Transcript chapt12_lecture_updated

Seventh Edition
CHAPTER
12
VECTOR MECHANICS FOR ENGINEERS:
DYNAMICS
Ferdinand P. Beer
E. Russell Johnston, Jr.
Kinetics of Particles:
Newton’s Second Law
Lecture Notes:
J. Walt Oler
Texas Tech University
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Seventh
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Vector Mechanics for Engineers: Dynamics
Contents
Introduction
Newton’s Second Law of Motion
Linear Momentum of a Particle
Systems of Units
Equations of Motion
Dynamic Equilibrium
Sample Problem 12.1
Sample Problem 12.3
Sample Problem 12.4
Sample Problem 12.5
Sample Problem 12.6
Angular Momentum of a Particle
Equations of Motion in Radial &
Transverse Components
Conservation of Angular Momentum
Newton’s Law of Gravitation
Sample Problem 12.7
Sample Problem 12.8
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Vector Mechanics for Engineers: Dynamics
Introduction
• 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 (changes in velocity magnitude or direction),
Newton’s second law is required to relate the motion of the body to the
forces acting on it.
• Newton’s second law:
- A particle will have an acceleration proportional to the magnitude of
the 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.
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Vector Mechanics for Engineers: Dynamics
Newton’s Second Law of Motion
• Newton’s Second Law: 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,
F1 F2 F3


   constant  mass, m
a1 a2 a3

• When a particle of mass m is acted upon by a force F ,
the acceleration of the particle must satisfy


F  ma
• Acceleration must be evaluated with respect to a
Newtonian frame of reference, i.e., one that is not
accelerating or rotating.
• 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.
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Vector Mechanics for Engineers: Dynamics
Linear Momentum of a Particle
• Replacing the acceleration by the derivative of the
velocity yields


dv
F  m
dt

d
 dL
 m v  
dt
dt

L  linear momentum of the particle
• 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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Vector Mechanics for Engineers: Dynamics
Systems of Units
• Of the units for the four primary dimensions (force,
mass, length, and time), three may be chosen arbitrarily.
The fourth must be compatible with Newton’s 2nd Law.
• International System of Units (SI Units): base units are
the units of length (m), mass (kg), and time (second).
The unit of force is derived,
kg  m
 m
1 N  1 kg 1 2   1 2
 s 
s
• U.S. Customary Units: base units are the units of force
(lb), length (m), and time (second). The unit of mass is
derived,
1lb
1lb
lb  s 2
1lbm 
1slug 
1
2
2
ft
32.2 ft s
1ft s
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Vector Mechanics for Engineers: Dynamics
Equations of Motion
• Newton’s second law provides


F

m
a

• Solution for particle motion is facilitated by resolving
vector equation into scalar component equations, e.g.,
for rectangular components,






 Fx i  Fy j  Fz k   ma x i  a y j  a z k 
 Fx  max  Fy  ma y  Fz  maz
 Fx  mx  Fy  my  Fz  mz
• For tangential and normal components,
 F t  mat
dv
F

m
 t
dt
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 F n  man
Fn  m
v2

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Vector Mechanics for Engineers: Dynamics
Dynamic Equilibrium
• Alternate expression of Newton’s second law,


F

m
a
0


 ma  inertial vector
• With the inclusion of the inertial vector, the system
of forces acting on the particle is equivalent to
zero. The particle is in dynamic equilibrium.
• Methods developed for particles in static
equilibrium may be applied, e.g., coplanar forces
may be represented with a closed vector polygon.
• Inertia vectors are often called inertial forces as
they measure the resistance that particles offer to
changes in motion, i.e., changes in speed or
direction.
• Inertial forces may be conceptually useful but are
not like the contact and gravitational forces found
in statics.
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Vector Mechanics for Engineers: Dynamics
Sample Problem 12.1
SOLUTION:
• Resolve the equation of motion for the
block into two rectangular component
equations.
A 200-lb block rests on a horizontal
plane. Find the magnitude of the force
P required to give the block an acceleration or 10 ft/s2 to the right. The coefficient of kinetic friction between the
block and plane is mk  0.25.
• Unknowns consist of the applied force
P and the normal reaction N from the
plane. The two equations may be
solved for these unknowns.
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Vector Mechanics for Engineers: Dynamics
Sample Problem 12.1
SOLUTION:
• Resolve the equation of motion for the block
into two rectangular component equations.
 Fx  ma :

y
O

P cos 30  0.25N  6.21lb  s 2 ft 10 ft s 2
 62.1lb
x
W
200 lb
m 
g 32.2 ft s 2
lb  s 2
 6.21
ft
F  mk N
 0.25N

