Transcript VECTOR MECHANICS FOR ENGINEERS: DYNAMICS Eighth

Eighth Edition
CHAPTER
13
VECTOR MECHANICS FOR ENGINEERS:
DYNAMICS
Ferdinand P. Beer
E. Russell Johnston, Jr.
Lecture Notes:
J. Walt Oler
Texas Tech University
Kinetics of Particles:
Energy and Momentum
Methods
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
Eighth
Edition
Vector Mechanics for Engineers: Dynamics
Contents
Introduction
Work of a Force
Principle of Work & Energy
Applications of the Principle of Work
& Energy
Power and Efficiency
Sample Problem 13.1
Sample Problem 13.2
Sample Problem 13.3
Sample Problem 13.4
Sample Problem 13.5
Potential Energy
Conservative Forces
Conservation of Energy
Motion Under a Conservative Central
Force
Sample Problem 13.6
Sample Problem 13.7
Sample Problem 13.9
Principle of Impulse and Momentum
Impulsive Motion
Sample Problem 13.10
Sample Problem 13.11
Sample Problem 13.12
Impact
Direct Central Impact
Oblique Central Impact
Problems Involving Energy and Momentum
Sample Problem 13.14
Sample Problem 13.15
Sample Problems 13.16
Sample Problem !3.17
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13 - 2
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Vector Mechanics for Engineers: Dynamics
Introduction
• Previously, problems dealing with the motion of particles
 were

solved through the fundamental equation of motion, F  ma.
Current chapter introduces two additional methods of analysis.
• Method of work and energy: directly relates force, mass,
velocity and displacement.
• Method of impulse and momentum: directly relates force,
mass, velocity, and time.
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13 - 3
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Vector Mechanics for Engineers: Dynamics
Work of a Force

• Differential vector dr is the particle displacement.
• Work of the force is
 
dU  F  dr
 F ds cos 
 Fx dx  Fy dy  Fz dz
• Work is a scalar quantity, i.e., it has magnitude and
sign but not direction.
• Dimensions of work are length  force. Units are
1 J  joule  1 N 1 m
1ft  lb  1.356 J
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13 - 4
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Vector Mechanics for Engineers: Dynamics
Work of a Force
• Work of a force during a finite displacement,
U12 
A2 

F

d
r

A1


s2
s2
s1
s1
 F cos ds   Ft ds
A2
 Fx dx  Fy dy  Fz dz 
A1
• Work is represented by the area under the
curve of Ft plotted against s.
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13 - 5
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Edition
Vector Mechanics for Engineers: Dynamics
Work of a Force
• Work of a constant force in rectilinear motion,
U12  F cos  x
• Work of the force of gravity,
dU  Fx dx  Fy dy  Fz dz
 W dy
y2
U12    W dy
y1
 W  y 2  y1   W y
• Work of the weight is equal to product of
weight W and vertical displacement y.
• Work of the weight is positive when y < 0,
i.e., when the weight moves down.
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13 - 6
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Vector Mechanics for Engineers: Dynamics
Work of a Force
• Magnitude of the force exerted by a spring is
proportional to deflection,
F  kx
k  spring constant N/m or lb/in. 
• Work of the force exerted by spring,
dU   F dx  kx dx
x2
U12    kx dx  12 kx12  12 kx22
x1
• Work of the force exerted by spring is positive
when x2 < x1, i.e., when the spring is returning to
its undeformed position.
• Work of the force exerted by the spring is equal to
negative of area under curve of F plotted against x,
U12   12 F1  F2  x
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13 - 7
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Vector Mechanics for Engineers: Dynamics
Work of a Force
Work of a gravitational force (assume particle M
occupies fixed position O while particle m follows path
shown),
Mm
dU   Fdr  G 2 dr
r
r2
Mm
r1
r2
U12    G
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dr  G
Mm
Mm
G
r2
r1
13 - 8
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Vector Mechanics for Engineers: Dynamics
Work of a Force
Forces which do not do work (ds = 0 or cos   0:
• reaction at frictionless pin supporting rotating body,
• reaction at frictionless surface when body in contact
moves along surface,
• reaction at a roller moving along its track, and
• weight of a body when its center of gravity moves
horizontally.
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13 - 9
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Vector Mechanics for Engineers: Dynamics
Particle Kinetic Energy: Principle of Work & Energy

• Consider a particle of mass m acted upon by force F
dv
Ft  mat  m
dt
dv ds
dv
m
 mv
ds dt
ds
F t ds  mv dv
• Integrating from A1 to A2 ,
s2
v2
s1
v1
 Ft ds  m  v dv  12 mv2  12 mv1
2
2
U12  T2  T1
T  12 mv 2  kinetic energy

• The work of the force F is equal to the change in
kinetic energy of the particle.
• Units of work and kinetic energy are the same:
2
m
 m


2
T  12 mv  kg    kg 2 m  N  m  J
s
 s 
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13 - 10
Eighth
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Vector Mechanics for Engineers: Dynamics
Applications of the Principle of Work and Energy
• Wish to determine velocity of pendulum bob
at A2. Consider work & kinetic energy.

