Transcript Slide 1
Lecture 7
Ch5 . Newton’s Law of Motion
University Physics: Mechanics Dr.-Ing. Erwin Sitompul
http://zitompul.wordpress.com
2012
What Causes an Acceleration?
Out of common experience, we know that any change in velocity must be due to
an interaction
between
an object
(
a body
) and something in
its surroundings
.
An interaction that can cause an acceleration of a body is called a
force
on the body.
. Force can be loosely defined as a push or pull The relation between a force and the acceleration it causes was first understood by Isaac Newton. The study of that relationship is called
Newtonian mechanics
.
We shall now focus on its
three
primary laws of motion.
Erwin Sitompul Sir Isaac Newton (1642 —1727) University Physics: Mechanics 7/2
Force in Various Forms
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Newton’s First Law
Newton’s First Law
: “If no force acts on a body, then the body’s velocity cannot change, that is the body cannot accelerate.” In other words, if the body is at rest, it stays at rest. If the body is moving, it will continue to move with the same velocity (same magnitude and same direction).
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Mass, Motion & Force
Experimental Observations
: An object acted on by a net force accelerates in the same direction as the net force.
Some objects accelerate slower or faster than others when subjected to the same force
inertial mass
.
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Force
A force can cause the acceleration of a body.
As the standard body, we shall use the standard kilogram. It is assigned, exactly and by definition, a mass of
1 kg
.
We put the standard body on a horizontal frictionless surface and pull the body to the right, so that it eventually experiences an acceleration of
1 m/s 2
.
We can now declare, as a matter of definition, that the force we are exerting on the standard body has a magnitude of
1 newton
(
1 N
).
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Force
Forces are vector quantities. They have magnitudes and directions.
Principle of Superposition for Forces
A single force with the magnitude and direction of the net force acting on a body has the same effect as all the individual forces acting together.
:
Newton’s First Law
“If no
net
: (proper statement) force acts on a body (
F
net = 0) , then the body’s velocity cannot change, that is the body cannot accelerate.”
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Net Force Calculation using Vector Sum
F
net
F A
F B m a
F C
0
Erwin Sitompul
F F
F
F
F
0 0
F C
F
ˆ i (
F
F
) i ˆ
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Checkpoint
Which of the following six arrangements correctly show the vector addition of forces →
F
1 and →
F
2 to yield the third vector, which is meant to represent their net force →
F
net ?
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Mass
Mass is a scalar quantity.
Mass is an intrinsic characteristic of a body.
The mass of a body is the characteristic that
relates
on the body to the resulting acceleration.
a force A
physical sensation
of a mass can only be obtained when we attempt to accelerate the body.
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Newton’s Second Law
Newton’s Second Law
: “The net force on a body is equal to the product of the body’s mass and its acceleration.
” The Newton’s second law in equation form
F
net
ma
It the net force acceleration →
a
→
F
net on a body is zero, then the body’s is zero If the body is at rest, it stays at rest.
If it is moving, it continues to move at constant velocity.
1 N 2 (1 kg)(1 m s ) 2
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Newton’s Second Law
The vector equation
F
net =
ma
is equivalent to three component equation, one written for each axis of an
xyz
coordinate system:
F
net,
x
ma x
,
F
net,
y
ma y
,
F
net,
z
ma z
.
The acceleration component along a given axis is caused
only
by the sum of the force components along that
same axis
, and not by force components along any other axis.
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3 Astronauts and an Asteroid
Three astronauts, propelled by jet backpacks, push and guide a 120 kg asteroid toward a processing dock, exerting the forces shown in the figure, with
F
1 = 32 N,
F
2 = 55 N, and
F
3 = 41 N. What is the (a) magnitude and (b) angle of the asteroid's acceleration?
F
1
F F
2 3
(32 N)(cos30 i ˆ (55 N) i ˆ
ˆ
F net
m a F net m
a
(0.860) 2 0.88 m s 2
Erwin Sitompul
2 ˆ 0.860 i ˆ 120 tan 1 0.16
0.86
10.73
2
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Free-Body Diagram
The most important step in solving problems involving Newton’s Laws is to draw the free-body diagram.
Only the forces acting on the object of interest should be included in a free-body diagram.
F
on book from hand
F
on book from earth Erwin Sitompul University Physics: Mechanics 7/14
Free-Body Diagram
The system of interest is the cart Erwin Sitompul The free-body diagram of the cart University Physics: Mechanics 7/15
Puck (Ice Hockey “Ball”)
Three situations in which one or two forces act on a puck that moves over frictionless ice along an
x
axis, in one-dimensional motion, are presented here.
The puck’s mass is → Force
F
3
F
3 = 1 N.
m
→ = 0.2 kg. Forces
F
1 along the axis and have magnitudes is directed at angle
θ F
1 → and
F
2 are directed = 4 N and
F
2 = 2 N. = 30 ° and has magnitude In each situation, what is the acceleration of the puck?
F
1
ma x a x
F
1
m
4 N 0.2 kg
20 m s
2
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Puck (Ice Hockey “Ball”)
The puck’s mass is
m
= 0.2 kg. Forces
F
1 along the axis and have magnitudes
F
1 Force
F
3
F
3 = 1 N.
is directed at angle
θ
and
F
2 are directed = 4 N and
F
2 = 2 N. = 30 ° and has magnitude
F
1
F
2
ma x a x
0.2 kg
10 m s
2
F
3,
x a x
F
2
F
3,
x
ma x
F
2
F
3
m
cos
m
F
2
0.2 kg
5.67 m s
2
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Some Particular Forces
The gravitational force toward the ground.
