Upload
gracee-dee
View
223
Download
0
Embed Size (px)
Citation preview
8/12/2019 ES 12 Chapter 13
1/68
Seventh Edition
VECTOR MECHANICS FOR ENGINEERS:
DYNAMICSCHAPTER
.
E. Russell Johnston, Jr. Kinetics of Particles:
Energy and MomentumJ. Walt Oler
Texas Tech University
Methods
2003 The McGraw-Hill Companies, Inc. All rights reserved.
8/12/2019 ES 12 Chapter 13
2/68
Vector Mechanics for Engineers: DynamicsSeventh
Edition
ContentsIntroduction
Work of a Force
Princi le of Work & Ener
Sample Problem 13.6
Sample Problem 13.7
Sam le Problem 13.9
Applications of the Principle of Work& Energy
Power and Efficienc
Principle of Impulse and MomentumImpulsive Motion
Sample Problem 13.1
Sample Problem 13.2
.
Sample Problem 13.11
Sample Problem 13.12
.
Sample Problem 13.4
Sample Problem 13.5
Direct Central Impact
Oblique Central Impact
o en a nergyConservative Forces
Conservation of Energy
Pro ems Invo v ng Energy an MomentumSample Problem 13.14
Sample Problem 13.15
2003 The McGraw-Hill Companies, Inc. All rights reserved. 13 - 2
Motion Under a Conservative Central
Force
Sample Problems 13.16
Sample Problem !3.17
8/12/2019 ES 12 Chapter 13
3/68
Vector Mechanics for Engineers: DynamicsSeventh
Edition
Introduction
Previously, problems dealing with the motion of particles were
solved through the fundamental equation of motion, .amF
= .
Method of work and energy: directly relates force, mass,
.
Method of impulse and momentum: directly relates force,mass, velocity, and time.
2003 The McGraw-Hill Companies, Inc. All rights reserved. 13 - 3
8/12/2019 ES 12 Chapter 13
4/68
Vector Mechanics for Engineers: DynamicsSeventh
Edition
Work of a Force
Differential vector is theparticle displacement.rd
or o e orce s
dsF
rdFdU
=
=
cos
dzFdyFdxF zyx ++=
, . .,
sign but not direction.
force.len th Dimensions of work are Units are( ) ( )( ) J1.356lb1ftm1N1J1 ==joule
2003 The McGraw-Hill Companies, Inc. All rights reserved. 13 - 4
8/12/2019 ES 12 Chapter 13
5/68
Vector Mechanics for Engineers: DynamicsSeventh
Edition
Work of a Force
Work of a force during a finite displacement,
2A =
22
1
21
ss
A
r
=
==
2
11
A
s
t
s
1A
Work is re resented b the area under the
curve of Ftplotted against s.
2003 The McGraw-Hill Companies, Inc. All rights reserved. 13 - 5
8/12/2019 ES 12 Chapter 13
6/68
Vector Mechanics for Engineers: DynamicsSeventh
Edition
Work of a Force
Work of a constant force in rectilinear motion,
( ) xFU = cos21
Work of the force of gravity,
dzFdyFdxFdU zyx ++=
dyWU
dyW
y
=
=
212
( ) yWyyW
y
== 12
1
Work of the weightis equal to product ofweight Wand vertical displacement y.
2003 The McGraw-Hill Companies, Inc. All rights reserved. 13 - 6
or o t e we g t s pos ve w en y < ,
i.e., when the weight moves down.
8/12/2019 ES 12 Chapter 13
7/68
8/12/2019 ES 12 Chapter 13
8/68
Vector Mechanics for Engineers: DynamicsSeventh
Edition
Work of a Force
Work of a gravitational force (assume particleM
shown),
2 dr
MmGFdrdU ==
12221
2
r
MmG
r
MmGdr
r
MmGU
r
r
==
2003 The McGraw-Hill Companies, Inc. All rights reserved. 13 - 8
8/12/2019 ES 12 Chapter 13
9/68
Vector Mechanics for Engineers: DynamicsSeventh
Edition
Work of a Force
Forces which do notdo work (ds = 0 or cos = 0):
reaction at frictionless surface when body in contact
reaction at frictionless pin supporting rotating body,
reaction at a roller moving along its track, and
moves along surface,
weight of a body when its center of gravity moves
horizontally.
2003 The McGraw-Hill Companies, Inc. All rights reserved. 13 - 9
8/12/2019 ES 12 Chapter 13
10/68
Vector Mechanics for Engineers: DynamicsSeventh
Edition
Particle Kinetic Energy: Principle of Work & Energy
dt
dvmmaF tt ==
Consider a particle of mass m acted upon by force F
dvmvdsF
ds
dv
mvdt
ds
ds
dv
m
t ===
Integrating fromA1 toA2 ,
mvmvdvvmdsF
vs
t == 2121222122
energykineticmvTTTU
vs
===2
21
1221
11
e wor o t e orce s equa to t e c ange nkinetic energy of the particle.