 Fy  0 :
N  P sin 30  200 lb  0
• Unknowns consist of the applied force P and
the normal reaction N from the plane. The two
equations may be solved for these unknowns.
N  P sin 30  200 lb
P cos 30  0.25 P sin 30  200 lb  62.1lb
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P  151lb
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Vector Mechanics for Engineers: Dynamics
Sample Problem 12.3
SOLUTION:
• Write the kinematic relationships for the
dependent motions and accelerations of
the blocks.
• Write the equations of motion for the
blocks and pulley.
• Combine the kinematic relationships
with the equations of motion to solve for
the accelerations and cord tension.
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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Vector Mechanics for Engineers: Dynamics
Sample Problem 12.3
O
x
y
SOLUTION:
• Write the kinematic relationships for the dependent
motions and accelerations of the blocks.
y B  12 x A
a B  12 a A
• Write equations of motion for blocks and pulley.
 Fx  m Aa A :
T1  100 kg a A
 Fy  m B a B :
mB g  T2  mB a B
300 kg 9.81m s 2  T2  300 kg a B
T2  2940N - 300 kg a B
 Fy  mC aC  0 :
T2  2T1  0
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Vector Mechanics for Engineers: Dynamics
Sample Problem 12.3
• Combine kinematic relationships with equations of
motion to solve for accelerations and cord tension.
O
x
y
y B  12 x A
a B  12 a A
T1  100 kg a A
T2  2940N - 300 kg a B

 2940N - 300 kg  12 a A

T2  2T1  0
2940 N  150 kg a A  2100 kg a A  0
a A  8.40 m s 2
a B  12 a A  4.20 m s 2
T1  100 kg a A  840 N
T2  2T1  1680 N
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Vector Mechanics for Engineers: Dynamics
Sample Problem 12.4
SOLUTION:
• The block is constrained to slide down
the wedge. Therefore, their motions are
dependent. Express the acceleration of
block as the acceleration of wedge plus
the acceleration of the block relative to
the wedge.
The 12-lb block B starts from rest and
slides on the 30-lb wedge A, which is
supported by a horizontal surface.
• Write the equations of motion for the
wedge and block.
• Solve for the accelerations.
Neglecting friction, determine (a) the
acceleration of the wedge, and (b) the
acceleration of the block relative to the
wedge.
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Vector Mechanics for Engineers: Dynamics
Sample Problem 12.4
SOLUTION:
• The block is constrained to slide down the
wedge. Therefore, their motions are dependent.



aB  a A  aB A
• Write equations of motion for wedge and block.
 Fx  m Aa A :
N1 sin 30  m A a A
y
0.5 N1  W A g a A
x
 Fx  mB a x  mB a A cos 30  aB A :
 WB sin 30  WB g a A cos 30  a B
aB
A
A

 a A cos 30  g sin 30
 Fy  mB a y  mB  a A sin 30 :
N1  WB cos 30  WB g a A sin 30
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Vector Mechanics for Engineers: Dynamics
Sample Problem 12.4
• Solve for the accelerations.
0.5 N1  W A g a A
N1  WB cos 30  WB g a A sin 30
2W A g a A  WB cos 30  WB g a A sin 30
aA 
gWB cos 30
2W A  WB sin 30

32.2 ft s 2 12 lb cos 30
aA 
230 lb  12 lb sin 30
a A  5.07 ft s 2
aB
A
 a A cos 30  g sin 30
aB
A
 5.07 ft s 2 cos 30  32.2 ft s 2 sin 30




a B A  20.5 ft s 2
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Vector Mechanics for Engineers: Dynamics
Sample Problem 12.5
SOLUTION:
• Resolve the equation of motion for the
bob into tangential and normal
components.
• Solve the component equations for the
normal and tangential accelerations.
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.
• Solve for the velocity in terms of the
normal acceleration.
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Vector Mechanics for Engineers: Dynamics
Sample Problem 12.5
SOLUTION:
• Resolve the equation of motion for the bob into
tangential and normal components.
• Solve the component equations for the normal and
tangential accelerations.
mg sin 30  mat
 Ft  mat :
at  g sin 30
 Fn  man :
at  4.9 m s 2
2.5mg  mg cos 30  man
an  g 2.5  cos 30
an  16.03 m s 2
• Solve for velocity in terms of normal acceleration.
an 
v2

