cap 16 beer.pdf

COSMOS: Complete Online Solutions Manual Organization System

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell © 2007 The McGraw-Hill Companies.

Chapter 16, Solution 1.

( ) ( )( )( )1.5 m sin 60 294.3 N 0.75 m cos60A BM N∑ = ° − °

( )( )( )( )230 kg 4 m/s sin 60 0.75 m= °

144.96 NBN =

( )230 kg 4 m/sx BF N F∑ = − =

24.96 NF =

(a) 2 2 295.36 NA AR N F= + =

or 295 NA =R 85.2° !

1 294.3tan 85.2

24.96α −= = °

and 145.0 NB = !

(b) 24.96

0.08481294.3A

F

Nμ = = =

or 0.0848μ = !




Alternative

eff :A AM M∑ = ∑

cos60 sin 602 2

L Lmg ma° = °

So tan 60

ga =

o

( )( )( ) ( )( )( )( )294.3 N 0.75 m cos60 30 kg 0.75 m sin 60AM a∑ = ° = °

25.664 m/sa =

or 25.66 m/s=a !

(a) ( )( )230 kg 5.664 m/s 169.9 NxF F∑ = = =

(b) 169.9 N

0.5773294.3 NA

F

Nμ = = =

or 0.577μ = !




xF ma∑ =

(a) ( )cos60xF mg ma∑ = ° =

2

ga =

216.10 ft/sa = !

( )effC CM M∑ = ∑

(b) ( )( ) ( ) ( ) ( )( )1.8 lb 2 in. cos30 8 in. sin 30 8 in.C A AM R R∑ = − + ° − °

( )( )( ) ( )( )( )1.8 1.8cos30 6 in. sin 30 2 in.a a

g g

⎛ ⎞ ⎛ ⎞−= ° − °⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

( ) ( )3.6 2.9282 11.1531A

aR

g− + = −

13.6 11.1531

20.675 lb

2.9282AR

⎛ ⎞− ⎜ ⎟⎝ ⎠= = −

or 0.675 lbA =R 60° !

( )1.8 lb cos30 0y A CF R R∑ = + − °=

1.5588A CR R+ =

( )1.5588 0.675 2.234 lbCR = − − =

or 2.23 lbC =R 60° !




See the free body diagram for problem 16.3

sinxF mg maθ∑ = =

sina g θ=

( ) ( )0 cos 2 in. sin in.G C CM R Rθ θ∑ = = − 6

(a) tanθ 1=3

or θ = °!18.43

(b) ( ) ( )2 232.2 ft/s sin18.43 10.180 ft/sa = ° =

or 210.18 ft/s=a 18.43° !




(a) If rear-wheel brakes fail to operate:

( ) ( ) ( ) ( )eff : 12 ft 5 ft 4 ftA A BM M N W maΣ = Σ − =

5 112 3B

WN W ag

= +

( )eff : , x x B k BWF F F ma N ag

µΣ = Σ = =

5 10.69912 3

W WW a ag g

+ =

( )2

2

50.699 32.2 ft/s12 12.227 ft/s

1 0.233a a

= =

−

Uniformly accelerated motion

( ) ( )2 2 2 20 2 0 30 ft/s 2 12.227 ft/sv v ax x= + = −

36.8 ftx = ! (b) If front-wheel brakes fail to operate:

( ) ( ) ( ) ( )eff : 7 ft 12 ft 4 ftB B AM M W N maΣ = Σ − =

7 112 3A

WN W ag

= −

continued



( )eff : , x x A k AWF F F ma N ag

µΣ = Σ = =

7 10.69912 3

W WW a ag g

− =

( )2

2

70.699 32.2 ft/s12 10.648 ft/s

1 0.233a a

= =

+

Uniformly accelerated motion

( ) ( )2 2 2 20 2 0 30 ft/s 2 10.648 ft/sv v ax x= + = −

42.3 ftx = !




(a) Four-wheel drive:

0: 0 y A B A BF N N W N N W mgΣ = + − = + = =

Thus: ( ) 0.80A B k A k B k A B kF F N N N N W mgµ µ µ µ+ = + = + = =

( )eff :x x A BF F F F maΣ = Σ + =

0.80mg ma=

( )2 20.80 0.80 9.81 m/s 7.848 m/sa g= = =

27.85 m/sa = ! (b) Rear-wheel drive:

( ) ( ) ( ) ( )eff : 1 m 1.5 m 0.5 mB B AM M W N maΣ = Σ − = −

0.4 0.2AN W ma= +

Thus: ( )0.80 0.4 0.2 0.32 0.16A k BF N W ma mg maµ= = + = +

( )eff :x x AF F F maΣ = Σ =

0.32 0.16mg ma ma+ =

0.32 0.84g a=

( )2 20.32 9.81 m/s 3.7371 m/s0.84

a = =

or 23.74 m/sa = !

continued



(c) Front-wheel drive:

( ) ( ) ( ) ( )eff : 2.5 m 1.5 m 0.5 mA A BM M N W maΣ = Σ − = −

0.6 0.2BN W ma= −

Thus: ( )0.80 0.6 0.2 0.48 0.16B k BF N W ma mg maµ= = − = −

( )eff :x x BF F F maΣ = Σ =

0.48 0.16mg ma ma− =

0.48 1.16g a=

( )2 20.48 9.81 m/s 4.0593 m/s1.16

a = =

or 24.06 m/s=a !




(a) Sliding impends:

( )eff: cos30y xF F F maΣ = Σ = °

( )eff

: sin 30y yF F N mg maΣ = Σ − = °

( )sin 30N m g a= + °

( )cos30

; 0.25 ; sin 30 3.3333 cos30sin 30s

F mag a a

N m g aμ °= = + ° = °

+ °

2.3867g a= 0.419g=a 30° !

(b) Tipping impends:

( )eff: 0

2 2G Gh d

M M F N⎛ ⎞ ⎛ ⎞Σ = Σ − =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

F d

N h=

; 0.3 ; 3.33F d h

N h dμ = = = !




(a) Sliding impends:

( )eff: cos30x xF F F maΣ = Σ = °

( )eff

: sin 30y yF F N mg maΣ = Σ − = − °

( )sin 30N m g a= − °

( )cos30

; 0.3sin 30s

F ma

N m g aμ °= =

− °

sin 30 3.3333 cos30g a a− ° = °

1

0.295273.3333cos30 sin 30

a

g= =

° + ° 0.295=a !