• Force P acts normal to path and does no
work.
T1  U12  T2
0  Wl 
1W 2
v2
2 g
v2  2 gl
• Velocity found without determining
expression for acceleration and integrating.
• All quantities are scalars and can be added
directly.
• Forces which do no work are eliminated from
the problem.
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13 - 11
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Vector Mechanics for Engineers: Dynamics
Applications of the Principle of Work and Energy
• Principle of work and energy cannot be
applied to directly determine the acceleration
of the pendulum bob.
• Calculating the tension in the cord requires
supplementing the method of work and energy
with an application of Newton’s second law.
• As the bob passes through A2 ,
 Fn  m an
v22
l
W 2 gl
P W 
 3W
g l
W
P W 
g
v2  2 gl
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13 - 12
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Vector Mechanics for Engineers: Dynamics
Power and Efficiency
• Power  rate at which work is done.
 
dU F  dr


dt
dt
 
 F v
• Dimensions of power are work/time or force*velocity.
Units for power are
J
m
1 W (watt)  1  1 N 
s
s
or 1 hp  550
ft  lb
 746 W
s
•   efficiency
output work

input work
power output

power input
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13 - 13
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Vector Mechanics for Engineers: Dynamics
Sample Problem 13.1
SOLUTION:
• Evaluate the change in kinetic energy.
• Determine the distance required for the
work to equal the kinetic energy change.
An automobile weighing 4000 lb is
driven down a 5o incline at a speed of
60 mi/h when the brakes are applied
causing a constant total breaking force
of 1500 lb.
Determine the distance traveled by the
automobile as it comes to a stop.
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13 - 14
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Vector Mechanics for Engineers: Dynamics
Sample Problem 13.1
SOLUTION:
• Evaluate the change in kinetic energy.
 mi  5280 ft  h 
v1   60 

  88 ft s
h  mi  3600 s 

T1  12 mv12  12 4000 32.2882  481000 ft  lb
v2  0
T2  0
• Determine the distance required for the work
to equal the kinetic energy change.
U12   1500 lb x  4000 lb sin 5x
 1151lb x
T1  U12  T2
481000 ft  lb  1151lb x  0
x  418 ft
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13 - 15
Eighth
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Vector Mechanics for Engineers: Dynamics
Sample Problem 13.2
SOLUTION:
• Apply the principle of work and
energy separately to blocks A and B.
• When the two relations are combined,
the work of the cable forces cancel.
Solve for the velocity.
Two blocks are joined by an inextensible
cable as shown. If the system is released
from rest, determine the velocity of block
A after it has moved 2 m. Assume that the
coefficient of friction between block A
and the plane is mk = 0.25 and that the
pulley is weightless and frictionless.
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Eighth
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Vector Mechanics for Engineers: Dynamics
Sample Problem 13.2
SOLUTION:
• Apply the principle of work and energy separately
to blocks A and B.


W A  200 kg  9.81 m s 2  1962 N
FA  m k N A  m k W A  0.251962 N   490 N
T1  U12  T2 :
0  FC 2 m   FA 2 m   12 m Av 2
FC 2 m   490 N 2 m   12 200 kg v 2


WB  300 kg  9.81 m s 2  2940 N
T1  U12  T2 :
0  Fc 2 m   WB 2 m   12 m B v 2
 Fc 2 m   2940 N 2 m   12 300 kg v 2
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Vector Mechanics for Engineers: Dynamics
Sample Problem 13.2
• When the two relations are combined, the work of the
cable forces cancel. Solve for the velocity.
FC 2 m  490 N 2 m  12 200 kg v 2
 Fc 2 m  2940 N2 m  12 300 kgv 2
2940 N 2 m   490 N 2 m   12 200 kg  300 kg v 2
4900 J  12 500 kg v 2
v  4.43 m s
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Vector Mechanics for Engineers: Dynamics
Sample Problem 13.3
SOLUTION:
• Apply the principle of work and energy
between the initial position and the
point at which the spring is fully
compressed and the velocity is zero.
The only unknown in the relation is the
friction coefficient.
A spring is used to stop a 60 kg package
which is sliding on a horizontal surface.
The spring has a constant k = 20 kN/m
and is held by cables so that it is initially • Apply the principle of work and energy
for the rebound of the package. The
compressed 120 mm. The package has a
only unknown in the relation is the
velocity of 2.5 m/s in the position shown
and the maximum deflection of the spring velocity at the final position.
is 40 mm.
Determine (a) the coefficient of kinetic
friction between the package and surface
and (b) the velocity of the package as it
passes again through the position shown.
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Vector Mechanics for Engineers: Dynamics
Sample Problem 13.3
SOLUTION:
• Apply principle of work and energy between initial
position and the point at which spring is fully compressed.
T1  12 mv12  12 60 kg 2.5 m s 2  187.5 J
U12  f
  m kW x