Units of work and kinetic energy are the same:
2003 The McGraw-Hill Companies, Inc. All rights reserved. 13 - 10
JmNms
mkg
s
mkg
2
2
2
1 ==
=
== mvT
8/12/2019 ES 12 Chapter 13
11/68
Vector Mechanics for Engineers: DynamicsSeventh
Edition
Applications of the Principle of Work and Energy
Wish to determine velocity of pendulum bob
atA2. Consider work & kinetic energy.
Force acts normal to path and does no
work.
P
TUT 2211 =+
vg
WWl
2
10 22
=
=+
Velocity found without determining
expression for acceleration and integrating.
All quantities are scalars and can be added
directly.
2003 The McGraw-Hill Companies, Inc. All rights reserved. 13 - 11
Forces which do no work are eliminated from
the problem.
8/12/2019 ES 12 Chapter 13
12/68
Vector Mechanics for Engineers: DynamicsSeventh
Edition
Applications of the Principle of Work and Energy
Principle of work and energy cannot be
applied to directly determine the acceleration
of the endulum bob.
Calculating the tension in the cord requires
supplementing the method of work and energy
with an application of Newtons second law.
As the bob passes throughA2 ,
vWWP
amF nn
22=
=
Wl
gl
g
WWP 3
2=+=
glv 22=
2003 The McGraw-Hill Companies, Inc. All rights reserved. 13 - 12
8/12/2019 ES 12 Chapter 13
13/68
Vector Mechanics for Engineers: DynamicsSeventh
Edition
Power and Efficiency
rate at which work is done.
rdFdU
Power
==
=
vF
tt
=
mens ons o power are wor me or orce ve oc y.
Units for power are
W746lbft550h1ormN1J1wattW1 ====s
ss
efficiency=
outputpower
input workkoutput wor=
2003 The McGraw-Hill Companies, Inc. All rights reserved. 13 - 13
inputpower
8/12/2019 ES 12 Chapter 13
14/68
Vector Mechanics for Engineers: DynamicsSeventh
Edition
Sample Problem 13.1
SOLUTION:
Evaluate the change in kinetic energy.
Determine the distance required for thework to equal the kinetic energy change.
driven down a 5o incline at a speed of
60 mi/h when the brakes are applied
of 1500 lb.
Determine the distance traveled by the
2003 The McGraw-Hill Companies, Inc. All rights reserved. 13 - 14
au omo e as comes o a s op.
8/12/2019 ES 12 Chapter 13
15/68
Vector Mechanics for Engineers: DynamicsSeventh
Edition
Sample Problem 13.1
SOLUTION:
Evaluate the change in kinetic energy.
lbft481000882.324000
sft88s3600mi
t
h
m
60
2121
1
=====
mvT
v
00 22 == Tv
equal the kinetic energy change.
( ) ( )( )xxU 5sinlb4000lb150021 +=
2211 =+ TUT( )xlb1151=
2003 The McGraw-Hill Companies, Inc. All rights reserved. 13 - 15
= x
ft418=x
8/12/2019 ES 12 Chapter 13
16/68
Vector Mechanics for Engineers: DynamicsSeventh
Edition
Sample Problem 13.2
SOLUTION:
Apply the principle of work and
energy separately to blocks andB.
When the two relations are combined,
Two blocks are joined by an inextensible
.
Solve for the velocity.
ca e as s own. e sys em s re ease
from rest, determine the velocity of block
A after it has moved 2 m. Assume that the
and the plane is k= 0.25 and that the
pulley is weightless and frictionless.
2003 The McGraw-Hill Companies, Inc. All rights reserved. 13 - 16
8/12/2019 ES 12 Chapter 13
17/68
Vector Mechanics for Engineers: DynamicsSeventh
Edition
Sample Problem 13.2SOLUTION:
Apply the principle of work and energy separately
to blocksA andB.
( )( )
2
N490N196225.0N1962sm81.9kg200
WNF
W
AkAkA
A
======
( ) ( )
2
2
1
2211
m2m20
:
vmFF
TUT
AAC =+
=+
( ) ( )( ) ( )21 kg200m2N490m2 vFC =
2 ==
( ) ( ) 212211
m2m20
:
.
vmWF
TUT
BBc =+
=+
2003 The McGraw-Hill Companies, Inc. All rights reserved. 13 - 17
( ) ( )( ) ( ) 2
2
1 kg300m2N2940m2 vFc =+
8/12/2019 ES 12 Chapter 13
18/68
8/12/2019 ES 12 Chapter 13
19/68
Vector Mechanics for Engineers: DynamicsSeventh
Edition
Sample Problem 13.3SOLUTION:
Apply the principle of work and energy
A spring is used to stop a 60 kg package
point at which the spring is fullycompressed and the velocity is zero.
The onl unknown in the relation is thewhich 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
friction coefficient.
Apply the principle of work and energy
compresse mm. e pac age as a
velocity of 2.5 m/s in the position shown
and the maximum deflection of the spring
.
only unknown in the relation is the
velocity at the final position.
.