(b) Tipping impends:

( )effG GM MΣ = Σ

; 2 2

h d F dF W

N h⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

; 0.30 ; 3.33F d h

N h dμ = = = !




(a) Acceleration

( )eff:x xF FΣ = Σ

25 lb ma=

2

50 lb25 lb

32 ft/sa=

216.10 ft/sa = !

(b) For tipping to impend ; 0A =

( )eff:B BM MΣ = Σ

( ) ( )( ) ( )25 lb 50 lb 12 in. 36 in.h ma− =

For tipping to impend ; 0B =

( )eff:A AM MΣ = Σ

( ) ( )( ) ( )25 lb 50 lb 12 in. 36h ma+ = or 12 in.h =

cabinet will not tip for 12 in. 60 in.h≤ ≤ !




(a) Acceleration

(b) Tipping

0 0y A BF N N WΣ = + − =

50 lbA BN N+ =

But ,F Nμ= Thus

( )50 lb , 0.25A B s sF F μ μ+ = =


( )25 lb A BF F ma− + =

( )( ) 2

50 lb25 lb 0.25 50 lb

32.2 ft/sa

⎛ ⎞⎡ ⎤− = ⎜ ⎟⎣ ⎦ ⎝ ⎠

28.05 ft/sa =

For tipping to impend : 0AN =


( ) ( )( ) ( )( )22

50 lb25 lb 50 lb 12 in. 8.05 ft/s 36 in.

32.2 ft/sh − =

42 in.h =

For tipping to impend : 0BN =

( )eff:A AM MΣ = Σ

( ) ( )( ) ( )( )22

50 lb25 lb 50 lb 12 in. 8.05 ft/s 36 in.

32.2 ft/s+ =

6 in.h = − impossible

cabinet will not tip if 42 in.h ≤ !




0 0G AM FΣ = ⇒ =

( )( )sin 30BGF mg ma⊥Σ = ° =

2or 16.1 ft/s2

ga a= =

( )( )16 lb cos30 0BG BF F∑ = − ° =

13.856 lbBF =

(a) 216.10 ft/s=a 30°

(b) 0, 13.86 lb compressionA BF F= =



(c) Front-wheel drive:

( ) ( ) ( ) ( )eff: 2.5 m 1.5 m 0.5 mA A BM M N W maΣ = Σ − = −

0.6 0.2BN W ma= −

Thus: ( )0.80 0.6 0.2 0.48 0.16B k BF N W ma mg maμ= = − = −

( )eff:x x BF F F maΣ = Σ =

0.48 0.16mg ma ma− =

0.48 1.16g a=

( )2 20.489.81 m/s 4.0593 m/s

1.16a = =

or 24.06 m/s=a !




Analysis of linkage

Since members ACE and DCB are of negligible mass, their effective forces may also be neglected and the methods of statics may be applied to their analysis.

Free body: Entire linkage:

0:DMΣ =

( ) ( ) ( )30 sin 30 30 sin 30 0y xB E B− ° − ° =

( ) cos30 sin 30 0y xB E B− °− °= (1)

Free body: member ACE

0:CMΣ =

( ) ( )15 cos30 15 cos30 0 ;A E E A° − ° = =

carrying into eq. (1):

( ) cos30 sin 30 0y xB A B− °− °= (2)

Equations of motion for member AB



(a) + 60° ( )eff

:t tF FΣ − ∑

( ) cos30 sin 30 cos30y xB A B W ma− °− ° + °=

Recalling equation (2), we have,

cos30 cos30 cos30WW ma a gm

°= = °= °

( )232.2 ft/s cos30a = ° 227.9 ft/s=a 60° !

(b) ( )eff :B BM MΣ = ∑

( ) ( ) ( ) ( ) ( ) ( )7.5 in. 30 in. cos30 sin 60 7.5 in. cos60 2.5 in.W A ma ma− °= ° − °

But cos30ma W= ° As found above. Thus:

( )7.5 30 cos30 cos30 7.5 sin 60 2.5 cos60W A W− ° = ° ° − °

1 1 1sin 60 cos60 0.113844 cos30 4 12

A W W

= − °+ ° = °

Recalling that 20 lbW =

( )0.11384 20 lb 2.2768 lbA = =

2.28 lb=A !




Bar AC

( )10 6 N m 0.45 m 0CM RΣ = ⇒ ⋅ − =

16 N m 40

N0.45 m 3

R⋅= =

Bar AB

(a) ( )( )40N 4 9.81 sin 30 4

3xF aΣ = − ° = −

21.5717 m/sa =

or 21.572 m/sa = 30°!

(b) ( )eff:A AM M∑ = ∑

( )( ) ( )( ) ( )( )39.24 N 0.3 m cos30 0.6 m sin 30 0.3 mA EBM T ma∑ = − ° = °

( )( )( )

( )( )4 1.5717 0.15 11.772

20.841 N0.866 0.6ABT

− += =

or 20.8 NABT = !




Bar AC

0CMΣ =

( )1 10 0.45 , so0.45

MM R R= − =

( )( ) ( )( )239.24 N sin 30 4 kg 4 m/s0.45 mx

MFΣ = − ° = −

(a) 1.629 N mM = ⋅ !

(b) ( )( ) ( )( ) ( )( )39.24 N 0.3 m cos30 0.6 m sin 30 0.3 mA EBM T maΣ = − ° = °

18.036 NEBT =

or 18.04 NEBT = !




We will need the acceleration of the center of gravity of the bar and since it is translating let’s find the acceleration of point B.

180 rpm 18.85 rad/sω = =

Since ω is constant

( )( )22 0.2 m 18.85 rad/sB Bna a rω= = =

271.061 m/s=

Since AB is translating 271.061 m/sG Ba a= = 60°

Apply Newton’s second law to the system = Bar BC

( ) ( ) ( ) ( )eff0.75 0.375 sin 60 0.375B B y GM M C mg maΣ = Σ ⇒ − = − °

( )0.5 sin 60 0.5y GC mg ma= − °

( )( ) ( ) ( )0.5 7.5 9.81 7.5 71.061 sin 60 0.5= − °

193.99 N= −

194.0 NyC = !

( )eff

sin 60y y y y GF F B mg C maΣ = Σ ⇒ − + = − °

( ) ( )sin 60 7.5 9.81 193.99 7.5 71.061 sin 60y y GB mg C ma= − − °= + − °

193.99 N= −

194.0 NyB = !




( )( ) ( )( ) ( )( )0 200 0.6 0.866 300 0.6 0.866 0.6 0.5GM P= = − +∑

(a) 173.2 N=P !

