chapter 12 -...
TRANSCRIPT
Chapter 12 Kinetics of Particles:
Newton’s Second Law
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Dr. Mohammad Suliman Abuhaiba, PE
Introduction
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• Newton’s first and third laws are sufficient for the
study of bodies at rest (statics) or bodies in
motion with no acceleration.
• When a body accelerates, Newton’s 2nd law is
required to relate the motion of the body to the
forces acting on it.
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Introduction
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Newton’s second law:
• A particle will have an acceleration proportional to
magnitude of resultant force acting on it and in the
direction of the resultant force.
• The resultant of the forces acting on a particle is
equal to the rate of change of linear momentum of
the particle.
• The sum of the moments about O of the forces
acting on a particle is equal to the rate of change of
angular momentum of the particle about O.
3
Newton’s 2nd of Motion
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• Newton’s 2nd: If the resultant force
acting on a particle is not zero, the
particle will have an acceleration
proportional to the magnitude of
resultant and in the direction of the
resultant.
• Consider a particle subjected to
constant forces,
ma
F
a
F
a
F mass,constant
3
3
2
2
1
1
4
• If force acting on particle is zero, particle will not
accelerate, i.e., it will remain stationary or continue
on a straight line at constant velocity.
Newton’s 2nd of Motion
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• When a particle of mass m is acted upon by a
force the acceleration of the particle must
satisfy
,F
amF
• Acceleration must be evaluated wrt a
Newtonian frame of reference (one
that is not accelerating or rotating).
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Linear Momentum of a Particle • Replacing the acceleration by the
derivative of the velocity yields
dt
Ldvm
dt
d
dt
vdmF
• Linear Momentum Conservation Principle:
If the resultant force on a particle is zero, the
linear momentum of the particle remains constant
in both magnitude and direction.
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Systems of Units
• International System of Units (SI
Units): length (m), mass (kg), and
time (second). The unit of force is
derived,
22 s
mkg1
s
m1kg1N1
ft
slb1
sft1
lb1slug1
sft32.2
lb1lbm1
2
22
• U.S. Customary Units: force (lb),
length (ft), and time (second). The
unit of mass is derived,
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Equations of Motion • Newton’s second law provides
amF
zmFymFxmF
maFmaFmaF
kajaiamkFjFiF
zyx
zzyyxx
zyxzyx
• For tangential & normal
components,
2vmF
dt
dvmF
maFmaF
nt
nntt
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Dynamic
Equilibrium
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• Alternate expression of Newton’s
second law,
ectorinertial vam
amF
0
• Particle is in dynamic equilibrium.
• Methods developed for particles in
static equilibrium may be applied
• Inertial forces = measure the
resistance that particles offer to
changes in motion, i.e., changes in
speed or direction.
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Sample Problem 12.1
A 200-lb block rests on a
horizontal plane. Find the
magnitude of the force P
required to give the block
an acceleration of 10 ft/s2
to the right. The
coefficient of kinetic
friction between the block
and plane is mk 0.25.
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Sample
Problem 12.1
N
NF
g
Wm
k
25.0
ft
slb21.6
sft2.32
lb200
2
2
m
x
y
O
SOLUTION:
• Resolve the equation of motion for
the block into two rectangular
component equations.
:maFx
lb1.62
sft10ftslb21.625.030cos 22
NP
0yF 0lb20030sin PN
• The two equations may be solved
for the applied force P and the
normal reaction N.
lb1.62lb20030sin25.030cos
lb20030sin
PP
PN
lb151P
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Sample Problem 12.3
The two blocks shown start
from rest. The horizontal
plane and the pulley are
frictionless, and the pulley
is assumed to be of
negligible mass.
Determine the acceleration
of each block and the
tension in the cord.
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Solution:
• Write the kinematic relationships
for the dependent motions and
accelerations of the blocks.
ABAB aaxy21
21
x
y
Sample Problem 12.3
• Write equations of motion for
blocks and pulley.
:AAx amF AaT kg1001
O
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Sample Problem 12.3
:BBy amF
B
B
BBB
aT
aT
amTgm
kg300-N2940
kg300sm81.9kg300
2
22
2
:0 CCy amF
02 12 TT
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Sample Problem 12.3
N16802
N840kg100
sm20.4
sm40.8
12
1
221
2
TT
aT
aa
a
A
AB
A
• Combine kinematic relationships with
equations of motion to solve for
accelerations and cord tension.
ABAB aaxy21
21 AaT kg1001
A
B
a
aT
21
2
kg300-N2940
kg300-N2940
0kg1002kg150N2940
02 12
AA aa
TT
x
y
O
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Sample Problem 12.4
The 12-lb block B starts from
rest and slides on the 30-lb
wedge A, which is supported
by a horizontal surface.
Neglecting friction, determine:
a. Acceleration of the wedge
b. Acceleration of the block
relative to the wedge.
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Sample Problem 12.4
Solution:
• The block is constrained to slide down the
wedge. Therefore, their motions are
dependent. ABAB aaa
:30cos ABABxBx aamamF
30sin30cos
30cos30sin
gaa
aagWW
AAB
ABABB
:30sin AByBy amamF
30sin30cos1 ABB agWWN
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• Write equations of motion for wedge and
block.
x
y
:AAx amF
AA
AA
agWN
amN
1
1
5.0
30sin
17
Sample
Problem 12.4 AA agWN 15.0
• Solve for the accelerations.
