introduction to biomechanics for human …health.uottawa.ca/biomech/watbiom/solution...
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INTRODUCTION to BIOMECHANICS for HUMAN MOTION ANALYSIS, THIRD EDITION
SOLUTIONS to ODD-NUMBERED PROBLEMS
by D. Gordon E. Robertson, PhD, FCSB
School of Human Kinetics, University of Ottawa © Copyright 2013 (revised 11 December 2013)
INTRODUCTION (p. 12)
Conversion factors are taken from Table 1.3 on page 8. 1. (a) 350 × 0.4536 × 9.81 = 1557 N (b) 6.50 / 0.4536 = 14.33 lbs. (c) 168.5 × 2.54 = 428 cm (d) (10 × 100) /2.54 = 394 inches
(e) m/s 3.31km1
m 1000s3600
hour 1mile 1
km 609.1hourmiles0.70 =×××
(f) m/s 7.128mile 1
km 609.1hourmiles0.80 =×
(g) 8.35 × 12 × 2.54 = 255 cm (h) 440 × 0.9144 = 402 m (i) (800 / 0.9144) × 3 = 2620 feet (j) 50.0 × 1.609 = 80.5 km (k) 25.0 / 1.609 = 15.54 miles (l) 3.00 × 9.81 = 29.4 newtons 3.
N 5.111281.94.113
kg 4.113lbs 1
kg 4536.01
lbs250
=×==
=×
mgW
Thus, the 1200 N person weighs more than the 250 lbs. person. 5. m 716.13cm 6.137154.2540in. 5401245ft. 45 ==×==×= Thus, 13.75 m is longer than 45 feet.
2FUNDAMENTAL CONCEPTS (p. 26)
⎟⎠⎞
⎜⎝⎛==
+==
−
xyry
yxrrx
1
22
tanθ θsin
θcos
1.
(a) o5.54)250/350(tanθ
cm 4303502501
22
==
=+=−
r (b) o6.26)00.2/000.1(tanθ
kN 24.2000.100.21
22
==
=+=−
r
(c) o6.116)00.10/0.20(tanθ
m/s 22.40.2000.101
22
−=−−=
=−+−=−
r (d) o4.63)1000/2000(tanθ
N 2240200010001
22
==
=+=−
r
(e) o5.54)250/350(tanθ
m/s 0.43)0.35(0.251
222
−==
=−+=−
r (f) o4.108)00.5/00.15(tanθ
kN 81.1500.1500.51
22
=−=
=+−=−
r
3.
(a) m/s 57.100.25sin0.25
m/s 7.220.25cos0.25
==
==o
o
y
x (b)
N 00.50.30sin00.10
N 66.80.30cos00.10
==
==o
o
y
x
(c) 2
2
m/s 7.700.45sin0.100
m/s 7.700.45cos0.100
−=−=
=−=o
o
y
x (d)
m 00.5)2/πsin(00.5m 0.0)2/πcos(00.5
====
yx
(e) N 55.80.160sin0.25
N 5.230.160cos0.25
==
−==o
o
y
x (f)
m/s 68.17)4/πsin(0.25m/s 17.68)4/πcos(0.25
====
yx
5. Sum of sides is always greater than length of hypotenuse. 7. (a) cm )00.10 ,00.10()0.400.50 ,0.200.30( =−− (b) N )300 ,0.100()400700 ,500400( −=−− 9. R = (750, 1000) N = 1250 N at 53.1 deg 11. km 93.525.575.2 22 =+=r
3
SPATIAL (3D) COORDINATE SYSTEMS (p. 30)
zzxy
yxr
PP
PPPP
PPPP
z
y
x
zyx
==θ
+=
=γ+β+α
=γ
=β=α
++=
− )/(tan
1coscoscos
/cos
/cos/cos
1
22
222
222
1.