Determine (a) the coefficient of kinetic
friction between the package and surface
2003 The McGraw-Hill Companies, Inc. All rights reserved. 13 - 19
and ( ) the velocity of the package as it
passes again through the position shown.
8/12/2019 ES 12 Chapter 13
20/68
Vector Mechanics for Engineers: DynamicsSeventh
Edition
Sample Problem 13.3SOLUTION:
Apply principle of work and energy between initial
position and the point at which spring is fully compressed.
( )( ) 0J5.187sm5.2kg60 22
2
121
2
11 ==== TmvT
( ) kf xWU 21 =
( )( )( ) ( ) kk J377m640.0sm81.9kg60 2 ==
( )( ) N2400m120.0mkN200min ===kxP( ) ( )( )
( ) ( )
N3200m160.0mkN20
1
maxmin21
21
0max
==
+=
==+=
xPPU
xxkP
e
..2
( ) ( ) ( ) J112J377212121 =+= kef UUU
2003 The McGraw-Hill Companies, Inc. All rights reserved. 13 - 20( ) 0J112J377-J5.187
:2211
=
=+
k
TUT
20.0=k
8/12/2019 ES 12 Chapter 13
21/68
Vector Mechanics for Engineers: DynamicsSeventh
Edition
Sample Problem 13.3
Apply the principle of work and energy for the rebound
of the package.
2121 ===22
( ) ( ) ( ) J112J377323232
=
+=+= kef UUU
( ) 2313322
kg60J5.360
:
v
TUT
=+
=+
sm103.13=v
2003 The McGraw-Hill Companies, Inc. All rights reserved. 13 - 21
8/12/2019 ES 12 Chapter 13
22/68
Vector Mechanics for Engineers: DynamicsSeventh
Edition
Sample Problem 13.4
SOLUTION:
Apply principle of work and energy to
eterm ne ve oc ty at po nt 2.
Apply Newtons second law to find
normal force by the track at point 2.A 2000 lb car starts from rest at point 1
and moves without friction down the
track shown.
Apply principle of work and energy to
determine velocity at point 3.
Determine:
a) the force exerted by the track on
pp y ewton s secon aw to n
minimum radius of curvature at point 3
such that a positive normal force is
,
b) the minimum safe value of the
radius of curvature at point 3.
.
2003 The McGraw-Hill Companies, Inc. All rights reserved. 13 - 22
8/12/2019 ES 12 Chapter 13
23/68
Vector Mechanics for Engineers: DynamicsSeventh
Edition
Sample Problem 13.4SOLUTION:
Apply principle of work and energy to determine
velocity at point 2.
( )ft40
2
1
0
21
2
2
2
22
1
21
+====
WU
vg
W
mvTT
( )
sft8.50ft2.32ft402ft402
2
1ft400:
22
222211
===
=+=+
vsv
vg
WWTUT
Apply Newtons second law to find normal force by
the track at point 2.
:nn amF =+ ( )gWvW
amNW nft40222 ===+
2003 The McGraw-Hill Companies, Inc. All rights reserved. 13 - 23
WN 5
2
= lb10000=N
8/12/2019 ES 12 Chapter 13
24/68
Vector Mechanics for Engineers: DynamicsSeventh
Edition
Sample Problem 13.4 Apply principle of work and energy to determine
velocity at point 3.
1 2W
( ) ( )( ) sft1.40sft2.32ft252ft252
2
323
33311
===
==
vgv
g
Apply Newtons second law to find minimum radius of
curvature at point 3 such that a positive normal force is
exerted by the track.
:nn amF =+
amW n=
( )33
23 ft252
g
g
Wv
g
W == ft503=
2003 The McGraw-Hill Companies, Inc. All rights reserved. 13 - 24
8/12/2019 ES 12 Chapter 13
25/68
Vector Mechanics for Engineers: DynamicsSeventh
Edition
Sample Problem 13.5
SOLUTION:
Force exerted by the motor
the dumbwaiter velocity.Power delivered by motor is
e ual to Fv , v = 8 ft/s.
The dumbwaiterD and its load have a In the first case, bodies are in uniformcombined weight of 600 lb, while the
counterweight C weighs 800 lb.
Determine the ower delivered b the
.
motor cable from conditions for static
equilibrium.
electric motorMwhen the dumbwaiter(a) is moving up at a constant speed of
8 ft/s and (b) has an instantaneous
n t e secon case, ot o es areaccelerating. Apply Newtons
second law to each body to
2003 The McGraw-Hill Companies, Inc. All rights reserved. 13 - 25
velocity of 8 ft/s and an acceleration of
2.5 ft/s2, both directed upwards.
force.
8/12/2019 ES 12 Chapter 13
26/68
SE
8/12/2019 ES 12 Chapter 13
27/68
Vector Mechanics for Engineers: DynamicsSeventh
Edition
Sample Problem 13.5
In the second case, both bodies are accelerating. Apply
Newtons second law to each body to determine the required
motor cable force.
