INTRODUCTION to BIOMECHANICS for HUMAN …health.uottawa.ca/biomech/watbiom/Solution...

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