(a)
723.0622/450cos562.0622/350cos402.0622/250cos
5.622450350250 222
==γ==β==α
=++=P
(b)
802.074.3/3cos267.074.3/1cos535.03.74/2cos
74.3312 222
==γ==β==α
=++=P
(c)
836.07.40/34cos492.07.40/20cos246.07.40/10cos
7.4034)20(10 222
==γ−=−=β−=−=α
=+−+−=P
(d)
913.05477/5000cos365.05477/2000cos1826.05477/1000cos
5477500020001000 222
==γ==β==α
=++=P
(e)
788.069.8/55cos501.069.8/35cos
358.069.8/25cos8.6955)35(25 222
==γ−=−=β
==α
=+−+=P
(f)
785.05.25/20cos588.05.25/15cos
1961.025.5/5cos5.2520155 222
==γ==β
−=−=α=++−=P
3. (a) cm 450
5.54)250/350(tancm 430350250
1
22
===θ
=+=−
z
ro (b)
m 00.36.26)2/1(tanm 24.212
1
22
===θ
=+=−
z
ro
(c)
m 0.344.243
1804.63)10/20(tan
m 4.22)20(101
22
==
+=−−=θ
=−+−=−
z
r
o
oo
(d) mm 5000
4.63)1000/2000(tanmm 224020001000
1
22
===θ
=+=−
z
ro
4
(e) cm 0.55
5.54)25/35(tan
cm 0.43)35(251
22
=−=−=θ
=−+=−
z
ro (f)
cm 0.204.108
0.1806.71)5/15(tancm 81.15155
1
22
==
+−=−=θ
=+−=−
z
r
o
oo
5. (a) cm 0.45
cm 34.140.35sin0.25cm 5.200.35cos0.25
===
==
zyx
o
o
(b) m 500.0
m 034.10.35sin100.1m 376.00.35cos100.1
===
−==
zyx
o
o
(c) cm 0.20
cm 7.470.45sin5.67cm 7.470.45cos5.67
−===
==
zyx
o
o
(d) cm 0.55
cm 6.730.120sin0.85cm 5.420.120cos0.85
=−==
−=−=
zyx
o
o
5RESOLUTION of FORCES into COMPONENTS (p. 37)
θsin θcos
weight 221
FFFFr
mGmFmgW
yx
G
==
===
1. N 49581.95.50 =×== mgW
3.
N 6699.8168.2mgW
kg 2.68lbs 2.2045
kg 1lbs. 150
=×==
=×
5.
kg 7.6381.9
625===
gWm
7.
)6.21 ,03.2()10)5(256.1 ,0)25.1(253.4( )()00.7 ,25.11()51210 ,25.1100( )(
)68.9 ,34.12()556.13 ,25.153.43( )()4.40 ,5.39()56.11230 ,53.41025( )(
−=−−+−−−+=−−=−−−−=
−=−−×−×==−+++=
RdRcRbRa
)00.18 ,00.15()]3012( ),2510[( )()00.7 ,25.11()10125 ,01025.1( )(
)50.10 ,75.8()1230(41 ),1053.2(
41 )(
)78.2 ,87.7()256.1(21 ,6.5)53.4(
21 )(
=−−−=−−=+−−+−−=
=⎥⎦⎤
⎢⎣⎡ ++=
−=⎥⎦⎤
⎢⎣⎡ −−+=
RhRg
Rf
Re
9.
N 00.0
kg 0.60
N 1.980.60)81.9(61
=
=
==
space
space
moon
F
m
F
11.
2
26
2311
2
m/s 71.3
)10396.3(10419.610673.6
=
××
××=
×=
−
mars
mars
rmGg
6MOMENT of FORCE (p. 46–7)
θsinrFMFdM
==
1.
N 303650.1
500===
dMF
3. (a) N 00.00.500.1250.250.50 =−+−−=Σ= FR (b)
N.m00.550.1275.1825.10
)25.0(0.50)15.0(0.125)05.0(0.25)0.0(0.50=−+−=
−+−=Σ= iiR dFM
(c)
0.1100.012.50)/125(1.25d50.0(0.25)25.0(0.05)00)125.0(d
050.0(0.25)25.0(0.05))00(050)125.0(d0
C
C
C
A
=+=++=
=−−−∴=Σ
...
M
The force at C should be moved 11.00 cm to the right of A. 5. (a) ΣM = 0.50)20.2(250)50.2(200 −=− Therefore, Cathy will go up. (b) Jill must move 20.0 cm towards Cathy so the two moments are equal. (c)
333.00.1500.50
0)150.0(250(2.20)200(2.50)0M
=−−
=
=−−=Σ
d
d
Therefore, Ian must sit 33.3 cm from fulcrum on Cathy’s side. 7. (a) N.m 0.75)25.0(300 −=−== FdM A (b) N.m 6.35)(0.20)20cos(200θ(0.20)cos O −=−== CA FM Note, vertical component at C cancels vertical component at D and horizontal component at D has no moment about axis at A. 9. N.m 7.36)650.0(5.56 === FdM
7LAWS of STATICS (p. 62–4)
0)( 0
0 0
)(F 0
=×Σ=Σ=Σ
=Σ=Σ
−=×==Σ
FrMM
FF
kFrFrrMF
A
y
xyyxx
1. (a) N.m 00.18)0.5020.00.8035.0( kM =×−×= (b) N.m 00.13)0.3020.00.2035.0( kM =−×−×= (c) N.m 00.16)]0.300.50(30.0)0.200.80(10.0[ kM −=−×−+×−= (d) N.m 00.7)0.2030.00.2010.0( =×−×−=M (e) N.m 275)0.8550.20.50250.1( −=×−−×=M (f) N.m 5.20)0.8530.00.5010.0( kM −=×−−×−= 3.