===2
212
sft25.1sft5.2 DCD aaa
ree- o y :
:CCy amF =+ ( ) lb5.38425.12.32
8002800 == TT
Free-body D:
:DDy amF =+ ( )5.22.32
600600=+ TF
lb1.2626.466005.384 ==+ FF( )( ) slbft2097sft8lb1.262 === DFvPower
2003 The McGraw-Hill Companies, Inc. All rights reserved. 13 - 27
( ) hp81.3
slbft550
hp1slbft2097 =
=Power
V M h i f E i D iSE
8/12/2019 ES 12 Chapter 13
28/68
Vector Mechanics for Engineers: DynamicsSeventh
Edition
Potential Energy
2121 yWyWU =
Work of the force of gravity ,W
Work is independent of path followed; dependsonly on the initial and final values of Wy.
=
= yg
potential energy of the body with respect
toforce of gravity.
2121 gg VVU =
measured is arbitrary.
2003 The McGraw-Hill Companies, Inc. All rights reserved. 13 - 28
JmN === WyVg
V t M h i f E i D iSE
8/12/2019 ES 12 Chapter 13
29/68
Vector Mechanics for Engineers: DynamicsSeventh
Edition
Potential Energy
Previous expression for potential energy of a body
with respect to gravity is only valid when the
wei ht of the bod can be assumed constant.
For a space vehicle, the variation of the force of
ravit with distance from the center of the earth
should be considered.
Work of a ravitational force
1221
r
GMm
r
GMmU =
Potential energy Vg when the variation in the
force of gravity can not be neglected,
2003 The McGraw-Hill Companies, Inc. All rights reserved. 13 - 29r
WR
r
GMmVg ==
V t M h i f E i D iSE
8/12/2019 ES 12 Chapter 13
30/68
Vector Mechanics for Engineers: Dynamicseventh
dition
Potential Energy
Work of the force exerted by a spring depends
only on the initial and final deflections of the
s rin
2
2212
121
21 kxkxU =
The potential energy of the body with respect
to the elastic force,
21=
( ) ( )2121
2
ee
e
VVU =
Note that the preceding expression for Ve isvalid only if the deflection of the spring is
measured from its undeformed position.
2003 The McGraw-Hill Companies, Inc. All rights reserved. 13 - 30
V t M h i f E i D iSe
Ed
8/12/2019 ES 12 Chapter 13
31/68
Vector Mechanics for Engineers: Dynamicseventh
dition
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.
( ) ( )22211121 ,,,, zyxVzyxVU =Such forces are described as conservative forces.
,
0= rdF
Elementary work corresponding to displacementbetween two neighboring points,
( ) ( )
zxdV
dzzdyydxxVzyxVdU
,,
,,,,
=
+++=
dzz
Vdy
y
Vdx
x
VdzFdyFdxF zyx
+
+
=++
2003 The McGraw-Hill Companies, Inc. All rights reserved. 13 - 31
Vz
V
y
V
x
VF grad=
+
+
=
Vector Mechanics for Engineers: DynamicsSe
Ed
8/12/2019 ES 12 Chapter 13
32/68
Vector Mechanics for Engineers: Dynamicseventh
dition
Conservation of Energy
Work of a conservative force,
2121 VVU =
,
1221 TTU = Follows that
constant
2211
=+=
+=+
VTE
VTVT
When a article moves under the action ofWVT == 0 conservative forces, the total mechanical
energy is constant.WVT =+ 11
W2 1
WVT
gg
mv
=+
====
22
222
22
r c on orces are no conserva ve. o a
mechanical energy of a system involving
friction decreases.
2003 The McGraw-Hill Companies, Inc. All rights reserved. 13 - 32
Mechanical energy is dissipated by friction
into thermal energy. Total energy is constant.
Vector Mechanics for Engineers: DynamicsSe
Ed
8/12/2019 ES 12 Chapter 13
33/68
Vector Mechanics for Engineers: Dynamicseventh
dition
Motion Under a Conservative Central Force When a particle moves under a conservative central
force, both the principle of conservation of angular
momentum
and the principle of conservation of energy
sinsin000
rmvmvr =
r
GMm
mvr
GMm
mv
VTVT
=
+=+
2
2
12
02
1
00
may be applied.
, .
At minimum and maximum r, = 90o. Given the
launch conditions the e uations ma be solved for
2003 The McGraw-Hill Companies, Inc. All rights reserved. 13 - 33
rmin, rmax, vmin, and vmax.
Vector Mechanics for Engineers: DynamicsSe
Ed
8/12/2019 ES 12 Chapter 13
34/68
Vector Mechanics for Engineers: Dynamicseventh
dition
Sample Problem 13.6
SOLUTION:
Apply the principle of conservation of
energy between positions 1 and 2.
The elastic and gravitational potential
the given information. The initial kinetic
energy is zero.
co ar s es w ou r c on
along a vertical rod as shown. The
spring attached to the collar has an
Solve for the kinetic energy and velocity
at 2.
.
constant of 3 lb/in.