( ) ( ) 2300 5 9.81 200 5 0.4yF ω= − + =∑

(b) = 15.02 rad/sω !

( )5 0.4 173.2xF P α= = =∑

(c) 2= 86.6 rad/sα !




Bar AB

Free body diagram Kinetic diagram

0;na = Released from rest

( )effx xF F=∑ ∑

8 cos3032.2 1.5 2

u gV − = °

2.3094V u− =

= 2.3094V u− At 1.5 ftu = 3.464 lb;V = − + 2p tuM M ma = =

∑

8 cos3032.2 1.5 2 2

u g uM = °

21.1547M u=

At 1.5 ft; 2.598 ft lbu M= = ⋅

Shear Moment

"



Bar BE

8= 832.2 1.5 1.5

u umg g =

8 cos30 2.309432.2 1.5 2x

u gm a u = ° =

( )effx xF F=∑ ∑

8 cos30 2.30941.5uV u − + ° =

2.3094V u=

+ 8 cos30 2.30941.5 2 2pu u uM M u = + ° =

∑

21.1547M u= −

Shear Moment

"

"




From Problem 16.15 271.061 m/sGa = 60°

261.541 m/sya =

Distributed mass per unit length 7.5 kg

10 kg m0.75 m

m

L= = =

Distributed weight per unit length ( )( )210 kg/m 9.81 m/s 98.1 N/m= =

Vertical component of effective forces ( )( )210 kg/m 61.541 m/s 615.41 N/my= ma = =

( ) ( )( ) ( )( )eff

: 98.1 0.75 0.75 615.41 N/my y y yF F B C= + + =∑ ∑

387.98y yB C+ =

From symmetry 194.0 Ny yB C= = (Same as what we got for Prob. 16.15)

Max value of bending moment occurs at G where V = 0

maxM = Area under V – Diagram from B to G

( )( )1194.0 N 0.375 m

2= 194 NBV = − !

36.4 N mmaxM = ⋅ !




Since slab is in translation, each particle has same acceleration as G, namely .a The effective forces consist of

( ) .imΔ a

The sum of these vectors is: ( ) ( )i im mΣ Δ = ΣΔa a

or since ,im mΣΔ =

( )im mΣ Δ =a a

The sum of the moments about G is: ( ) ( )i i i ir m m r′ ′Σ × Δ = ΣΔ ×a a (1)

But, 0,i im r mr′ ′ΣΔ = = because G is the mass center. It follows that the right-hand member of Eq. (1) is zero.

Thus, the moment about G of ma must also be zero, which means that its line of action passes through G and that it may be attached at G.




For centroidal rotation: ( ) ( ) 2i i i i it n

ω′ ′= + = × −a a a r rα

Effective forces are: ( ) ( )( ) ( ) ( )( )2i i i i i i i im a m m r mω α′Δ = Δ × − Δ Δ ×r rα

( ) ( )( ) ( ) 2i i i i i im m m rω′ ′Σ Δ = Σ Δ × − Σ Δa rα

( ) ( )2i i i im m rα ω′ ′= × Σ Δ − Σ Δr

Since G is the mass center, ( ) 0i im r ′Σ Δ =

∴ effective forces reduce to a couple, summing moments about G

( ) ( )( ) ( ) 21 i i i i i i i im m m rω⎡ ⎤′′ ′ ′ ′Σ × Δ = Σ × Δ × − Σ × Δ⎢ ⎥⎣ ⎦

r a r r rα

But, ( ) ( )( )2 2 0i i i i i im mω ω′ ′ ′ ′× Δ = Δ × =r r r r

and, ( )( ) ( ) 21i i i im r m r ′′ ′× Δ × = Δr α α

Thus, ( ) ( ) ( )2 21i i i i i im m r m rα α⎡ ⎤′ ′Σ × Δ = Σ Δ = Σ Δ⎣ ⎦r a

Since ( ) 2 ,im r I′Σ Δ =

the moment of the couple is I α




20 ,M I mkα α= =∑ MF from friction

( )2300 200 0.4 32FM α α− = =

(a) Kinematics 1 1 12

, 2400 80 rad/s60

tπω α ω π⎛ ⎞= = =⎜ ⎟

⎝ ⎠

1 28 st =

∴ ( )2 80300 200 0.4 287.23

28FMπ⎛ ⎞− = =⎜ ⎟

⎝ ⎠

12.77 N mFM = ⋅ !

(b) 22 2

12.768712.77 N m, rad/s

32Iα α −= − ⋅ =

Kinematics 0

2 0 80

12.7687, 80

32T

dt d Tπα ω π−= = −∫ ∫

630 s,T = after 28 st =

658 st =∴ !




( )20 200 12.5 lb ftM I mkα α= = = − ⋅∑

(a) 2500187.5 lb.ft

32.2k α⎛ ⎞ =⎜ ⎟

⎝ ⎠

Kinematics 1 1, 0d

tdt

ωα α ω= = −

( ) ( )( )3000 2 /60 100 2

30 60

π πα = =

210.4720 rad/sα =

∴ ( )( )2 2187.5

1.15308 ft10.4720 500/32.2

k = =

1.074 ftk = !

(b) 20 212.5M mk α= − =∑

( )( )2

212.5

0.69813 rad/s500/32.2 1.15308

α = − = −

Kinematics 2d

d

ωα ωθ

=

( )2

0

0 100

1000.69813

2d d

θπ

πθ ω ω

−− = =∫ ∫

1 rev

= 70,686 radians,2 rad

θπ

⎛ ⎞⎜ ⎟⎝ ⎠

= 11,250 revθ !