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v  an 
2 m 16.03 m s 2 
v  5.66 m s
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Vector Mechanics for Engineers: Dynamics
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.The forces acting on the car
are its weight and a normal reaction
from the road surface.
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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• Resolve the equation of motion for
the car into vertical and normal
components.
• Solve for the vehicle speed.
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Vector Mechanics for Engineers: Dynamics
Sample Problem 12.6
• Resolve the equation of motion for
the car into vertical and normal
components.
R cosq  W  0
 Fy  0 :
W
R
cosq
 Fn  man : R sin q 
SOLUTION:
• The car travels in a horizontal circular
path with a normal component of
acceleration directed toward the center
of the path.The forces acting on the
car are its weight and a normal
reaction from the road surface.
W
an
g
W
W v2
sin q 
cosq
g 
• Solve for the vehicle speed.
v 2  g tan q


 32.2 ft s 2 400 ft  tan 18
v  64.7 ft s  44.1 mi h
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Vector Mechanics for Engineers: Dynamics
Angular Momentum of a Particle



• H O  r  mV  moment of momentum or the angular
momentum of the particle about O.



• H O is perpendicular to plane containing r and mV



H O  rmV sin 
i
j
k

 rm vq
HO  x
y
z
mv x mv y mvz
 mr 2q
• Derivative of angular momentum with respect to time,

 
 
 


H O  r  mV  r  mV  V  mV  r  ma

 rF

  MO
• It follows from Newton’s second law that the sum of
the moments about O of the forces acting on the
particle is equal to the rate of change of the angular
momentum of the particle about O.
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Vector Mechanics for Engineers: Dynamics
Eqs of Motion in Radial & Transverse Components
• Consider particle at r and q, in polar coordinates,
 Fr  mar  mr  rq 2 
 Fq  maq  mrq  2rq 
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Vector Mechanics for Engineers: Dynamics
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,  M O  H O  0 and
 

r  mV  H O  constant
• Position vector and motion
 of particle are in a
plane perpendicular to H O .
• Magnitude of angular momentum,
H O  rmV sin   constant
 r0 m V0 sin 0
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Vector Mechanics for Engineers: Dynamics
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 and m attract each other with equal and opposite
force directed along the line connecting the particles,
Mm
F G 2
r
G  constant of gravitatio n
3
4
m
ft
9
 66.73  1012

34
.
4

10
kg  s 2
lb  s 4
• For particle of mass m on the earth’s surface,
MG
m
ft
W  m 2  mg g  9.81 2  32.2 2
R
s
s
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Vector Mechanics for Engineers: Dynamics
Sample Problem 12.7
SOLUTION:
• Write the radial and transverse
equations of motion for the block.
• Integrate the radial equation to find an
expression for the radial velocity.
A block B of mass m can slide freely on
a frictionless arm OA which rotates in a
horizontal plane at a constant rate q0 .
• Substitute known information into the
transverse equation to find an
expression for the force on the block.
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, and
b) the magnitude of the horizontal force
exerted on B by the arm OA.
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Vector Mechanics for Engineers: Dynamics
Sample Problem 12.7
SOLUTION:
• Write the radial and transverse
equations of motion for the block.
 Fr  m ar :

vr
r
2
 vr dvr  q0  r dr
0
r0

0  m r  rq
 F q  m aq : F  mrq  2rq 
2
• Integrate the radial equation to find an
expression for the radial velocity.
dv
dv dr
dv
 vr r
r  vr  r  r
dt
dr dt
dr
dv
dv dr
dv
 vr r
r  vr  r  r
dt
dr dt
dr
vr dvr  rq 2 dr  rq02 dr

vr2  q 02 r 2  r02

• Substitute known information into the
transverse equation to find an expression
for the force on the block.
F
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 2mq 02
r
2

2 12
 r0
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Vector Mechanics for Engineers: Dynamics
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.
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 it
maximum altitude of 2340 mi. The
radius of the earth is 3960 mi.
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Vector Mechanics for Engineers: Dynamics
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.
rm v sin   H O  constant
rA m v A  rB m v B
r
vB  v A A
rB
 18820mi h 
3960  240mi
3960  2340mi
v B  12550 mi h
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