If the collar is released from rest at
2003 The McGraw-Hill Companies, Inc. All rights reserved. 13 - 34
pos t on 1, eterm ne ts ve oc ty a ter
it has moved 6 in. to position 2.
Vector Mechanics for Engineers: DynamicsSev
Ed
8/12/2019 ES 12 Chapter 13
35/68
Vector Mechanics for Engineers: Dynamicsventh
ition
Sample Problem 13.6SOLUTION:
Apply the principle of conservation of energy between
positions 1 and 2.
Position 1: ( )( )
lbft20lbin.24
lbin.24in.4in.8in.lb3
1
2
2
12
12
1
=+=+=
===
VVV
kxV
ge
e
1=
Position 2: ( )( )
( )( )
2
212
221
lbin.120in.6lb20
lbin.54in.4in.01in.lb3
WyV
kxVe
===
===
22
22
222
12
2
311.0201
lbft5.5lbin.6612054
vvmvT
VVV ge
===
===+=
.
Conservation of Energy:
2211 +=+ VTVT
2003 The McGraw-Hill Companies, Inc. All rights reserved. 13 - 35
.. 2 =+ v
= sft91.42v
8/12/2019 ES 12 Chapter 13
36/68
Vector Mechanics for Engineers: DynamicsSev
Edi
8/12/2019 ES 12 Chapter 13
37/68
Vector Mechanics for Engineers: Dynamicsventh
tion
Sample Problem 13.7SOLUTION:
Setting the force exerted by the loop to zero, solve for the
minimum velocity atD.
:nn maF =+
( )( ) 222 sft4.64sft32.2ft2 ===
==
rgv
rvmmgmaW
D
Dn
Apply the principle of conservation of energy between
pointsA andD.
( ) 18ftlb360 221211 ==+=+= xxkxVVV e
01=T
( )( ) lbft2ft4lb5.002 ==+=+= ge WyVVV
( ) lbft5.0sft4.64sft2.32lb5.0
21 22
22
21
2 === DmvT
2003 The McGraw-Hill Companies, Inc. All rights reserved. 13 - 37
25.0180 22211
+=+
+=+
x in.47.4ft3727.0 ==x
Vector Mechanics for Engineers: DynamicsSev
Edit
8/12/2019 ES 12 Chapter 13
38/68
Vector Mechanics for Engineers: Dynamicsventh
tion
Sample Problem 13.9SOLUTION:
For motion under a conservative central
energy and conservation of angular
momentum may be applied simultaneously.
A satellite is launched in a direction
parallel to the surface of the earth
minimum and maximum altitude to
determine the maximum altitude.
with a velocity of 36900 km/h from
an altitude of 500 km.
Determine a the maximum altitude
Apply the principles to the orbit insertion
point and the point of minimum altitude to
determine maximum allowable orbit
reached by the satellite, and (b) themaximum allowable error in the
direction of launching if the satellite
nser on ang e error.
2003 The McGraw-Hill Companies, Inc. All rights reserved. 13 - 38
is to come no closer than 200 km to
the surface of the earth
Vector Mechanics for Engineers: DynamicsSev
Edit
8/12/2019 ES 12 Chapter 13
39/68
Vector Mechanics for Engineers: Dynamicsenth
tion
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:
12121
02021 r
GMmmv
r
GMmmvVTVT AAAA =+=+
onserva on o angu ar momen um:
1
0011100
r
rvvmvrmvr ==
Combinin
2001
0
1
0
02
1
202
021 2111
vr
GM
r
r
r
r
r
GM
r
rv =+
=
23122622
60
0
sm10398m1037.6sm81.9
sm1025.10hkm36900
===
==
RGM
v
2003 The McGraw-Hill Companies, Inc. All rights reserved. 13 - 39
km60400m104.60 6
1
==r
Vector Mechanics for Engineers: DynamicsSeve
Edit
8/12/2019 ES 12 Chapter 13
40/68
Vector Mechanics for Engineers: Dynamicsenth
ion
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:
min
2max21
0
202100 r
GMmmvr
GMmmvVTVT AA =+=+
Conservation of angular momentum:
min
000maxmaxmin000 sinsin
r
rvvmvrmvr ==
Combining and solving for sin0,
=
=
5.1190
9801.0sin 0
= 5.11errorallowable
2003 The McGraw-Hill Companies, Inc. All rights reserved. 13 - 40
Vector Mechanics for Engineers: DynamicsSeve
Editi
8/12/2019 ES 12 Chapter 13
41/68
Vector Mechanics for Engineers: Dynamicsenth
on
Principle of Impulse and Momentum
From Newtons second law,
( ) == vmvmd
F
linear momentumt
( )2
vmddtFt
=
2t
12
1
vmvmdtF
t
=
Dimensions of the impulse of
2211
21 forcetheofimpulse
1
vmvm
FdtF
t
=+
==
Imp
Impa orce are
force*time.
Units for the impulse of a
The final momentum of the particle can be
obtained by adding vectorially its initial
force aresmkgssmkgsN 2 ==
2003 The McGraw-Hill Companies, Inc. All rights reserved. 13 - 41
momen um an e mpu se o e orce ur ng
the time interval.