Kinematics: 0 360 rpm 37.699 rad/sω = =

25 rev 157.08 radθ = =

( ) ( )22 20 2 ; 37.699 2 157.08ω ω αθ θ α= + = +

2= 4.524 rad/sα − (sense opposite to θ):

2= 4.524 rad/sα

Equation of motion for the flywheel

15 in. 1.25 ftr = =

21

2I m r=

( )21 401.25

2 32.2⎛ ⎞= ⎜ ⎟⎝ ⎠

25.823 lb ft s= ⋅ ⋅

( )eff:G GM M Tr Pr Iα= − =∑ ∑

( )5.823 4.524

1.25

IT P

r

α− = =

21.075T P− = (1)

Belt Friction: We recall equation (8.14), page 351 of Statics. Using μk insted of μs, since the band brake is slipping, and noting that T1 = P and T2 = T:

( )0.35 22

1

: 1.7329kT T

e e T PT P

μ β π= = =

Substituting into (1):

1.7329 P – P = 21.075 P = 28.8 lb !




Equilibrium of lever AD

(Dimensions in mm)

0;CM =∑

( )( ) ( ) ( )2 150N 200 40 160 0T T+ − =

1 24 250 NT T− = (1)

Equation of Motion for flywheel and drum

2

I m k=

( )( )2300 kg 0.600 m=

2108 kg m= ⋅

0.100 mr =

( ) 2 1eff :C CM M T r T r Iα= − =∑ ∑

( )( )2 1 0.100 m 108T T α− =

( )( )62 1925.93 10 T Tα −= × − (2)

Belt Friction We recall equation (8.14), page 351 of Statics. Using µk instead of µs since the brake band is slipping:

2

1

kT eT

µ β= or 2 1kT T eµ β= (3)

Making 0.30 and = 180 rad in (3):kµ β π= ° =

0.302 1T T e π= 2 12.5663T T= (4)



Substituting for T2 from (4) into (1):

1 1 14 2.5663 250 N 174.38 NT T T− = =

From (1): ( )2 24 174.38 250 447.51 NT T= − =

Substituting for 1 2and into (2):T T

( )( )6 2925.93 10 447.51 174.38 , 0.2529 rad/sα −= × − =αααα

Kinematics

0 180 rpm=ωωωω 0 18.850 rad/sω = +

20.2529 rad/s=αααα 20.2529 rad/sα = −

0 : 0 18.850 0.2529t tω ω α= + = −

= 74.5 st "




( )( ) ( )2 2110 kg 0.225 m 0.25313 kg m

2GI α α α= = ⋅

( )effB BM M=∑ ∑

( )( )11 N m 0.4 0.225 0.25313BM N α= ⋅ − =∑

0.4 cos30 cos 60 98.1 0; 115.901 NyF N N N= ° + ° − = =∑

( )( )( )11 0.4 115.901 0.225 mNα = −

20.568892.2474 rad/s

0.25313= =

(a) 22.25 rad/sα = !

( ) ( )cos30 0.4 sin 30 0xF C N N= − ° + ° =∑

( )( )0.6660 0.6660 115.901C N= =

= 77.190 N

(b) 77.2 NC = (compression) !




( )( ) ( )2 2110 kg 0.225 m 16 rad/s 4.05 N m

2GI α = = ⋅

( )effB BM M= ⇒∑ ∑ ( )( )0.25 0.225 4.05 N mBM M N= − = ⋅∑

( )( ) ( )0.25 cos30 sin 30 98.1 24.525 0yF N N= ° + ° − − =∑

= 171.143 NN

(a) Now ( )( )( )4.05 N m 0.25 171.143 0.225 mM N= ⋅ +

13.677 N m= ⋅

or 13.68 N mM = ⋅ !

(b) 1 0.25= tan 14.04

1.00β − ⎛ ⎞ = °⎜ ⎟

⎝ ⎠

( )22 0.25 176.4 NF N N= + =

or 176.4 NF = 44°!




Case 1:

Case 2:

Case 3:

Ca

(a) ( )0 0 eff :M MΣ = Σ

( )( ) ( )2785 N 0.2 m 20 kg m α= ⋅

27.85 rad/s=αααα """"

(b) 3 m 15 rad0.2 m

θ = =

( )( )2 22 2 7.85 rad/s 15 radω αθ= =

215.35 rad/s=ω """"

(a) ( )0 0 eff :M MΣ = Σ

( )( )( ) ( )280 kg 9.81 m/s 0.2 m 20 0.2 mmaα= +

( )( )( )157.0 20 80 0.2 0.2α α= +

26.767 rad/sα =

26.77 rad/s=αααα """"

(b) 3 m 15 rad0.2 m

θ = =

( )( )2 22 2 6.767 rad/s 15 radω αθ= =

214.25 rad/s=ω """"

(a) ( )0 0 eff :M MΣ = Σ

( )( )( ) ( )( )( )2 2230 kg 9.81 m/s 0.2 m 150 kg 9.81 m/s 0.2 m−

( ) ( )20 230 0.2 m 150 0.2 ma aα= + +

( ) ( )2 2157.0 20 230 0.2 150 0.2α α α= + +

24.460 rad/sα = 2or 4.46 rad/s=αααα """"



Case 4:

(b) 3 m 15 rad0.2 m

θ = =

( )( )2 22 2 4.46 rad/s 15 radω αθ= =

11.57 rad/s=ωωωω """"

(a) ( )0 0 eff :M MΣ = Σ

( )( )( ) ( )240 kg 9.81 m/s 0.4 m 20 40 0.4 maα= +

( )2 2157.0 20 40 0.4 , 5.947 rad/sα α α= + =

25.95 rad/s=αααα """"

(b) 3 m 7.5 rad0.4 m

θ = =

( )( )2 22 2 5.947 rad/s 7.5 radω αθ= =

9.44 rad/s=ω """"




Kinematics Kinetics


( ) ( )A Am g r I m a rα= +

2A F A

am gr m k m ar

r⎛ ⎞= +⎜ ⎟⎝ ⎠

2

A

A F

m ga

km m

r

=⎛ ⎞+ ⎜ ⎟⎝ ⎠

(a) ( )( )

( )

22

2

18 kg 9.81 m/s1.7836 m/s

450 mm18 kg 144 kg

600 mm

a = =⎛ ⎞+ ⎜ ⎟⎝ ⎠

2or 1.784 m/sA =a !

(b) 2 2 2A BV V as+ +

For 1.8 ms =

( )( )2 2 2 20 2 1.7836 m/s 1.8 m 6.42096 m /sAV = + =

2.5339 m/sAV =

or 2.53 m/sAV = !




Kinematics Kinetics


( ) ( )A f Am g r M I m a rα− = +

A f Aa

m gr M I m arr

− = + (1)

( )221 1Case 1: 12 ft, 4.5 s; 12 ft 4.5 s

2 2y t y at a= = = ⇒ =

21.1852 ft/sa =

From (1) ( )( ) ( )( )2

22

1.1852 ft/s 20 lb20 lb 1.5 ft 1.1852 ft/s 1.5 ft

1.5 ft 32.2 ft/sfM I

⎛ ⎞ ⎛ ⎞− = +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

30 0.79013 1.1042fM I− = + (2)

( )221 1Case 2: 12 ft, 2.8 s; 12 ft 2.8 s

2 2y t y at a= = = ⇒ =

23.0612 ft/sa =

From (1) ( )( ) ( )( )2

22

3.0612 ft/s 40 lb40 lb 1.5 ft 3.0612 ft/s 1.5 ft

1.5 ft 32.2 ft/sfM I

⎛ ⎞ ⎛ ⎞− = +⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

60 2.0408 5.7041fM I− = + (3)

Subtract (2) from (3) 30 1.25067 4.5999I= +

220.309 lb ft sI = ⋅ ⋅ 2or 20.3 lb ft sI = ⋅ ⋅ !