F g
on a body is a force that pulls on the body, directly toward the center of Earth (that is, directly down
F F g g
mg mg
The Weight
The weight body is equal to the magnitude on the body
W F
g of a of the gravitational force
W
mg
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Some Particular Forces
The Normal Force
When a body presses against a surface, the surface (even a seemingly rigid one) deforms and pushes back on the body with a normal force →
F
N In mathematics,
normal
that is perpendicular to the surface.
means
perpendicular
.
F
N
F
N According to Newton’s second law,
F
N
F
N
F
N
F
g
mg
mg
ma y m
(0)
•
Why?
Forces on a Body, Resting on a Table Erwin Sitompul University Physics: Mechanics 7/19
Some Particular Forces
Friction
→ The frictional force or simply friction is a force surface.
f
that resists the motion when we slide or attempt to slide a body over a Friction is directed along the surface, opposite the direction of the intended motion.
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Some Particular Forces
Tension
When a cord (or a rope, cable, or other such object) is attached to a body and tensed, the cord pulls on the body with a force
T
directed away from the body.
The force is often called a tension force. The tension in the cord is the magnitude →
T
of the force on the body.
• •
A cord is considered as massless and unstretchable A pulley is considered as massless and frictionless Erwin Sitompul University Physics: Mechanics 7/21
Beware of High-Energy Tension
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Various Forms of Energy
• •
Energy can convert from one form to the other.
Energy in higher level (more concentrated) tends to convert to lower (less concentrated) condition.
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Newton’s Third Law
Newton’s Third Law
: “When two bodies interact, the forces on the bodies from each other are always equal in magnitude an opposite in direction.”
F BC
F CB
(Equal Magnitudes)
F
F BC CB
(Equal Magnitudes and Opposite Directions)
F BC
: The force on the book
B
from the box
C F CB
: The force on the box
C
from the book
B
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Applying Newton’s Law: Problem 1
A block
S
(the
sliding block
) with mass
M
=3.3 kg is free to move along a horizontal frictionless surface. It is connected by a cord that wraps over a frictionless pulley, to a second block
H
(the
hanging block
) with mass
m
are considered to be “massless”. = 2.1 kg. The cord and pulley The hanging block
H
falls as the sliding block
S
the right. Find: (a) the acceleration of block
S
(b) the acceleration of block
H
(c) the tension in the cord accelerate to
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F
N
Applying Newton’s Law: Problem 1
F
N
The Forces Acting On The Two Blocks Erwin Sitompul Free-Body Diagram for Block S and Block H University Physics: Mechanics 7/26
F
N
Erwin Sitompul
Applying Newton’s Law: Problem 1
F
net,
x T
Ma Ma x x F
N
F
net,
y F
g
S
Ma y
0
2
F
N
F
g
S T
F
net,
y F
g
H
ma y
ma y
1
The cord does not stretch, so
a x
of
M
and
a y
of
m
have the same magnitude.
a x
of
M
a y
of
m
a
1
The tension at
M
and the tension at
m
also have the same magnitude.
T T
F
g
H
mg
ma
ma
•
Why?
2
Ma
mg
ma a
M m
m g T
M Mm
m g
University Physics: Mechanics 7/27
Applying Newton’s Law: Problem 1
a
M m
m g
2.1
9.8
3.81m s 2 (a) the acceleration of block
S a S
2 (b) the acceleration of block
H a H
2 (c) the tension in the cord
T
Ma
(3.3)(3.81) 12.573 N
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Applying Newton’s Law: Problem 2
A cord pulls on a box of sea biscuits up along a frictionless plane inclined at
θ
= 30 °. The box has mass
m
= 5 kg, and the force from the cord has magnitude
T
= 25 N.
What is the box’s acceleration component a along the inclined plane?
Free-Body Diagram of the Box Erwin Sitompul Free-Body Diagram,
F
g in components University Physics: Mechanics 7/29
Applying Newton’s Law: Problem 2
What is the force exerted by the plane on the box?
T
mg F
net,
x
sin
a x
ma x
ma x T
mg
sin
m
25 (5)(9.8) sin 30 0.1 m s 2 5 •
What is the meaning of this value?
F
N
mg F
net,
y
cos
F
N
mg
cos
F
N
ma y
ma y
m
(0) •
No motion in y direction
mg
cos (5)(9.8) cos 30 42.435 N
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Homework 6: Two Boxes and A Pulley
A block of mass angle
θ m
1 = 3.7 kg on a frictionless plane inclined at = 30 ° is connected by a cord over a massless, frictionless pulley to a second block of mass
m
2 = 2.3 kg. What are: (a) the magnitude of the acceleration of each block, (b) the direction of the acceleration of the hanging block, and (c) the tension in the cord?
Hint
: Draw the free-body diagram of
m
1 and
m
2 first.
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Homework 6A
Two blocks of mass 3.5 kg and 8.0 kg are connected by a massless string that passes over a frictionless pulley (see figure below). The inclines are frictionless.
When the assembly is released from rest, find (a) the magnitude of the acceleration of each block; and (b) the tension in the string.
(c) the increase of elevation of the climbing block (which one?) after 0.4 s.
Hint
: Draw the free-body diagram of the two blocks.
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Homework 6B
1. A 2.75-kg cat moves in a straight line (the
x
-axis). The figure below shows a graph of the
x
component of this cat’s velocity as a function of time. (a) Find the maximum net force on this cat. When does this force occur?; (b) When is the net force on the cat equal to zero?; (c) What is the net force at time 8.5 s?
2. Three rainbow boxes are attached by cords, one of which wraps over a frictionless pulley with negligible mass. The three masses are
m A
= 30 kg,
m B
= 40 kg, and
m C
= 10 kg. When the assembly is released from rest, determine (a) the tension in the cord connecting the pulley)?
B
and
C
, and (b) how far
A
moves in the first 0.25 s (assuming it does not reach
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