forces. The.](https://reader035.fdocuments.in/reader035/viewer/2022081518/5514ef8b550346935c8b5c9d/html5/thumbnails/1.jpg)
Chapter 4
Forces and Newton’s Laws of Motion
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4.1 The Concepts of Force and Mass
A force is a push or a pull.Arrows are used to represent forces. The length of the arrowis proportional to the magnitude of the force.
15 N
5 N
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4.3 Newton’s Second Law of Motion
SI Unit for Force
This unit of force is called a newton (N).
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4.1 The Concepts of Force and Mass
Mass is a measure of the amount of “stuff” contained in an object.
SI Unit of Mass: kilogram (kg)
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4.2 Newton’s First Law of Motion
An object continues in a state of restor in a state of motion at a constant Speed unless changed by a net force
The net force is the SUM of allof the forces acting on an object.
Newton’s First Law (Law of Inertia)
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4.2 Newton’s First Law of Motion
The net force on an object is the sum of all forces acting on that object.
Individual Forces Net Force
10 N4 N 6 N
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4.2 Newton’s First Law of Motion
Individual Forces Net Force
3 N
4 N
5 N36
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4.3 Newton’s Second Law of Motion
F
Mathematically, the net force is written as
where the Greek letter sigma denotes the vector sum.
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4.3 Newton’s Second Law of Motion
Newton’s Second Law
When a net force acts on an objectof mass m, the acceleration that results is directly proportional to the net force and hasa magnitude that is inversely proportional tothe mass. The direction of the acceleration isthe same as the direction of the net force.
mF
a
aF
m
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4.4 The Vector Nature of Newton’s Second Law
xx maFyy maF
The direction of force and acceleration vectorscan be taken into account by using x and ycomponents.
aFm
is equivalent to
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4.4 The Vector Nature of Newton’s Second Law
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4.4 The Vector Nature of Newton’s Second Law
Force x component y component
+17 N
+(15 N) cos67
0 N
+(15 N) sin67
+23 N +14 N
The net force on the raft can be calculatedin the following way:
P
A
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4.4 The Vector Nature of Newton’s Second Law
2sm 018.0kg 1300
N 23
m
Fa
x
x
2sm 011.0kg 1300
N 14
m
Fa
y
y
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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.
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4.5 Newton’s Third Law of Motion
Suppose that the magnitude of the force is 36 N. If the massof the spacecraft is 11,000 kg and the mass of the astronautis 92 kg, what are the accelerations?
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4.5 Newton’s Third Law of Motion
.astronaut On the
. spacecraft On the
PF
PF
2sm0033.0kg 11,000
N 36
ss m
Pa
2sm39.0kg 92
N 36
AA m
Pa
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4.7 The Gravitational Force
Newton’s Law of Universal Gravitation
Every particle in the universe exerts an attractive force on everyother particle.
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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
221
r
mmGF
2211 kgmN10673.6 G
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4.7 The Gravitational Force
N 104.1
m 1.2
kg 25kg 12kgmN1067.6
8
22211
221
r
mmGF
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4.7 The Gravitational Force
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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.
SI Unit of Weight: newton (N)
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4.8 The Normal Force
Definition of the Normal Force
The normal force is the force that a surfaceexerts on an object with which it is in contact – It is ALWAYS perpendicular to the surface.
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4.8 The Normal Force
N 4
0N 15N 11
N
N
F
F
A hand pushes down on a block with a force of 11 N. The block weighs 15N. What is the magnitude of the normal force?
A rope is now used to lift the block with a force of 11 N. What is the magnitude of the normal force?
N 26
0N 15N 11
N
N
F
F
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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.
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4.8 The Normal Force
mamgFF Ny
mamgFN
apparent weight
trueweight
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4.9 Static and Kinetic Frictional Forces
When an object is in contact with a surface there is a forceacting on that object. The component of this force that is parallel to the surface is called the frictional force.
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4.9 Static and Kinetic Frictional Forces
When the two surfaces are not sliding across one anotherthe friction is called static friction.
Nss Ff
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4.9 Static and Kinetic Frictional Forces
Note that the magnitude of the frictional force doesnot depend on the contact area of the surfaces.
What does it depend on???
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4.9 Static and Kinetic Frictional Forces
Static friction opposes the impending relative motion betweentwo objects.