N.m 20.5)51.46205.073.311366.0(
)73.31 ,51.46()3.34sin3.56 ,3.34cos3.56()θsin,θcos(kM
FFF−=×−×=
=== oo
5. (a) N.m 600.0N.cm 0.60)0.2500.80.3500.4( −=−=×−×=M (b) N.m 40.3N.cm 340)0.2500.80.3500.4( −=−=×−×−=M (c) N.m 40.3N.cm 340)0.25)00.8(0.3500.4( −==×−−×=M (d) N.m 600.0N.cm 0.60)0.25)00.8(0.3500.4( ==×−−×−=M 7.
cm 101.9m 019.1)81.9(0.60
30000.2==
×=
×=
mgFLC A
9.
0)()(
0
0
=×+×+=Σ
=−+=Σ
=+=Σ
ggkneekneekneecg
ykneeygy
xkneexgx
FrFrMM
mgFFF
FFF
8 15.
cableper N 392
2)81.9(0.80
02 :0
==
=−=Σ
cable
cabley
F
mgFF
17. (a) jacking up a car, prying with a bottle opener, shoveling (b) throwing a dart, kicking, jumping 21.
N 30040030sin200
0 :0
N 2.17330cos200
0 :0
21
21
2
21
21
=+−=+=
=−+=Σ
==
=
=+−=Σ
o
o
mgFF
WFFF
F
FF
FFF
xx
yyy
x
xx
xxx
23.
N 1650.4030sin250
0 :0
N 21730cos250
0 :0
=+=+=
=−−=Σ
==
=
=−=Σ
o
o
mgFF
WFFF
F
FF
FFF
xloadxknee
yloadykneey
xknee
xloadxknee
xloadxkneex
9SPATIAL (3D) LAWS of STATICS (p. 67)
kFrFrjFrFriFrFrFFFrrrkji
Fr
MMMFFF
FrMF
xyyxxzzxyzzy
zyx
zyx
zAyAxA
Zyx
)()()(
000 00 0
0)(0
−+−−−==×
=Σ=Σ=Σ=Σ=Σ=Σ
=×Σ=Σ=Σ
1. (a)
N.m 18.00 N.cm 1800 0.500.200.800.35
N.m 50.11N.cm 1150)0.500.400.900.35(N.m 00.14N.cm 14000.800.400.900.20
==×−×=
−=−=×−×−=−=−=×−×=
z
y
x
M
MM
(b)
N.m 13.00 N.cm 1300 0.300.200.200.35
N.m 8.27N.cm 2780)0.300.400.450.35(N.m 000.1N.cm 0.1000.200.400.450.20
==−×−×=
−=−=−×−×−===×−×=
z
y
x
M
MM
(c)
N.m 16.00 N.cm 16000.200.300.10000.10
N.m 50.13N.cm 1350)0.200.0)0.13500.10(N.m 5.40N.cm 40500.1000.00.1350.30
)0.135 ,0.100 ,0.20()0.450.90 ,0.200.80 ,0.300.50(
−=−=×−×−=
==×−×−−===×−×==++−=+
z
y
x
M
MM
DC
(d)
N.m 115.0 N.cm 11500 0.500.1500.800.50
N.m 0.45N.cm 4500)0.500.00.900.50(N.m 0.135N.cm 135000.800.00.900.150
)0.0 ,0.150 ,0.50()0.0 ,0.30 ,00.10(55
−=−=×−×−=
==×−×−−===×−×=
−=−=
z
y
x
M
MMb
(e)
N.m 5.00 N.cm 500 0.500.500.800.25
N.m 50.2N.cm 250)0.500.400.900.25(N.m 00.13N.cm 13000.800.400.900.50
)0.04 ,0.50 ,0.25()0.040.0 ,0.3020.0 ,00.100.35(
−=−=×−×=
−=−=×−×−===×−×==++−=+
z
y
x
M
MM
ba
(f)
N.m 70.0 N.cm 70003000.3020000.10
N.m 0.45N.cm 4500)3000.0)45000.10(N.m 0.135N.cm 135002000.04500.30
)450 ,200 ,300()0.45 ,0.20 ,0.30(1010
==−×−×−=
==−×−×−−===×−×=
−=−=
z
y
x
M
MM
D
103.