T2  0

  m k 60 kg  9.81m s 2 0.640 m   377 J m k
Pmin  kx0  20 kN m 0.120 m   2400 N
Pmax  k  x0  x   20 kN m 0.160 m   3200 N
U12 e   12 Pmin  Pmax x
  12 2400 N  3200 N 0.040 m   112.0 J
U12  U12  f  U12 e  377 J mk  112 J
T1  U12  T2 :
187.5 J - 377 J m k  112 J  0
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m k  0.20
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Vector Mechanics for Engineers: Dynamics
Sample Problem 13.3
• Apply the principle of work and energy for the rebound
of the package.
T 3 12 mv32  12 60kg v32
T2  0
U 23  U 23  f  U 23 e  377 J m k  112 J
 36.5 J
T2  U 23  T3 :
0  36.5 J  12 60 kg v32
v3  1.103 m s
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13 - 21
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Vector Mechanics for Engineers: Dynamics
Sample Problem 13.4
SOLUTION:
• Apply principle of work and energy to
determine velocity at point 2.
• Apply Newton’s second law to find
normal force by the track at point 2.
A 2000 lb car starts from rest at point 1
• Apply principle of work and energy to
and moves without friction down the
determine velocity at point 3.
track shown.
• Apply Newton’s second law to find
Determine:
minimum radius of curvature at point 3
such that a positive normal force is
a) the force exerted by the track on
exerted by the track.
the car at point 2, and
b) the minimum safe value of the
radius of curvature at point 3.
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13 - 22
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Vector Mechanics for Engineers: Dynamics
Sample Problem 13.4
SOLUTION:
• Apply principle of work and energy to determine
velocity at point 2.
T1  0
T2  12 mv22 
U1 2  W 40 ft 
T1  U1 2  T2 :
1W 2
v2
2g
0  W 40 ft  

v22  240 ft g  240 ft  32.2 ft s 2

1W 2
v2
2g
v2  50.8 ft s
• Apply Newton’s second law to find normal force by
the track at point 2.
   Fn  m an :
W v22 W 240 ft g
 W  N  m an 

g  2 g 20 ft
N  5W
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N  10000 lb
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Vector Mechanics for Engineers: Dynamics
Sample Problem 13.4
• Apply principle of work and energy to determine
velocity at point 3.
T1  U13  T3
0  W 25 ft  
v32  225 ft g  225 ft 32.2 ft s 
1W 2
v3
2g
v3  40.1ft s
• Apply Newton’s second law to find minimum radius of
curvature at point 3 such that a positive normal force is
exerted by the track.
   Fn  m an :
W  m an
W v32 W 225 ft g


g 3 g
3
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3  50 ft
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Vector Mechanics for Engineers: Dynamics
Sample Problem 13.5
SOLUTION:
Force exerted by the motor
cable has same direction as
the dumbwaiter velocity.
Power delivered by motor is
equal to FvD, vD = 8 ft/s.
The dumbwaiter D and its load have a
combined weight of 600 lb, while the
counterweight C weighs 800 lb.
• In the first case, bodies are in uniform
motion. Determine force exerted by
motor cable from conditions for static
equilibrium.
Determine the power delivered by the
electric motor M when the dumbwaiter
(a) is moving up at a constant speed of
8 ft/s and (b) has an instantaneous
velocity of 8 ft/s and an acceleration of
2.5 ft/s2, both directed upwards.
• In the second case, both bodies are
accelerating. Apply Newton’s
second law to each body to
determine the required motor cable
force.
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13 - 25
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Vector Mechanics for Engineers: Dynamics
Sample Problem 13.5
• In the first case, bodies are in uniform motion.
Determine force exerted by motor cable from
conditions for static equilibrium.
Free-body C:
   Fy  0 :
2T  800 lb  0
T  400 lb
Free-body D:
   Fy  0 : F  T  600 lb  0
F  600 lb  T  600 lb  400 lb  200 lb
Power  Fv D  200 lb 8 ft s 
 1600 ft  lb s
Power  1600 ft  lb s 
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
1 hp
 2.91 hp
550 ft  lb s
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Vector Mechanics for Engineers: Dynamics
Sample Problem 13.5
• In the second case, both bodies are accelerating. Apply
Newton’s second law to each body to determine the required
motor cable force.
a D  2.5 ft s 2 
aC   12 a D  1.25 ft s 2 
Free-body C:
   Fy  mC aC : 800  2T 
800
1.25
32.2
T  384.5 lb
Free-body D:
600
2.5
32.2
F  384.5  600  46.6
   Fy  m D a D : F  T  600 
F  262.1 lb
Power  FvD  262.1 lb 8 ft s   2097 ft  lb s
Power  2097 ft  lb s 
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
1 hp
 3.81 hp
550 ft  lb s
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Vector Mechanics for Engineers: Dynamics
Potential Energy