8/12/2019 ES 12 Chapter 13
42/68
Vector Mechanics for Engineers: DynamicsSeve
Editio
8/12/2019 ES 12 Chapter 13
43/68
g ynth
on
Sample Problem 13.10
SOLUTION:
Apply the principle of impulse and
momentum. The impulse is equal to the
product of the constant forces and thetime interval.
An automobile weighing 4000 lb is
driven down a 5o
incline at a speed of60 mi/h when the brakes are applied,
causing a constant total braking force of
1500 lb.
Determine the time required for theautomobile to come to a stop.
2003 The McGraw-Hill Companies, Inc. All rights reserved. 13 - 43
8/12/2019 ES 12 Chapter 13
44/68
Vector Mechanics for Engineers: DynamicsSeven
Editio
8/12/2019 ES 12 Chapter 13
45/68
g ynth
on
Sample Problem 13.11
SOLUTION:
A l the rinci le of im ulse and
momentum in terms of horizontal and
vertical component equations.
velocity of 80 ft/s. After the ball is hit
by the bat, it has a velocity of 120 ft/s
in the direction shown. If the bat and
ball are in contact for 0.015 s,determine the average impulsive force
exerted on the ball during the impact.
2003 The McGraw-Hill Companies, Inc. All rights reserved. 13 - 45
8/12/2019 ES 12 Chapter 13
46/68
Vector Mechanics for Engineers: DynamicsSeven
Editio
8/12/2019 ES 12 Chapter 13
47/68
th
n
Sample Problem 13.12
SOLUTION:
Apply the principle of impulse and
momen um o e pac age-car sys em
to determine the final velocity. Apply the same principle to the package
A 10 kg package drops from a chute
into a 24 kg cart with a velocity of 3
a one o e erm ne e mpu se exer e
on it from the change in its momentum.
m/s. Knowing that the cart is initially at
rest and can roll freely, determine (a)
the final velocity of the cart, (b) the
impulse exerted by the cart on the
package, and (c) the fraction of the
initial energy lost in the impact.
2003 The McGraw-Hill Companies, Inc. All rights reserved. 13 - 47
Vector Mechanics for Engineers: DynamicsSeven
Edition
8/12/2019 ES 12 Chapter 13
48/68
th
n
Sample Problem 13.12SOLUTION:
Apply the principle of impulse and momentum to the package-cart
system to determine the final velocity.
y
x
2211 vmmvm cpp
+=+ Imp
=
( )( ) ( ) 2kg25kg1030cosm/s3kg10 v
cpp
+=
m/s742.02=v
2003 The McGraw-Hill Companies, Inc. All rights reserved. 13 - 48
Vector Mechanics for Engineers: DynamicsSevent
Edition
8/12/2019 ES 12 Chapter 13
49/68
th
n
Sample Problem 13.12
Apply the same principle to the package alone to determine the impulse
exerted on it from the change in its momentum.
x
y
2211 vmvm pp
=+ Imp
x components:
( )( ) ( ) 2
21
kg1030cosm/s3kg10
cos
vtF
vmtvm
x
pxp
=+
=+
sN56.18 =tFx
com onents: 030sin =+ tFvm
( )( ) 030sinm/s3kg10 =+ tFy sN15 =tFy
2003 The McGraw-Hill Companies, Inc. All rights reserved. 13 - 49
s.ss.21 =+== ttmp
Vector Mechanics for Engineers: DynamicsSevent
Edition
8/12/2019 ES 12 Chapter 13
50/68
th
n
Sample Problem 13.12
To determine the fraction of energy lost,
( )( ) J45sm3kg102
212
121
1 === vmT p
( ) ( )( ) J63.9sm742.0kg25kg10 2212
221
1 =+=+= vmmT cp
786.0J45
.
1
21 =
=
T
2003 The McGraw-Hill Companies, Inc. All rights reserved. 13 - 50
Vector Mechanics for Engineers: DynamicsSevent
Edition
8/12/2019 ES 12 Chapter 13
51/68
h
Impact
Impact: Collision between two bodies which
occurs during a small time interval and during
which the bodies exert large forces on each other.
Line of Impact: Common normal to the surfacesin contact during impact.
Central Impact: Impact for which the mass
centers of the two bodies lie on the line of impact;
otherwise, it is an eccentric impact..
Direct Central Impact
Direct Impact: Impact for which the velocities of
the two bodies are directed along the line of
mpac .
Oblique Impact: Impact for which one or both of
2003 The McGraw-Hill Companies, Inc. All rights reserved. 13 - 51
Oblique Central Impact
impact.
Vector Mechanics for Engineers: DynamicsSeventh
Edition
8/12/2019 ES 12 Chapter 13
52/68
h
Direct Central Impact
Bodies moving in the same straight line,
vA > vB .
period of deformation, at the end of which,
they are in contact and moving at acommon velocit .
Aperiod of restitution follows during
which the bodies either regain their
deformed.