N 250N 600N 350 0
N 0.65
N 65.0 0N 0.150
N 0.1500N 6000
1
1
1
1
1
1
21
=∴−+==Σ
−=∴
+==Σ−=∴
+==Σ−+==Σ
z
zz
y
yy
x
xx
FFF
F
FFF
FFkFFF
DRY FRICTION (pp. 78–9)
normalkinetickinetic
normalstaticstatic
FFFF
µµ
==
1.
N 4415.49090.0µN 4665.49095.0µ
5.49081.9500 :0
=×===×==
=×===−=Σ
normalkinetickinetic
normalstaticstatic
normal
normaln
FFFF
mgFmgFF
3.
direction negativein N 8.1884.34390.0µ:motionin isobject Since
4.34381.9350 :0
=×==
=×===−=Σ
normalkinetickinetic
normal
normaln
FF
mgFmgFF
5.
incline theup N 0.12725882.081.95015sin: thatsee we triangle thengconstructi be
0 :0:rule trianglethe
applycan you staticsfor zero toaddmust they and forces only three are thereSince
=××==
=++=Σ
WF
WFFF
friction
normalfriction
7.
2.1174.14680.0µ4.146sin45 20012cos81.930
0sin4512cos
:0
=×===−××=
=+−
=Σ
normalstaticstatic
normal
appliednormal
n
FFF
FmgF
F
oo
oo
incline. down the N 2.80 isfriction an smaller th is valueabsulute thissince
2.80cos45 20012sin81.930
0cos4512sin
:0 Assume
static
equilbrium
appliedequilbrium
t
F
F
FmgF
F
−=−××=
=+−
=Σ
oo
oo
119.
).(friction moving isbody thean greater th is valueabsolute since
2.487cos13 500
0cos13 :0 Assume
N 4.48260380.0µ0.603sin13 50081.950
0sin13 :0
kineticstatic
equilbrium
appliedequilbriumt
normalstaticstatic
normal
appliednormaln
FF
F
FFF
FFF
FmgFF
=
−=−=
=+=Σ
=×===+×=
=−−=Σ
o
o
o
o
11.
N 5.16641.4310.12310 sin250
010 cos :0
. equalmust friction themovingbox get the To
N 10.1232.246500.0µ2.246cos10 250
010 cos :0
=+=+=
=+−−=Σ
=×====
=−=Σ
o
o
o
o
staticapplied
appliedstatict
static
normalstaticstatic
normal
normaln
FF
FWFF
F
FFF
WFF
13.
N 1324)81.90.150(900.0
N 1398)81.90.150(950.0=×=µ=
=×=µ=
normalkinetickinetic
normalstaticstatic
FFFF
15.
204.01.980.20
255.01.980.25
N 1.9881.900.10
===µ
===µ
=×=
normal
kinetickinetic
normal
staticstatic
normal
FF
FF
F
17.
731.03.294
215
765.03.294
225N 3.29481.90.30
===µ
===µ
=×=
normal
kinetickinetic
normal
staticstatic
normal
FF
FF
F
19. No. By definition the coefficient of static friction only occurs at the instant the slipping
occurs when the maximum friction also occurs.
12LINEAR KINEMATICS (pp. 94–5)
tvvss
ssavv
attvss
atvv
fiif
ifif
iif
if
)(½
(2
½
)22
2
++=
−+=
++=
+=
1. (a)
0 to 40 m: )(2
22
if
if
ssvv
a−
=−
2
22
m/s 0125.18081
)40(29
)040(209
==
=−
−=a
40 to 70 m: a = 0 m/s2
70 to 100 m: 2½atvtss if ++=
2
2
m/s 750.016
)3630(2
)4(½)4(970100
−=−
=
++=
a
a
(b)
s 22.16433.389.8s 00.4
s 33.39
4070
s 89.80125.1
09
10070
7040
400
=++==
=−
=
=−
=−
=
−
−
−
total
if
tt
t
avv
t
(c)
m 66.1258081½00 2 =⎟
⎠⎞
⎜⎝⎛++=fs
(d)
m/s 00.64439100 =⎟
⎠⎞
⎜⎝⎛−=v
3. (a)
s 11.7005.0
50.3536.3m/s 536.3
)025)(005.0(25.3 22
=−+
=−
=
±=
−+±=
avv
t
v
if
f
(b)
m 3.3525.035
10)005.0(½105.30
½22
2
=+=+×+=
++= attvss iif
5. (a)
s 33.7300.0
20.20=
−−
=−
=a
vvt if
(b)
m 07.80.3)(7.33)½(
(7.333)20.20
½(7.333)
2
2
=−
++=
++= atvss iif
7.
ms 24.0 s 2400.01736
67.410
1736
)0500.0(267.410
)(2
:catcher 2
ms 14.40 s 40014.02894
67.410
2894
)0300.0(267.410
)(2
:catcher 1
m/s 67.416.3km/h 150
222
nd
222
st
==−−
=−
=
−=
−−
=−
−=
==−−
=−
=
−=
−−
=−
−=
=÷=
avv
t
ssvv
a
avv
t
ssvv
a
v
if
if
if
if
if
if
The difference in the accelerations is 1158 m/s2. The difference in the times is 9.60 ms.