Disk A 21

2A A AI m r=

( )eff:A AM = M∑ ∑

2

2 11

2A A A AM F r F r m r α+ − −

2 11

2 A A AA

MF F m r

rα+ − = (1)

Disk B 21

2B B BI m r=

( )eff:B BM M=∑ ∑

21 2

1

2B B B B BF r F r m r α− =

1 21

2 B B BF F m r α− = (2)

Add (1) and (2): 1 1

2 2A A A B B BA

Mm r m y

rα α= +

( )1= , Thus:

2Belt A A B B A A A B B AA

MBut r r m r m r

rα α α α α= = +

( ) 2

2,

2A B

A A AA A B A

M m m Mr

r m m rα α+= =

+

With given data: ( )( )( )2

2 2.70 N m

2 kg + 4 kg 0.2 mAα

⋅=

222.5 rad/sA =α !

From ( )0.2 m: 22.5

0.3 mA

A A B B B AB

rr r

rα α α α= = =

215.00 rad/sB =α !




04 8

3.512 12A BM T T⎛ ⎞ ⎛ ⎞Σ = + −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

2

25 6

32.2 12I α α⎛ ⎞ ⎛ ⎞= = ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

4

12A Aa r α α⎛ ⎞= = ⎜ ⎟⎝ ⎠

8

12B Ba r α α⎛ ⎞= = ⎜ ⎟⎝ ⎠

:A 10

1032.2A AF T a⎛ ⎞Σ = − = ⎜ ⎟⎝ ⎠

:B5

532.2B BF T a⎛ ⎞Σ = − = ⎜ ⎟⎝ ⎠

Elimination: ( )210 4 5 8 253.5 10 5 0.5

12 12A Ba a

g g gα⎛ ⎞ ⎛ ⎞

+ − − + =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

3.5 3.333+ 1.111 3.333g

α⎛ ⎞− −⎜ ⎟

⎝ ⎠2.222 6.25

g g

α α⎛ ⎞ ⎛ ⎞− =⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

(a) 211.76 rad/s=α !

(b) 8.78 lb , 6.22 lbA BT T= = !




Kinematics: Since the tangential accelerations of the outside of the disks are equal,

A A B Br rα α=

AB A

B

rr

α α= (1)

Kinetics:

Disk A:

( )eff :A AM MΣ = Σ

A A AM Fr I α− = (2)

Disk B:

( )eff :B BM MΣ = Σ

B B BFr I α=

212B B B BFr m r α =

1

2 B B BF m r α= (3)



Substitute for F from Eq. (3) into Eq. (2):

12 B B B A A AM m r r Iα α − =

Substitute for F from Eq. (3), and for Bα from Eq. (1).

21 12 2

AB B A A A A A

B

rM m r r m rr

α α

− =

( ) 212 A B A AM m m r α= +

( ) ( )2 22 2

AA B A A B A

M Mgm m r W W r

α = =+ +

Data: 12 lb, 6 lbA BW W= =

6 in. 0.5 ft; 4 in. 0.33333 ftA Br r= = = =

7.5 lb in. 0.625 lb ftM = ⋅ = ⋅

(a) ( )( )( )( )

22

2

2 0.625 lb ft 32.2 ft/s8.9444 rad/s

12 lb 6 lb 0.5 ftAα

⋅= =

+

28.94 rad/sA =αααα "

( )2 26 in. 8.9444 rad/s 13.417 rad/s4 in.

AB A

B

rr

α α= = =

213.42 rad/sB =αααα "

(b) ( )( )22

1 1 6 lb 0.33333 ft 13.417 rad/s2 2 32.2 ft/sB B BF m r α = =

0.41667 lbF =

or 0.417 lb=F "




Kinematics: Since the tangential acceleration of the outside of the disks are equal.

A A B Br rα α=

BA B

A

rr

α α= (1)

Kinetics: Disk A:

212A A AI m r=

( )effA AM MΣ = Σ :

A A AFr I α=

21 1 2 2A A A A A A AFr m r F m rα α= = (2)

Disk B:


B B BM Fr I α− = (3)

Substitute for F from Eq. (2) into Eq. (3)

12 A A A B B BM m r r Iα α − =



Substitute for Aα from Eq. (1), and 212B B BI m r=

21 12 2

BA A B B B B B

A

rM m r r m rr

α α

− =

( ) 212 A B B BM m m r α= +

( ) ( )2 22 2

BA B B A B B

M Mgm m r W W r

α = =+ +

Data: 12 lb, 6 lb,A BW W= =

( ) ( )6 in. 0.5 ft , 4 in. 0.33333 ftA Br r= = = =

7.5 lb in. 0.625 lb ftm = ⋅ = ⋅

(a) ( )( )( )( )

22

2

2 0.625 lb ft 32.2 ft/s20.125 rad/s

12 lb 0.33333 ftBα

⋅= =

220.1 rad/sB =αααα !

( )2 24 in. 20.125 rad/s 13.417 rad/s6 in.

BA B

A

rr

α α = = =

213.42 rad/sA =αααα !

(b) ( )( )22

1 1 12 lb 0.5 ft 13.417 rad/s2 2 32.2 ft/sA A AF m r α = =

1.2500 lb=

Friction force on disk B: 1.250 lb=F !




2

2 21 1 16 40.027605 lb ft s

2 2 32.2 12GI mr⎛ ⎞ ⎛ ⎞= = = ⋅ ⋅⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

A: 14

3 0.02760512GM F α⎛ ⎞Σ = − =⎜ ⎟⎝ ⎠

1 16 0yF NΣ = − =

Note: A Bα α= Since the cylinders have the same radius and

there is no slipping.

B: ( )1 24 4

0.2 0.02760512 12GM F N⎛ ⎞ ⎛ ⎞Σ = − =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

2 16 16yF NΣ = − −

Now 1 1 24 4 0.8

312 12 12

F F N⎛ ⎞ ⎛ ⎞ ⎛ ⎞− = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

or 1 7.70 lbF =

(a) 215.70rad/sA =α !