• Work of the force of gravity W,
U12  W y1  W y2
• Work is independent of path followed; depends
only on the initial and final values of Wy.
V g  Wy
 potential energy of the body with respect
to force of gravity.
   
U12  V g  V g
1
2
• Choice of datum from which the elevation y is
measured is arbitrary.
• Units of work and potential energy are the same:
Vg  Wy  N  m  J
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13 - 28
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Vector Mechanics for Engineers: Dynamics
Potential Energy
• Previous expression for potential energy of a body
with respect to gravity is only valid when the
weight of the body can be assumed constant.
• For a space vehicle, the variation of the force of
gravity with distance from the center of the earth
should be considered.
• Work of a gravitational force,
GMm GMm
U12 

r2
r1
• Potential energy Vg when the variation in the
force of gravity can not be neglected,
GMm
WR 2
Vg  

r
r
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13 - 29
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Vector Mechanics for Engineers: Dynamics
Potential Energy
• Work of the force exerted by a spring depends
only on the initial and final deflections of the
spring,
U12  12 kx12  12 kx22
• The potential energy of the body with respect
to the elastic force,
Ve  12 kx 2
U12  Ve 1  Ve 2
• Note that the preceding expression for Ve is
valid only if the deflection of the spring is
measured from its undeformed position.
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Vector Mechanics for Engineers: Dynamics
Conservative Forces
• Concept of potential energy can be applied if the
work of the force is independent of the path
followed by its point of application.
U12  V  x1, y1, z1   V  x2 , y2 , z2 
Such forces are described as conservative forces.
• For any conservative force applied on a closed path,
 
 F  dr  0
• Elementary work corresponding to displacement
between two neighboring points,
dU  V  x, y, z   V  x  dx, y  dy, z  dz 
 dV  x, y, z 
V
V 
 V
Fx dx  Fy dy  Fz dz  
dx 
dy 
dz 
y
z 
 x

 V V V 
F  


  grad V
 x y z 
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Vector Mechanics for Engineers: Dynamics
Conservation of Energy
• Work of a conservative force,
U12  V1  V2
• Concept of work and energy,
U12  T2  T1
• Follows that
T1  V1  T2  V2
E  T  V  constant
T1  0 V1  W
T1  V1  W
T2  12 mv22 
T2  V2  W
1W
2 g   W V2  0
2g
• When a particle moves under the action of
conservative forces, the total mechanical
energy is constant.
• Friction forces are not conservative. Total
mechanical energy of a system involving
friction decreases.
• Mechanical energy is dissipated by friction
into thermal energy. Total energy is constant.
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
13 - 32
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Vector Mechanics for Engineers: Dynamics
Motion Under a Conservative Central Force
• When a particle moves under a conservative central
force, both the principle of conservation of angular
momentum
r0 mv0 sin 0  rmv sin 
and the principle of conservation of energy
T0  V0  T  V
1 mv 2
0
2