139.
m/s36.8500.1/546.12/
m 546.12232.700.12
22
===
=⎟⎠⎞
⎜⎝⎛+=
tsv
s
Ball speed must be greater than 8.36 m/s. 11. (a)
m/s 417.0s 60
min 1min 4
m 100=×==
tsv
(b)
m 2850.1007.266
m 7.266min 4.00km 1
m 1000min 60h 1
hkm00.4
22
=+=
=×××==
total
currentriverdown
s
tvs
(c) m/s 187.1)0.6000.4/(8.284/ =×== tsv total 13.
90.40.24
)000.6)(00.2(2)(2
)(202
2
±=±=
−−−=−−=
−+=
i
ifi
ifi
v
ssav
ssav
The initial velocity must be 4.90 m/s.
14PROJECTILE MOTION (pp. 101–2)
tvvss
ssgvv
gtss
gtvv
fyiyiyfy
iyfyiyfy
iyfy
iyfy
)(½
)(222
++=
−−=
−=
−=
1. (a)
m/s 179.25sin25θsinm/s 90.245cos25θcos
m/s 0.25s 3600
h 1km 1
m 100090
=°===°==
=××
vvvv
hkm
y
x
(b)
st
tvss xixfx
402.09.24010
=−
=
−=
(c)
cm 2.24m 242.00)81.9(2)179.2(0
2
)(2
2
22
22
==+−−
=
+−
−=
−−=
iyiyfy
fy
iyfyiyfy
sgvv
s
ssgvv
(d)
m/s 6.28350.0
010=
−=xv
3. (a)
2
22
m/s 81.19m/s 81.194.392
)020)(81.9(20
)(2
−=
±=±=
−−±=
−−=
f
fy
ifiyfy
v
v
ssgvv yy
(b)
s 02.281.9
081.19=
−−−
=−
−=
gvv
t if
(c) m 04.4)02.2(20 =+=−= tvss xixtx
5. (a)
m 835.1)81.9(2
00.62
00
2
22
22
==+
++=
−
−= +
gv
gvv
ss
iy
iyfyiyfy
(b)
s 16.281.9
624.15 velocitynegative select the
m/s 24.152.232
)100)(81.9(26
)(22
22
=−
−−=
±=±=
−−=
−−=
−
−=
t
v
ssgvv
gvv
t
f
ifif
if
(c) m 082.1)16.2(500.0 === tvs xx
7. θ vx vy ymax xmax time 30̊ 8.66 5.00 1.270 8.83 1.019 45̊ 7.07 7.07 2.55 10.19 1.442 60̊ 5.00 8.66 3.82 8.83 1.765
9.
m/s 3.31319.010
s 319.081.9
013.3
13.381.9
81.9)5.10.1(202
==
=−
−−=
−
−=
−=±=
=−−=
x
if
fy
fy
v
gvv
t
v
gv
11.
fps 8.294206.012
4026.0
m/s 94.3
)795.0)(81.9(202
==
==
−=
−−=
fps
f
f
f
v
gv
t
v
v
15
ANGULAR KINEMATICS (p. 106)
t
tt
t
fiif
ifif
iif
if
)(½θθ
θθ(2
½θθ
)22
2
ω+ω+=
−α+ω=ω
α+ω+=
αω=ω +
1. (a)
22
2if
rad/s 429rad/s 3
r/s 50012
25tωωα
.
.