215.70 rad/sB =α !

(b) 1min

1

7.70( )

16sF

Nμ= =

0.4813sμ =

or 0.481sμ = !




22 21 1 16 4

0.027605 lb ft s2 2 32.2 12GI mr

⎛ ⎞ ⎛ ⎞= = = ⋅ ⋅⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

A: 4

3 ft lb ft (7.2 lb)12GM⎛ ⎞Σ = ⋅ − ⎜ ⎟⎝ ⎠

0.027605 Aα=

221.735 rad/sAα =

or 221.7 rad/sA =α !

B: ( )4ft 7.2 lb

12BM⎛ ⎞Σ = ⎜ ⎟⎝ ⎠

4

ft (4.8 lb) = 0.02760512 Bα⎛ ⎞− ⎜ ⎟⎝ ⎠

228.98 rad/sBα =

or 229.0 rad/sB =α !




21

2G k BM N r mrμ αΣ = =

2 k BN

mr

μα =

sin 2 cos 2 sin 0x B k BF N N mgφ μ φ φΣ = − + + =

sin

sin 2 cos 2Bk

mgN

φφ μ φ

=−

or ( )2 sin

sin 2 cos 2k

k

g

r

μ φαφ μ φ

=−

!




21

sin2BM mgr mrφ αΣ = =

2 sing

r

φα = !




While slipping occurs: k kF N Pµ µ= =

Disk A:

( )eff :A A A A AM M Fr I αΣ = Σ =

212k A A A APr m rµ α=

2 kA

A A

Pm r

µ=αααα (1)

Disk B:

( )eff :B B B B BM M Fr I α+ Σ = Σ =

212k B B B BPr m rµ α=

2 kB

B B

Pm r

µ=αααα (2)

At any time t

20 k

A AA A

Pt tm r

µω α= + = (3)

0 02 k

B BB B

Pt tm r

µω ω α ω= − = − (4)

Slipping ends when A A B Br rω ω=

( )0A A B Btr t rα ω α= −

( ) 0A A B B Br r t rα α ω+ =



Substitute from Eqs (1) + (2): 02 2k k

A B BA A B B

P Pr r t rm r m r

µ µ ω

+ =

00

1 1 12 ; 1 12B

k BA B k

A B

rP t r tm m P

m m

ωµ ωµ

+ = = ⋅

+

0

2B A B

k A B

r m mtP m m

ωµ

= ⋅+

Eq (3): 022

k B A BA A

A A k A B

P r m mtm r P m m

µ ωω αµ

= = ⋅ ⋅ +

0B B

AA A B

r mr m m

ω ω= ⋅+

( )is independent of , QEDA kω µ

Eq (4): 00 0

22

k B A BB B

B B k A B

P r m mtm r P m m

µ ωω ω α ωµ

= − = − ⋅ ⋅ +

0 1 AB

A B

mm m

ω ω

= − +

0 0A B A B

BA B A B

m m m mm m m m

ω ω ω+ −= =+ +

0

1B

A

B

mm

ωω =+

Bω depends only upon 0ω and A

B

mm

(QED)




(a) Angular accelerations

While slipping occurs:

( )0.25 5 lb 1.25 lbk kF N Pµ µ= = = =

Disk A:

( ) 2eff

1:2A A A A A A AM M Fr I m rα αΣ = Σ = =

12 A A AF m r α=

21 12 lb 61.25 lb ft2 1232.2 ft/s Aα =

213.417 rad/sAα = 213.42 rad/sA =αααα !

Disk B:

( ) 2eff

1:2B B B B B B B BM M Fr I m rα αΣ = Σ = =

12 B B BF m r α=

21 30 lb 101.25 lb ft2 1232.2 ft/s Bα =

23.22 rad/sBα = 23.22 rad/sB =αααα !



(b) Angular velocities

( )02750 rpm 78.5398 rad/s60Aπω = =

; ( )0 0Bω =

When disks stop sliding

A BP P A A B Bv v r rω ω= =

( ) ( )0A A A B Bt r t rω α α − =

( )( ) ( ) ( )( ) ( )2 278.5398 rad/s 13.417 rad/s 6 in. 3.22 rad/s 10 in. 4.1813 st t t − = ⇒ =

( )( )278.5398 rad/s 13.417 rad/s 4.1813 s 22.439 rad/sAω = − =

6022.439 rad/s 214.2767 rpm2Aωπ

= =

or 214 rpmA =ωωωω !

( )6 0.6 214.2767 128.5660 rpm10B Aω ω= = =

or 128.6 rpmB =ωωωω !




Based on the solution for P16.39

( ) ( )0 0

20; 750 rpm 25 rpm

60A Bπω ω π⎛ ⎞= = =⎜ ⎟

⎝ ⎠

Eq. (1): ( )0; A A B B A A B B Br r tr t rω ω α ω α⎡ ⎤= = −⎣ ⎦

( )( ) [ ]( )13.417 6 in. 25 3.22 10 in.t tπ= −

6.9688 st =

( )( ) 6013.417 6.9688 93.5 rad/s 892.86 rpm

2A Atω απ

⎛ ⎞= = = =⎜ ⎟⎝ ⎠

or 893 rpmA =ω !

( )60.6 892.86 535.716 rpm

10B Bω ω= = =

or 536 rpmB =ω !




We know that the system of effective forces can be reduced to the vector ma at G and the couple .Iα We further know from Chapter 3 of statics that a force-couple system in a plane can be further reduced to a single force.

The perpendicular distance d from G to the line of action of the single vector ma is expressed by writing

( ) ( )eff: G GM M I ma dαΣ = Σ =

2I mk

dma ma

α α= =

2

k

da

α= !