Wish to determine the final velocities of the
.
two body system is preserved,
BBBBBBAA vmvmvmvm +=+
2003 The McGraw-Hill Companies, Inc. All rights reserved. 13 - 52
A second relation between the final
velocities is required.
Vector Mechanics for Engineers: DynamicsSeventh
Edition
8/12/2019 ES 12 Chapter 13
53/68
h
Direct Central Impact
= nrestitutiooftcoefficiene AAA
10
==
e
uv
vu
Pdt A
A
Period of restitution: AAA vmRdtum =
B uv
e
=Bvu
Combining the relations leads to the desired ( )BAAB vvevv = .
Perfectly plastic impact, e = 0: vvv AB == ( )vmmvmvm BABBAA +=+
2003 The McGraw-Hill Companies, Inc. All rights reserved. 13 - 53
Perfectly elastic impact, e = 1:
Total energy and total momentum conserved.BAAB vvvv =
Vector Mechanics for Engineers: DynamicsSeventh
Edition
8/12/2019 ES 12 Chapter 13
54/68
h
Oblique Central Impact
Final velocities are
unknown in magnitude
.
equations are required.
No tangential impulse component;
tangential component of momentum
( ) ( ) ( ) ( )tBtBtAtA vvvv ==
.
Normal component of total
momentum of the two particles is
( ) ( ) ( ) ( )nBBnAAnBBnAA vmvmvmvm +=+
conserve .
Normal components of relative
velocities before and after impact( ) ( ) ( ) ( )
nBnAnAnB vvevv =
2003 The McGraw-Hill Companies, Inc. All rights reserved. 13 - 54
are related by the coefficient of
restitution.
Vector Mechanics for Engineers: DynamicsSeventh
Edition
Obli C t l I t
8/12/2019 ES 12 Chapter 13
55/68
Oblique Central Impact
Block constrained to move along horizontal
sur ace.
Impulses from internal forces FF and
exerted by horizontal surface and directed
along the vertical to the surface.
ext
Final velocity of ball unknown in direction and
magnitude and unknown final block velocity
. .
2003 The McGraw-Hill Companies, Inc. All rights reserved. 13 - 55
Vector Mechanics for Engineers: DynamicsSeventh
Edition
Obli C t l I t
8/12/2019 ES 12 Chapter 13
56/68
Oblique Central Impact
Tangential momentum of ball is
conserved.
( ) ( )tBtB vv =
=and ball is conserved.
xx
Normal component of relative ( ) ( ) ( ) ( )nBnAnAnB vvevv =
by coefficient of restitution.
2003 The McGraw-Hill Companies, Inc. All rights reserved. 13 - 56
Note: Validity of last expression does not follow from previous relation for
the coefficient of restitution. A similar but separate derivation is required.
Vector Mechanics for Engineers: DynamicsSeventh
Edition
P bl I l i E d M t
8/12/2019 ES 12 Chapter 13
57/68
Problems Involving Energy and Momentum
Three methods for the analysis of kinetics problems:
- Direct application of Newtons second law
- e o o wor an energy
- Method of impulse and momentum
Select the method best suited for the problem or part of a problem
under consideration.
2003 The McGraw-Hill Companies, Inc. All rights reserved. 13 - 57
Vector Mechanics for Engineers: DynamicsSeventh
Edition
Sample Problem 13 14
8/12/2019 ES 12 Chapter 13
58/68
Sample Problem 13.14
SOLUTION:
Resolve ball velocity into components
normal and tan ential to wall.
Impulse exerted by the wall is normalto the wall. Component of ball
momen um angen a o wa s
conserved.
Assume that the wall has infinite mass,vertical wall. Immediately before the
ball strikes the wall, its velocity has a
ma nitude v and forms an le of 30o
so that wall velocity before and after
impact is zero. Apply coefficient of
restitution relation to find change in
with the horizontal. Knowing thate = 0.90, determine the magnitude and
direction of the velocity of the ball as
norma re a ve ve oc y e ween wa
and ball, i.e., the normal ball velocity.
2003 The McGraw-Hill Companies, Inc. All rights reserved. 13 - 58
it rebounds from the wall.
Vector Mechanics for Engineers: DynamicsSeventh
Edition
Sample Problem 13 14
8/12/2019 ES 12 Chapter 13
59/68
Sample Problem 13.14
SOLUTION: Resolve ball velocity into components parallel and
perpendicular to wall.
Component of ball momentum tangential to wall is conserved.
vvvvvv tn 500.030sin866.030cos ====
vvv tt .==
Apply coefficient of restitution relation with zero wall
t
ve ocity.
( )
( ) vvv
vev
n
nn
779.0866.09.0
00
==
=n
+= 500.0779.0 vvv tn
2003 The McGraw-Hill Companies, Inc. All rights reserved. 13 - 59
=
= 7.32500.0
.tan926.0 1vv
Vector Mechanics for Engineers: DynamicsSeventh
Edition
Sample Problem 13 15
8/12/2019 ES 12 Chapter 13
60/68
Sample Problem 13.15
SOLUTION:
Resolve the ball velocities into components
normal and tangential to the contact plane.