=π=
=−
=−
=
(b)
srevolution 00.72)5.1(½)2(20
½αωθθ2
2
=++=
++= ttiif
3.
srevolution 227
36)35.0(2100θ 2
=
+=
5.
rad/s 75.9)5(750.06
αωω=+=
+= tif
7.
rad/s 2.2189.0π6
89.0 3ω ===r
9.
rad/s 00.500.3)200.0(60.5αωω =−+=+= tinitialfinal 11.
s 33.13500.1
0.200αωω
=−
−=
−= initialfinalt
16RELATIONSHIP between LINEAR and ANGULAR MEASURES (pp. 111–2)
22
22
tr
tr
t
t
aaa
rvra
rarv
+=
=ω=
α=ω=
1. (a) m/s 25.11)15(75.0ω === rvt (b)
2
2222
22
m/s 203
75.1685.117
75.168)15(75.0ω
5.117)150(75.0α
=
+=+=
===
===
rt
r
t
aaa
ra
ra
3. (a)
m/s 00.9
deg180rad πdeg573900.0ω
=
×⎟⎠⎞
⎜⎝⎛==
srvt
(b)
2rad/s 67.65.1
010=
−=
ω−ω=α
tif
(c)
2
2222
22
m/s 4.90
0.9000.9
0.90)10(9.0ω
00.9)10(9.0α
=
+=+=
===
===
rt
r
t
aaa
ra
ra
5.
m/s 50.7
)00.10(75.0ω
==
= rvt
2
2222
2
222
m/s 0.75
5.175
m/s 500.1)00.2(75.0α
m/s 0.75)10(75.0ω
=
+=+=
===
===
rt
t
r
aaa
ra
ra
7. (a)
m/s 00.13)10(30.1ω
m/s 0.23)10(30.2ω
===
===
rv
rv
tcg
tfeet
(b)
rad/s 00.52
010ω =−
==rv
(c)
m 29.616.845.14
m 16.8)3.1(π2π2
m 45.14)30.2(π2π2
=−=
===
===
rs
rs
cg
feet
(d)
2
22
m/s 230
)00.10(30.2ω
=
== ra feetr
9. (a)
m 1714.07.175.8375.16
ω5.170.1
,ω
;ω
m/s 875.14)75.8(70.1ωm 700.170.00.1
=−=
+=+
==
====+=
r
vr
vrrv
rvr
t
t
total
(b)
2
22
m/s 12.13
)75.8(1714.0ω
=
== rar
11.
m/s 681.0
)81.6(100.0m/s 25.2)81.6(330.0
rad/s 81.6s 60
min 1r 1rad 2
minr65
=
=ω===ω=
=
×π
×=ω
rvrv
pushwheel
rim
17
LAW of ACCELERATION (p. 117)
yy
xx
maFmaF
amF
==
=
ΣΣΣ
1.
N 0.81)350.1(0.60
m/s 350.16081
)030(290
)(22
222
−=−==
−=−
=⎟⎟⎠
⎞⎜⎜⎝
⎛−
−=⎟
⎟⎠
⎞⎜⎜⎝
⎛
−
−=
maF
xxvv
aif
if
3.
N 472)81.9(40)2(40
00.21
02
=+=+=
=−
=⎟⎟⎠
⎞⎜⎜⎝
⎛ −=
−==Σ
mgmaFt
vva
mgFmaF
ylifter
yiyfy
lifteryy
5.
m 49.8)886.5(2
10002
Thus,
m/s 886.570
N 4126.07.6867.686)81.9(70
222
2
=⎟⎟⎠
⎞⎜⎜⎝
⎛−−
+=⎟⎟⎠
⎞⎜⎜⎝
⎛ −+=
−=−
=
−==Σ=×==
===
x
ixfxif
x
kineticxx
kineticnormalkinetic
normal
avv
xx
Fkinetica
FmaFFFmgF
µ
7.
s 70.2111.1
03
m/s 111.1900/1000
90010002
=−
=−
=
==
===Σ
avv
t
a
amaF
if
9.
N 00.12)67.66(180.0
m/s 67.663.0200 2
−=−==
−=−
=−
=
maFt
vva if
11.
2
2
m/s 41.820/2.168
m/s 05.220/41
N 2.1682.19628232025
N 0.41525835
−=−=
==
−=−=−+−==Σ
=−+==Σ
y
x
yy
xx
a
a
mgmaF
maF
18MOMENT of FORCE (p. 122)
[ ]
F dM IMM
FrFrFrM
aaR
xyyxza
=α=Σ=
−=×= )(
1.
2rad/s 50.172/35α ===I
M
3.
2rad/s 10.2/α
N.m 5.52)35.0(150
==
===
IM
FdM
5.
N.m 44.13)42.0(0.32α
420.000.5
45.435.2ωωα
−=−==
−=−
=−
=
IMt
if
7.
N.m 0.14035.0400 =×== FdM 9.
2kg.m 00.1020/200α/
N.m 20040.0500
===
=×==
MI
FdM
19MOMENT of INERTIA (p. 129)
2cgaxis
2cgcg
cg
mrII
mkI
/LkK
+=
=
= cg
1.