Kinetic friction opposes the relative sliding motion motions thatactually does occur.
Nkk Ff
10 k is called the coefficient of kinetic friction.
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4.9 Static and Kinetic Frictional Forces
Page 104
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4.9 Static and Kinetic Frictional Forces
The sled comes to a halt because the kinetic frictional forceopposes its motion and causes the sled to slow down.
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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?
N 20sm80.9kg4005.0 2
mgFf kNkk
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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.
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4.3 Free Body Diagrams
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.
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4.6 Types of Forces: An Overview
Examples of different forces:
Friction forceTension forceNormal force
Weight (Force due to gravity)Gravitational Force
Magnetic ForceElectric Force
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4.10 The Tension Force
Cables and ropes transmit forces through tension.
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This diagram shows four forces acting upon an object. There aren’t always four forces, For example, there could be one or more forces.
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Problem 1
A book is at rest on a table top. Diagram the forces acting on the book.
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Problem 1
In this diagram, there are normal and gravitational forces on the book.
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Problem 2
An egg is free-falling from a nest in a tree. Neglect air resistance. Draw a free-body diagram showing the forces involved.
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Gravity is the only force acting on the egg as it falls.
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Problem 3
A rightward force is applied to a book in order to move it across a desk. Consider frictional forces. Neglect air resistance. Construct a free-body diagram. Let’s see what this one looks like.
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Note the applied force arrow pointing to the right. Notice how friction force points in the opposite direction. Finally, there is still gravity and normal forces involved.
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Problem 4
• A block is at rest on an incline and stays in place because of static friction. Draw a free-body diagram
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• A skier is standing motionless on a horizontal tow rope, which is about to pull her forward. The skier’s mass is 59kg and the coefficient of static friction between the skis and the snow is 0.14. Draw a free body diagram. What is the magnitude of the maximum force that the tow rope can apply to the skier without causing her to move.
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Objects in Equilibrium• Objects that are either at rest or moving with
constant velocity are said to be in equilibrium• This means that the net force acting on the
object is zero• Equivalent to the set of component equations
given by 0 F
0 xx maF 0 yy maF
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4.11 Equilibrium Application of Newton’s Laws of Motion
Reasoning Strategy
• Draw a free-body diagram.
• Choose a set of x, y axes for each object and resolve all forcesin the free-body diagram into components that point along theseaxes.
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4.11 Equilibrium Application of Newton’s Laws of Motion
What is the Tension in Rope 1 and 2?
The engine has a weight of 3150 N
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4.11 Equilibrium Application of Newton’s Laws of Motion
N 3150W
Force x component y component
1T
2T
W
0.10sin1T
0.80sin2T
0
0.10cos1T
0.80cos2T
W
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4.11 Equilibrium Application of Newton’s Laws of Motion
00.80sin0.10sin 21 TTFx
00.80cos0.10cos 21 WTTFy
The first equation gives 21 0.10sin
0.80sinTT
Substitution into the second gives
00.80cos0.10cos0.10sin
0.80sin22
WTT
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4.11 Equilibrium Application of Newton’s Laws of Motion
0.80cos0.10cos0.10sin0.80sin
2
WT
N 5822 T N 1030.3 31 T
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4.12 Nonequilibrium Application of Newton’s Laws of Motion
xx maF
yy maF
When an object is accelerating, it is not in equilibrium.
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4.12 Nonequilibrium Application of Newton’s Laws of Motion
The acceleration is along the x axis so 0ya
The mass of the supertanker is 1.5 x 108 kg. There is a drive force with magnitude 75000 N and an opposing force of 40000 N. The tanker moves with an acceleration of 2.0 x 10-3 m/s2. Solve for the Tension in each of the ropes.
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4.12 Nonequilibrium Application of Newton’s Laws of Motion
Force x component y component
1T
2T
D
R
0.30cos1T
0.30cos2T
0
0
D
R
0.30sin1T
0.30sin2T
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4.12 Nonequilibrium Application of Newton’s Laws of Motion
00.30sin0.30sin 21 TTFy
21 TT
x
x
ma
RDTTF
0.30cos0.30cos 21
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4.12 Nonequilibrium Application of Newton’s Laws of Motion
TTT 21
N 1053.10.30cos2
5
DRmaT x