Kinematics The acceleration of iP is

/i P G= +i

a a a

( ) 2i i i i iω′ ′ ′ ′= + × + × × = + × −a a α r ω ω r a α r r

Note that i′×α r is ⊥ to i′r

Thus, the effective forces are as shown in Fig P16.47 (also shown above). We write

( ) ( ) ( )( ) ( ) 2i i i i i i im m m m rω′ ′∆ = ∆ + ∆ × − ∆a a α r

The sum of the effective forces is

( ) ( ) ( )( ) ( ) 2i i i i i i im m m m ω′ ′Σ ∆ = Σ ∆ + Σ ∆ × − Σ ∆a a α r r

( ) ( ) ( ) ( )2i i i i i i im m m mω′ ′Σ ∆ = Σ ∆ + × Σ ∆ − Σ ∆a a α r r

We note that ( ) .im mΣ ∆ = And since G is the mass center,

( )i i im mr′Σ ∆ =r ′ 0=

Thus, ( )i im mΣ ∆ =a a (1)

The sum of the moments about G of the effective forces is:

( ) ( )( ) ( ) 2i i i i i i i i i i im m m m ω′ ′ ′ ′ ′ ′Σ × ∆ = Σ × ∆ + Σ × ∆ × − Σ × ∆r a r a r α r r r

( ) ( ) ( ) ( )2i i i i i i i i i i im m m mω′ ′ ′ ′ ′ ′ Σ × ∆ = Σ ∆ + Σ × × ∆ − Σ × ∆ r a r a r α r r r



Since G is the mass center, 0i im′Σ ∆ =r

Also, for each particle, 0i i′ ′× =r r

Thus,

( ) ( )i i i i i i im m′ ′ ′ Σ × ∆ = Σ × × ∆ r a r α r

Since α ⊥ ,i′r we have ( ) 2i i ir α′ ′× × =r α r and

( ) ( ) ( ) 2 2i i i i i i im r m r mα′ ′ ′Σ × ∆ = Σ ∆ = Σ ∆r a α

Since 2i ir m I′Σ ∆ =

( )i i im I α′Σ × ∆ =r a (2)

From Eqs. (1) and (2) we conclude that system of effective forces reduce to ma attached at G and a couple .Iα




Hoop: 2I mr=


P ma=

P

m=a

( ) 2eff

:G GM M Pr I mrα αΣ = Σ = =

P

mr=α

(a) 2AP P P

a a r rm mr m

α ⎛ ⎞= + = + =⎜ ⎟⎝ ⎠

( )2 20.75 lb2 32.2 ft/s 8.05 ft/s

6Aa = =

28.05 ft/sA =a !

(b) 0BP P

a a r rm mr

α ⎛ ⎞= − = − =⎜ ⎟⎝ ⎠

0B =a !




Let 1 time required for 360 rotationt = °

2 21 1

1 1 2; 2

2 2

Pt g t

wrθ α π ⎛ ⎞= = ⎜ ⎟

⎝ ⎠

21

2 wrt

Pg

π=

1 distance moves during 360 rotationx G= °

21 1

1 1 2

2 2

P wrx at g r

w Pg

π π⎛ ⎞⎛ ⎞= = =⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

1x rπ= Q.E.D.!




( )( )22

23 kg 1 m0.25 kg m

12 12

mLI = = =

( )eff:x xF F P maΣ = Σ =

( )2 3 kgN a=

20.66667 m/sa =

GMΣ = ( )eff: ;

2GL

M P I α⎛ ⎞ =⎜ ⎟⎝ ⎠

( ) ( )212 N m 0.25 kg m

2α⎛ ⎞ = ⋅⎜ ⎟

⎝ ⎠

( )( )( )

22

2 N 0.5 m4 rad/s

0.25 kg mα = =

⋅

(a) ( )2 21 10.66667 m/s m 4 rad/s

2 2A a α ⎛ ⎞= + = + ⎜ ⎟⎝ ⎠

a

22.66667 m/s=

2or 2.67 m/sA =a !

(b) ( )2 2 21 10.66667 m/s m 4 rad/s 1.33333 m/s

2 2Ba a α ⎛ ⎞= + = − = −⎜ ⎟⎝ ⎠

2or 1.333 m/sB =a !




2 2120 kg, (120 kg)(0.600 m) 43.2 kg m2

ym = I = mk = = ⋅

(a) With all four rockets fired:

eff : 0 = mF F a=∑ ∑ 0=a !

eff( ) : 4G GM M Tr Iα= =∑ ∑

4(16.20 N)(0.8 m) = 43.2 α 21.200 rad/s=α

2(1.200 rad/s )= − jα !

(b) With all rockets except D:

2

eff( ) : 16.20 N = 120 = 0.1350 m/sx x xxF F a a= − −∑ ∑

eff( ) : 0 120 = 0z zzF F a a= =∑ ∑

2(0.1350 m/s )= −a i !

eff( ) : 3GGM M Tr I α= =∑ ∑

3(16.20 N)(0.8 m) = 43.2 α 2= (0.900 rad/s )− jα !




(a)

2

2 22

1 4 1 16

2 3 2 9

rI mr m mrα α α

π π

⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞= − = −⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

( )2 22

12 1.7777(1 ft) 0.5 0.11921 lb s ft

32.2α α

π⎛ ⎞ ⎛ ⎞= − = ⋅ ⋅⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

2121 lb , 2.6833 ft/s

32.2x G GF a aΣ = = =

2or 2.68 ft/sG =a i !

(b) ( )4(1)1 lb ft 0.11921

3GM απ

⎛ ⎞Σ = =⎜ ⎟⎝ ⎠

23.5603 rad/sα =

/ 2.6833C G C G= + =a a a + 4(1)

13

απ

⎛ ⎞−⎜ ⎟⎝ ⎠

2.6833C =a + 2.0493

0.634= 2ft/s

2or 0.634 ft/sC =a i !




23.5603 rad/sα =

22.6833 ft/sGa =

(a) /A G A G= + =a a a 2.6833 +

= 2.6833 + = 2.6833 + 4.194 3.560A = −a i k

2or 4.19 3.56 ft/sA = −a i k !

(b) /B G B G= + =a a a 2.6833 +

= 2.6833 +

2or 4.19 3.56 ft/sB = +a i k !

3.5601.5110

1.5110 3.5603

0.42441(3.5601(3.5603)




(a)

( ) ( )1 2/2 /2m x m x

xm

+=

( ) ( )1 0.15 1 0

0.075 m2

x+

= =

x z=

( ) ( ) ( )( )2 2 2 212 0.3 0.075 0.075 0.0375 kg m

12 2 2

m mI

⎡ ⎤⎛ ⎞= + + = ⋅⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

0 , 0x Gx GxF ma a= = =∑

24 , 2 m/sz Gz GzF ma a= − = = −∑

( )4 0.075 0.0375 ,GM α= − =∑ 28 rad/s= − jα !

(b)

( ) 20.075 2 8 0.6 2 m/s=

/B G B G= + =a a a + 45°

aB =

20.6 2.6 (m/s )B = −a i k !

Mass center at G

2

0.6 m/s2

2.6 m/s2




2

2

2800 lb lb sm 86.957

ft32.2 ft/s

⋅= = r = 2.25 in.