GMm 1 2 GMm
 2 mv 
r0
r
may be applied.
• Given r, the equations may be solved for v and j.
• At minimum and maximum r, j  90o. Given the
launch conditions, the equations may be solved for
rmin, rmax, vmin, and vmax.
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
13 - 33
Eighth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 13.6
SOLUTION:
• Apply the principle of conservation of
energy between positions 1 and 2.
• The elastic and gravitational potential
energies at 1 and 2 are evaluated from
the given information. The initial kinetic
energy is zero.
A 20 lb collar slides without friction
along a vertical rod as shown. The
spring attached to the collar has an
undeflected length of 4 in. and a
constant of 3 lb/in.
• Solve for the kinetic energy and velocity
at 2.
If the collar is released from rest at
position 1, determine its velocity after
it has moved 6 in. to position 2.
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
13 - 34
Eighth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 13.6
SOLUTION:
• Apply the principle of conservation of energy between
positions 1 and 2.
Position 1: Ve  12 kx12  12 3 lb in. 8 in.  4 in. 2  24 in.  lb
V1  Ve  Vg  24 in.  lb  0  2 ft  lb
T1  0
Position 2: Ve  12 kx22  12 3 lb in. 10 in.  4 in. 2  54 in.  lb
Vg  Wy  20 lb  6 in.   120 in.  lb
V2  Ve  Vg  54  120  66 in.  lb  5.5 ft  lb
T2  12 mv22 
1 20 2
v2  0.311v22
2 32.2
Conservation of Energy:
T1  V1  T2  V2
0  2 ft  lb  0.311v22  5.5 ft  lb
v2  4.91ft s 
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
13 - 35
Eighth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 13.7
SOLUTION:
• Since the pellet must remain in contact
with the loop, the force exerted on the
pellet must be greater than or equal to
zero. Setting the force exerted by the
loop to zero, solve for the minimum
velocity at D.
The 0.5 lb pellet is pushed against the
spring and released from rest at A.
Neglecting friction, determine the
smallest deflection of the spring for
which the pellet will travel around the
loop and remain in contact with the
loop at all times.
• Apply the principle of conservation of
energy between points A and D. Solve
for the spring deflection required to
produce the required velocity and
kinetic energy at D.
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
13 - 36
Eighth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 13.7
SOLUTION:
• Setting the force exerted by the loop to zero, solve for the
minimum velocity at D.
2
   Fn  man : W  man
mg  m vD
r
2
vD
 rg  2 ft 32.2 ft s   64.4 ft 2 s 2
• Apply the principle of conservation of energy between
points A and D.
V1  Ve  Vg  12 kx2  0  12 36 lb ft x 2  18 x 2
T1  0
V2  Ve  Vg  0  Wy  0.5 lb 4 ft   2 ft  lb
2
T2  12 mvD



1 0.5 lb
2 2
64
.
4
ft
s  0.5 ft  lb
2 32.2 ft s 2
T1  V1  T2  V2
0  18 x 2  0.5  2
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
x  0.3727 ft  4.47 in.
13 - 37
Eighth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 13.9
SOLUTION:
• For motion under a conservative central
force, the principles of conservation of
energy and conservation of angular
momentum may be applied simultaneously.
A satellite is launched in a direction
parallel to the surface of the earth
with a velocity of 36900 km/h from
an altitude of 500 km.
• Apply the principles to the points of
minimum and maximum altitude to
determine the maximum altitude.
• Apply the principles to the orbit insertion
point and the point of minimum altitude to
determine maximum allowable orbit
Determine (a) the maximum altitude
insertion angle error.
reached by the satellite, and (b) the
maximum allowable error in the
direction of launching if the satellite
is to come no closer than 200 km to
the surface of the earth
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
13 - 38
Eighth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 13.9
• Apply the principles of conservation of energy and
conservation of angular momentum to the points of minimum
and maximum altitude to determine the maximum altitude.
Conservation of energy:
TA  VA  TA  VA
1 mv2
0
2

GMm 1 2 GMm
 2 mv1 
r0
r1
Conservation of angular momentum:
r
r0 mv0  r1mv1
v1  v0 0
r1
Combining,
2

r0 2GM
1 v 2 1  r0   GM 1  r0 
1


 r 
2 0
2
r
r
r0v02
0 
1
1
 r1 
r0  6370 km  500 km  6870 km
v0  36900 km h  10.25  106 m s


2
GM  gR 2  9.81m s 2 6.37  106 m  398  1012 m3 s 2
r1  60.4 106 m  60400 km
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
13 - 39
Eighth
Edition
Vector Mechanics for Engineers: Dynamics
Sample Problem 13.9
• Apply the principles to the orbit insertion point and the point
of minimum altitude to determine maximum allowable orbit
insertion angle error.
Conservation of energy:
GMm
1 mv2  GMm  1 mv2
T0  V0  TA  VA

0
max
2
2
r0
rmin
Conservation of angular momentum:
r
r0 mv0 sin 0  rmin mvmax
vmax  v0 sin 0 0
rmin
Combining and solving for sin j0,
sin 0  0.9801
j0  90  11.5
© 2007 The McGraw-Hill Companies, Inc. All rights reserved.
allowable error  11.5
13 - 40