Tangential component of momentum for
The magnitude and direction of the
velocities of two identical
.
Total normal component of the momentum
of the two ball system is conserved.frictionless balls before they strike
each other are as shown. Assuming
e = 0.9, determine the magnitude
The normal relative velocities of the
balls are related by the coefficient of
an rect on o t e ve oc ty o eac
ball after the impact.
.
Solve the last two equations simultaneously
for the normal velocities of the balls after
2003 The McGraw-Hill Companies, Inc. All rights reserved. 13 - 60
the impact.
8/12/2019 ES 12 Chapter 13
61/68
Vector Mechanics for Engineers: DynamicsSeventh
Edition
Sample Problem 13 15
8/12/2019 ES 12 Chapter 13
62/68
Sample Problem 13.15
The normal relative velocities of the balls are related by the
coefficient of restitution.
( ) ( ) ( ) ( )= nBnAnBnA vvevv
( )[ ] 4.410.200.2690.0 ==
Solve the last two equations simultaneously for the normal
velocities of the balls after the impact.
( ) sft7.17= nAv ( ) sft7.23= nBv
+=
0.15
0.157.17
1
ntAv
n
+=
=
=
6.347.23
.
7.17
ans.
ntB
A
v
v
2003 The McGraw-Hill Companies, Inc. All rights reserved. 13 - 62
=
= 6.557.23
.tansft9.41Bv
Vector Mechanics for Engineers: DynamicsSeventh
Edition
Sample Problem 13 16
8/12/2019 ES 12 Chapter 13
63/68
Sample Problem 13.16
SOLUTION:
Determine orientation of impact line of
The momentum component of ballAtangential to the contact plane is
conserve .
The total horizontal momentum of the
two ball system is conserved.BallB is hanging from an inextensible
cord. An identical ballA is released
from rest when it is just touching the
The relative velocities along the line of
action before and after the impact are
cord and acquires a velocity v0before
striking ballB. Assuming perfectly
elastic impact (e = 1) and no friction,
.
Solve the last two expressions for the
velocity of ballA along the line of action
2003 The McGraw-Hill Companies, Inc. All rights reserved. 13 - 63
eterm ne t e ve oc ty o eac a
immediately after impact.
an t e ve oc ty o a w c s
horizontal.
8/12/2019 ES 12 Chapter 13
64/68
Vector Mechanics for Engineers: DynamicsSeventh
Edition
Sample Problem 13 16
8/12/2019 ES 12 Chapter 13
65/68
Sample Problem 13.16
The relative velocities along the line of action beforeand after the impact are related by the coefficient of
restitution.
( )( ) 0
0
866.05.0030cos30sin
vvvvvv
vvevv
AB
nAB
nBnAnAnB
==
=
Solve the last two expressions for the velocity of ball
A along the line of action and the velocity of ballB
.
( ) 00 693.0520.0 vvvv BnA ==
=
==
=
10
00
1.465.0
52.0tan721.0
..
vv
vvv
A
ntA
2003 The McGraw-Hill Companies, Inc. All rights reserved. 13 - 65
=
==
0693.0
..
vvB
Vector Mechanics for Engineers: DynamicsSeventh
Edition
Sample Problem 13.17
8/12/2019 ES 12 Chapter 13
66/68
Sample Problem 13.17
SOLUTION:
Apply the principle of conservation of
block at the instant of impact.
Since the impact is perfectly plastic, the
oc an pan move toget er at t e same
velocity after impact. Determine that
velocity from the requirement that the
A 30 kg block is dropped from a height
of 2 m onto the the 10 kg pan of a
conserved.
Apply the principle of conservation of.
perfectly plastic, determine themaximum deflection of the pan. The
constant of the s rin is k= 20 kN/m.
energy to determine the maximum
deflection of the spring.
2003 The McGraw-Hill Companies, Inc. All rights reserved. 13 - 66
8/12/2019 ES 12 Chapter 13
67/68
Vector Mechanics for Engineers: DynamicsSeventh
Edition
Sample Problem 13.17
8/12/2019 ES 12 Chapter 13
68/68
p
Apply the principle of conservation of energy todetermine the maximum deflection of the spring.
( ) ( )( )21231
3 J4427.41030vmmT BA =+=+=
( )( )233212
321
3
J241.01091.410200 kx
VVV eg
==+=
+=
Initial spring deflection due to ( )( )242
14
4 0
kxhWWVVV
T
BAeg ++=+=
=
( )( )m1091.4
1020
81.910 333
=
==k
Wx B ( ) ( ) 24321
34
4234
10201091.4392
1020392
xx
xxx
+=
+=
( ) ( )m230.0
10201091.43920241.0442 243
213
4
4433
=
+=+
+=+
x
xx
2003 The McGraw-Hill Companies, Inc. All rights reserved. 13 - 68
m1091.4m230.0 334== xxh m225.0=h