Chapter 12 Kinetics of Particles: Newton’s Second Law 2/25/2014 6:39 PM Dr. Mohammad Suliman Abuhaiba, PE
Chapter 12 Kinetics of Particles: Newton’s Second Law
2/25/2014 6:39 PM
Dr. Mohammad Suliman Abuhaiba, PE
Introduction
2/25/2014 6:39 PM
Dr. Mohammad Suliman Abuhaiba, PE
• 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.
2
Introduction
2/25/2014 6:39 PM
Dr. Mohammad Suliman Abuhaiba, PE
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.
• Resultant of forces acting on a particle = rate of
change of linear momentum of the particle.
• Sum of moments about O of forces acting on a
particle = rate of change of angular momentum of
the particle about O.
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Newton’s 2nd of Motion
2/25/2014 6:39 PM
Dr. Mohammad Suliman Abuhaiba, PE
• Newton’s 2nd: If resultant force acting
on a particle is not zero, particle will
have an acceleration proportional to
magnitude of resultant & in the
direction of 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
2/25/2014 6:39 PM
Dr. Mohammad Suliman Abuhaiba, PE
• When a particle of mass m is acted upon by a
force 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).
5
Linear Momentum of a Particle
dt
Ldvm
dt
d
dt
vdmF
• Linear Momentum Conservation Principle:
If resultant force on a particle is zero, the linear
momentum of particle remains constant in both
magnitude & direction.
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Dr. Mohammad Suliman Abuhaiba, PE
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Systems of Units
• International System of Units (SI
Units): length (m), mass (kg), and
time (second)
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)
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Dr. Mohammad Suliman Abuhaiba, PE
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Equations of Motion • Newton’s 2nd law: amF
zmFymFxmF
maFmaFmaF
kajaiamkFjFiF
zyx
zzyyxx
zyxzyx
• For tangential & normal
components,
2vmF
dt
dvmF
maFmaF
nt
nntt
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Dr. Mohammad Suliman Abuhaiba, PE
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Dynamic
Equilibrium
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Dr. Mohammad Suliman Abuhaiba, PE
• Alternate expression of Newton’s
2nd 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.
9
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.
2/25/2014 6:39 PM
Dr. Mohammad Suliman Abuhaiba, PE
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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 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
2/25/2014 6:39 PM
Dr. Mohammad Suliman Abuhaiba, PE
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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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Dr. Mohammad Suliman Abuhaiba, PE
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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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Dr. Mohammad Suliman Abuhaiba, PE
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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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Dr. Mohammad Suliman Abuhaiba, PE
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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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Dr. Mohammad Suliman Abuhaiba, PE
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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.
2/25/2014 6:39 PM
Dr. Mohammad Suliman Abuhaiba, PE
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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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Dr. Mohammad Suliman Abuhaiba, PE
• 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 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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Dr. Mohammad Suliman Abuhaiba, PE
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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.
2/25/2014 6:39 PM
Dr. Mohammad Suliman Abuhaiba, PE
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Sample Problem 12.5
Solution:
• Resolve equation of motion for
the bob into tangential &normal
components.
• Solve the component equations
for normal & tangential
accelerations.
:tt maF
30sin
30sin
ga
mamg
t
t
2sm9.4ta
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Dr. Mohammad Suliman Abuhaiba, PE
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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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Dr. Mohammad Suliman Abuhaiba, PE
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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.
2/25/2014 6:39 PM
Dr. Mohammad Suliman Abuhaiba, PE
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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 center of the
path.
• Resolve equation of motion
for the car into vertical &
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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Dr. Mohammad Suliman Abuhaiba, PE
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Howe Work Assignment #12-1
10, 17, 23, 30, 37, 44, 51
Due Saturday 1/3/2014
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Dr. Mohammad Suliman Abuhaiba, PE
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Angular Momentum of a Particle
• moment of momentum or
angular momentum of 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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Dr. Mohammad Suliman Abuhaiba, PE
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• Derivative of angular momentum
wrt time,
O
O
M
Fr
amrVmVVmrVmrH
• Newton’s 2nd law: sum of moments
about O of forces acting on the
particle is equal to rate of change of
angular momentum of the particle
about O.
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Dr. Mohammad Suliman Abuhaiba, PE
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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Dr. Mohammad Suliman Abuhaiba, PE
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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
2/25/2014 6:39 PM
Dr. Mohammad Suliman Abuhaiba, PE
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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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Dr. Mohammad Suliman Abuhaiba, PE
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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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Dr. Mohammad Suliman Abuhaiba, PE
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Newton’s Law of Gravitation • Gravitational force exerted by the sun on a planet or by 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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Dr. Mohammad Suliman Abuhaiba, PE
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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 velocity
of B along OA
b. magnitude of horizontal force
exerted on B by the arm OA.
.0q
2/25/2014 6:39 PM
Dr. Mohammad Suliman Abuhaiba, PE
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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
2/25/2014 6:39 PM
Dr. Mohammad Suliman Abuhaiba, PE
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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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Dr. Mohammad Suliman Abuhaiba, PE
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Sample Problem 12.8 Solution:
• Satellite is moving under a central force, its angular
momentum is constant.
mi23403960
mi2403960hmi18820
constantsin
B
AAB
BBAA
O
r
rvv
vmrvmr
Hvrm
hmi12550Bv
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Dr. Mohammad Suliman Abuhaiba, PE
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Home Work Assignment #12-2
66, 74, 81
Due Monday 3/3/2014
2/25/2014 6:39 PM
Dr. Mohammad Suliman Abuhaiba, PE
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Computer Problem
12.C1
Due Monday
3/3/2014
2/25/2014 6:39 PM
Dr. Mohammad Suliman Abuhaiba, PE
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