2
22
kg.m 000.1
)25.0(85.0
m 250.085.0
=
+=+=
===
mrIImI
k
cgaxis
cgcg
3.
22
2
222
kg.m 90.6)50.0(00.1290.3
kg.m 90.3)57.0(00.12
m 57.0)95.0(60.0
=+=
+=
===
===
mrII
mkI
KLk
cghip
cg
5. (a)
2
22
kg.m 0.1842.1688.15
)450.1(0.8080.15
=+=
+=+= mrII cgbar
(b)
m 444.01975.00.80
80.15===k
7.
m 1581.00.90/25.2/
kg/m 25.220/45
rad/s 0.201/20/)ωω(α
N.m 0.45)15.0(300
2
2
===
===
==−=
===
mIk
M/aI
t
FdM
if
9.
cm 2.41m 241.005.7/453.0/
kg.m 410.0
)1925.0(05.71489.02
22
====
=
+=+=
mIk
mrII
proximalproximal
cgproximal
20REACTION FORCES (p. 136)
rmvmrwFRF
lcentripeta /22 ==
−=
1.
2
2
m/s 19.158.53/)81.98.531345(
m/s 22.48.53/227/
=×−=
=−=Σ
===
==Σ
y
ygyy
gxx
xgxx
a
mamgFF
mFa
maFF
3.
deg 2.21)3875.0(tan
81.983.3111tanθ
83.31π/100θ/θ
1
221
==
×=⎟⎟
⎠
⎞⎜⎜⎝
⎛=
====
−
−
rgv
srrs
5.
km 37.6m 637181.9
25022
2
====
=
gvr
rvmmg
7.
N 686
)81.975.0(0.65
=
+=+=
−==Σ
mgmaF
mgFmaF
ygy
gyyy
9.
N 0.1755)00.2(50.3
ω/2
22tr
−=−=
−=−= mrrmvma
11.
N 720 14)81.920(75 =×== maF
21LINEAR IMPULSE and MOMENTUM (p. 150)
∫∫
−=
==
==
if mvmvFdt
tFFdt
mvp
impulse
momentum
1. N 655)909.10(60100.1
00.1200.60 −=−=⎟⎠⎞
⎜⎝⎛ −
== amF
3. N.s 5.1370)50.2(0.55Impulse =−=−== if mvmvFt
5. 7.
N 772)87.12(0.60
m/s 87.12400.0
)127.5(0
m/s 1247.5476.26
487.26)0350.1)(81.9(20
)(2
2
22
===
=−−
=−
=
−==
=−−−=
−−=
landinglanding
iflanding
fy
ifiyfy
maFt
vva
v
yygvv
9.
N 0.40050.0
)400(005.0−=
−=
−=
−=
tmvF
mvmvtF
i
if
11.
m/s 00.64/)1(0.240 =+=+=m
tFvv if
m/s 37.50.70/86.37586.3752.1)81.9(7012000
m/s 86.20.70/200
2000
===−+=
−+=
==
+=+=
∫
∫
fy
yiyfy
fx
xixfx
v
WtdtFmvmv
v
dtFmvmv
22ANGULAR IMPULSE and MOMENTUM (p. 155)
∫∫
−=
==
==
if IIMdt
tMMdt
IL
ωω
impulseangular
ωmomentumangular
1. 3. 5.
N.m.s 8.168
500.0)350.1(250)(=
== tdFtM
7.
N.m.s 8.108
5)320.0(0.68)(=
== tdFtM
9.
/skg.m 75.4)55.2(8641.1ω
kg.m 8641.1)4026.0(50.11
m4026.0)235.1(326.0
2
222
===
===
===
IL
mkI
Klk
/skg.m 6.27
)52.5(00.5ω2=
== IL
/skg.m 9.79)137.14(65.5
1rad π225.265.5ω
2==
⎟⎠⎞
⎜⎝⎛ ×==
rsrIL
23CONSERVATION of MOMENTUM (p. 163)
constantω
constant
===
===
ILL
mvpp
if
if
1. (a) /skg.m 0.40)20(2ω 2=== IL (b)
2
2
2
kg.m 504.16.26/
kg.m 333.10.30/
/kg.m 0.40ω
==
==
====
LI
LI
sLLLI
land
top
landtopstarttoptop
3. (a)
rad/s 19.1943.0/25.8ωω430.025.8
ω
===
= IL
(b)
N.m 50.165.0
019.1943.0α =⎟⎠⎞
⎜⎝⎛ −
== IM
5.