30sinlb12lb302
30coslb12sft2.32
30sin2
30cos
30sin30cos2
30sin30cos
2
1
A
BA
BA
ABBAA
ABB
a
WW
gWa
agWWagW
agWWN
2sft07.5Aa
30sinsft2.3230cossft07.5
30sin30cos
22AB
AAB
a
gaa
2sft5.20ABa
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Sample Problem 12.5
The bob of a 2-m
pendulum describes an arc
of a circle in a vertical
plane. If the tension in the
cord is 2.5 times the
weight of the bob for the
position shown, find the
velocity and acceleration
of the bob in that position.
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Sample Problem 12.5
Solution:
• Resolve the equation of motion
for the bob into tangential
&normal components.
• Solve the component equations
for the normal and tangential
accelerations.
:tt maF
30sin
30sin
ga
mamg
t
t
2sm9.4ta
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Sample Problem 12.5
:nn maF
30cos5.2
30cos5.2
ga
mamgmg
n
n
2sm03.16na
• Solve for velocity in terms of
normal acceleration.
22
sm03.16m2 nn avv
a
sm66.5v
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Sample Problem 12.6 Determine the rated speed
of a highway curve of
radius = 400 ft banked
through an angle q = 18o.
The rated speed of a
banked highway curve is
the speed at which a car
should travel if no lateral
friction force is to be
exerted at its wheels.
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Sample Problem
12.6
Solution:
• The car travels in a horizontal
circular path with a normal
component of acceleration
directed toward the center of the
path.
• Resolve the equation of
motion for the car into
vertical and normal
components.
:0 yF
q
q
cos
0cos
WR
WR
:nn maF
q
q
q
2
sincos
sin
v
g
WW
ag
WR n
• Solve for vehicle speed.
18tanft400sft2.32
tan
2
2 qgv
hmi1.44sft7.64 v
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Howe Work Assignment #12-1
3, 10, 17, 23, 30, 37, 44, 51
Due Monday 24/2/2014
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Angular Momentum of a Particle
• moment of momentum or
the angular momentum of the particle
about O.
VmrHO
zyx
O
mvmvmv
zyx
kji
H
• is perpendicular to plane containing OH
Vmr
and
q
q
2
sin
mr
vrm
rmVHO
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• Derivative of angular momentum wrt
time,
O
O
M
Fr
amrVmVVmrVmrH
• Newton’s 2nd law: sum of the moments
about O of the forces acting on the
particle is equal to rate of change of the
angular momentum of the particle about
O.
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Angular Momentum of a Particle
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Equations of Motion in Radial &
Transverse Components
qqq qq rrmmaFrrmmaF rr 2 ,2
Consider particle at r and q, in polar coordinates,
qqq
q
q
q
rrmF
rrrmmrdt
dFr
mrHO
2
222
2
This result may also be derived from
conservation of angular momentum,
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Conservation of Angular Momentum
• When only force acting on
particle is directed toward or
away from a fixed point O, the
particle is said to be moving
under a central force.
• Since the line of action of the
central force passes through O,
and 0 OO HM
constant OHVmr
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Conservation of Angular Momentum
• Position vector and motion of
particle are in a plane
perpendicular to .OH
• Magnitude of angular
momentum,
000 sin
constantsin
Vmr
VrmHO
massunit
momentumangular
constant
2
2
hrm
H
mrH
O
O
q
q
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Conservation of Angular Momentum
• Radius vector OP sweeps
infinitesimal area qdrdA 2
21
• Define qq 2
212
21 r
dt
dr
dt
dA areal
velocity
• Recall, for a body moving under a
central force, constant2 qrh
• When a particle moves under a central
force, its areal velocity is constant.
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Newton’s Law of Gravitation • Gravitational force exerted by the sun on a planet or by the
earth on a satellite is an important example of gravitational
force.
• Newton’s law of universal gravitation - two particles of
mass M & m attract each other with equal and opposite force
directed along the line connecting the particles,
4
49
2
312
2
slb
ft104.34
skg
m1073.66
ngravitatio ofconstant
G
r
MmGF
• For particle of mass m on the earth’s surface,
222 s
ft2.32
s
m81.9 gmg
R
MGmW
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Sample Problem 12.7 A block B of mass m can slide
freely on a frictionless arm OA
which rotates in a horizontal
plane at a constant rate
Knowing that B is released at a
distance r0 from O, express as a
function of r
a. the component vr of the
velocity of B along OA
b. the magnitude of the
horizontal force exerted on B
by the arm OA.
.0q
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Sample Problem
12.7
Solution:
• Write the radial and
transverse equations of
motion for the block.
:
:
qq amF
amF rr
q
rrmF
rrm
2
0 2
• Integrate the radial equation to
find an expression for the radial
velocity.
r
r
v
rr
rr
rr
rrr
drrdvv
drrdrrdvv
dr
dvv
dt
dr
dr
dv
dt
dvvr
r
0
2
0
0
2
0
2
q
20
220
2 rrvr q
• Substitute known information
into the transverse equation to
find an expression for the force
on the block. 2120
2202 rrmF q
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Sample Problem 12.8 A satellite is launched in a
direction parallel to the
surface of the earth with a
velocity of 18820 mi/h
from an altitude of 240 mi.
Determine the velocity of
the satellite as it reaches
its maximum altitude of
2340 mi. The radius of
the earth is 3960 mi.
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Sample Problem 12.8 Solution:
• Since the satellite is moving under a central force, its angular
momentum is constant. Equate the angular momentum at A
and B and solve for the velocity at B.
mi23403960
mi2403960hmi18820
constantsin
B
AAB
BBAA
O
r
rvv
vmrvmr
Hvrm
hmi12550Bv
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Home Work Assignment #12-2
66, 74, 81
Due Saturday 1/3/2014
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