s 42.115504.02
ω0θ
θθω
5504.026.16/95.8/ω
==−
=
−=
===
pt
t
IL
f
if
7. 2kg.m 95.1025.3/6.35ω/ === LI
24WORK-ENERGY THEOREM (p. 174)
22 ω½½energy
work
ImvmgyE
EEEW if
++==
−=∆==
1.
kJ 5.83joules 583050407.789
00.12)70(½15070(9.81)1.½ 22
==+=+=+= mvmgyE
3. J 120.0 00.2)0.60(½½ 22 === mvE 5.
J 987416.31)00.2(½60min1
12
min300)00.2(½½Iw
2
22
==
⎟⎠⎞
⎜⎝⎛ ××==
srrE π
7.
kJ 4.60)1.123(5.49001.123)81.9(50
m 1.123)s 0.60(20sin00.620sin
==−=
−=−====
if mgymgyEiEfWtvy oo
9. J 313½(25.0)5½00 22 −=−=−=−=−= mvEEEW iif 11. (a)
m 00.6
50.10.9
)75.0(230
2
750.04
30
222
=−−
=−−
=−
+=
−=−
=−
=
avv
ss
tvivfa
ifif
(b) N 00.15)75.0(0.20 −=−== amF (c) J 0.90)00.6(00.15 −=−== FsW
25WORK of a FORCE or MOMENT of FORCE (p. 180–1)
θφ==
+=⋅=
φ=
)(θ sinFrMW
sFsFsFWcosFsW
moment
yyxxforce
force
1. (a)
J 1.185517.0358 =×== FsW (b)
J 218
92.22.63325.13.25=
×+×=
+= yyxx sFsFW
3.
J 69730 cos (23.0) 35.0
cos
==
φ=o
FsWforce
5.
J 491
00.102.025.245J 1962
00.102.196N 2.196800.025.245
µN 25.245
)81.9(0.25
2.0
8.0
=
××==
×===×=
==
==
=
=
µ
µ
W
sFW
FF
mgF
kinetic
kineticnormalkinetic
normal
7.
kJ 77.11J 11772)0.60(2.196
)00.6(10)81.9(20)(
===
=== sLgFsW
9.
kJ 3.137J 340 137
4000)81.9(50.3)(
===
== sLgFsW
11.
J 206
350.0)81.9(60=
== mgyW
13. (a)
J441
1500.0)81.9(300=
== mgyWtotal
(b)
J 0.54)rad 1)(600.00.90(
θ)(θ=×=
== FdMWmoment
(c)
cycles 8.17 1/6 and 8
54441
/
===
= momenttotal WWn
15. (a)
J 09020.0 (0.450) ½
½02
2
.IEEW if
−=−=
ω−=−=
(b)
s 00.3667.6
0.200αωωrad/s 667.6
450.0300.000.10α
N 00.10)800.0(50.12
µ
2
=−
−=
−=
−=
×===
−=−=
=
if
kineticnormalfriction
t
IFd
IM
FF
26POWER (p. 185)
MωP
vFvFvFP
FvPtEtWP
moment
yyxxforce
force
=
+=⋅=
φ===cos
/∆/
1.
W687)602/()840(1.98/
joules 1.9881.900.10metres 84000.60.702
=×==
=×==××=
tsWPWs
L
L
3.
W382)606/(4000)81.95.3(/)( =×××== tsgWP L 5. 0=P An isometric contraction does no mechanical work. 7.
W5.17439.30.50ω
49.3deg360rad π2deg200ω
=×==
=×=
MPs
9.
W103785.180.55ω
rad/s 85.181rad π23ω
=×==
=×=
MPrs
r
11. W9.124555.0225 =×== FvP J 43750.39.124 =×== PtW
27CONSERVATION of MECHANICAL ENERGY (p. 190)
constant== if EE
1.
m/s 67.73)81.9(22
½
J 221000.381.90.752
===
=
=××==
gyv
mgymv
mgyWG
3.
m/s 61.144.2130.65
69362
J 69365.63767.559
00.10)81.9(0.655½(65.0)4.10½
½½22
22
==×
=
=+=
+=+
+=+
=
f
f
iiff
if
v
mv
mgymvmgymv
EE
5.
m 542.1
62.1925.30
)81.9(250.5
2
½22
2
====
=
gvy
mvmgy
7.
m/s 98.2868.80.60
2662
J 266 ½
J 266452.0)81.9(0.60
½
2
2
==×
=
=
==
==
takeoff
takeoff
takeofftop
v
mv
W
mvmgyW
9.
m/s 2.329.1036100.0
85.512
J 85.51 ½
J 85.51)35.0)(81.9)(100.000.15( ½
2
2
==×
=
=
=+==
v
mv
mgymvmgy