22 26

86.957 ft 21.739 lb ft s12

I mk⎛ ⎞= = = ⋅ ⋅⎜ ⎟⎝ ⎠

( )eff

:y yF F=∑ ∑

2 2A BT T W ma+ − =

( )2 2800 86.957A BT T a+ − = (1)

( ) ( )eff: 2 2G G B AM M T T r Iα= − =∑ ∑

( ) 2.252 ft 21.739

12B AT T α⎛ ⎞− =⎜ ⎟⎝ ⎠ (2)

Given data: TA = 690 lb, TB = 730 lb.

(a) Substitute in (2):

( ) 2.252 730 690 21.739

12α− =

15 21.739α= 20.690 rad/s=α !

(b) Substitute in (1):

( )2 690 730 2800 86.957a+ − =

40 86.957a= 20.460 ft/s=a !

W = 2800 lb




See solution of Prob. 16.50 for diagram and derivation of equations (1) and (2):

( )2 2800 86.957A BT T a+ − = (1)

( ) 2.252 21.739

12B AT T α⎛ ⎞− =⎜ ⎟⎝ ⎠ (2)

Given data: 2 2680 lb, 6 in./s 0.5 ft/sAT a= = − = −

(a) Substitute in (1):

( ) ( )2 680 lb 2 2800 86.957 0.5BT+ − = −

2 2800 1360 43.48 1396.48 lbBT = − − =

698.26 lbBT = = 698 lbBT !

(b) Substitute for TA and TB into (2):

( ) 2.252 698.26 680 21.739

12α− = 20.315 rad/s=α !




( ) ( ) 25 kg 35 25 49.05 ; 2.19 m/syF a N a= = + − =∑

( ) ( )35 0.15 m 25 0.2 m = 0.072GM N N α= −∑

23.4722 rad/sα =

A =a 22.19 m/s + ( )2 23.4722 rad/s 0.15 m 2.711 m/s=

or 22.71 m/sA =a !

B =a 22.19 m/s + ( )2 23.4722 rad/s 0.2 m 1.496 m/s=

or 21.496 m/sB =a !




0: 0.2Ba a α= =

( ) ( )5 40 20 49.05yF a NΣ = = + −

22.19 m/sa = 210.95 rad/sα =

( ) ( ) ( ) ( )2 240 0.14 m 20 0.2 m 5 kg 10.95 rad/sGM N N kΣ = − =

2 20.02922 m 0.17095 mk k= =

or 170.9 mmk = !




( ) 212

2G kM P r mrμ αΣ = =

(a) 4 kP

mr

μ=α !

(b) ( )2y kF mg P maμΣ = + =

2 k P

gm

μ= +a !




See Problem 16.54

(a) ( )( )

2

4 0.12 11 lb4

12 lb 3.2ft

1232.2 ft/s

k P

mr

μα = =⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠

or 253.1 rad/s=α !

(b) ( )( )2

2

2 0.12 11 lb232.2 ft/s

12 lb

32.2 ft/s

k Pa g

m

μ= + = +⎛ ⎞⎜ ⎟⎝ ⎠

or 239.3 ft/s=a !




(a)

110 110 196.2y GF ma= + − =∑

223.8 20 , 1.19 m/sG Ga a= =

( ) ( )110 0.9 110 0.3GM I α= − =∑

( )66 N mI α= ⋅

( ) ( ) ( )( )2 2 2 21Where 2 1.2 0.3 0.3 kg m

12 2 2

m mI

⎡ ⎤⎛ ⎞= + + ⋅⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

6,I = 211 rad/s=α !

(b)

/B G B G= + =a a a 1.19 m/s2 +

B =a 23.92 m/sB =a 32.6° !

2.11 m/s2

3.3

11(0.3) = 3.3

11(0.3) = 3.3




(a)

120 100 196.2y GF ma= + − =∑

223.8 20 , 1.19 m/sG Ga a= =

( )100 0.6GM I α= =∑

26 kg m ,I = ⋅ 210 rad/s=α !

(b)

/B G B G= + =a a a 1.19 m/s2 +

B =a 23.50 m/sB =a 31.1° !

10(0.3) = 3

10(0.3) = 3 m/s2

1.81 m/s2,

3 m/s2




2y Gmg

F maΣ = =

21

2 4 12Gmg L

M mL α⎛ ⎞= =⎜ ⎟⎝ ⎠

(a) 2Gg

a = , 3

2

g

Lα = !

(b) A =a 2

g + 2 4

L gα =

4Ag=a !

(c) B =a 2

g + 0.75g

5

4Bg=a !




2

3y Gmg

F maΣ = =

21

3 2 12Gmg L

M mL α⎛ ⎞Σ = =⎜ ⎟⎝ ⎠

(a) 2

3Gg

a = , 2g

Lα = !

(b) A =a 2 2

3 2

g g L

L+

3

g= ,

3Ag=a !

(c) B =a 2 2

3 2

g g L

L+

5

3

g=

5

3Bg=a !




Before the spring breaks:

1 20: sin 30 sin 30 0yF T T W= ° + ° − =∑

12 sin 30T W mg° = =

1T mg= 30°

Immediately after spring 2 breaks, elongation of spring 1 is unchanged. Thus we still have T1 = mg 30°

(a) ( ) ( )1eff : sin 302GGLM M T I α= ° =∑ ∑

( ) 21sin 302 12Lmg mL α° =

3gL

=αααα !

( ) 1eff : cos30x xxF F T ma= ° =∑ ∑

0.866x g=a

( )eff: sin 30y yyF F W T ma= − ° =∑ ∑

0.5 ymg mg ma− =

12y g=a

continued



Accelerations of A and B

(b) Acceleration of A:

( ) ( )

/30.866 + 0.5

2

0.866

A G A Gg Lg g

L

g g

= + = → ↓ + ↑

= → + ↑

a a a

1.323A g=a 49.1° !

(c) Acceleration of B:

( ) ( )

/30.866 + 0.5

2

0.866 2

B G B Gg Lg g

L

g g

= + = → ↓ + ↓

= → + ↓

a a a

2.18B g=a 66.6° !

Translation with G + Rotation about G




( )2 2112

I m b b= +

216

I mb=

Statics: 1 21 12 2

T T W mg= = =

Kinetics: ( )effG GM MΣ = Σ

2bT I α=

21 12 2 6

bmg mb α =

32gb

=α

( ) 1eff:y yF F W T maΣ = Σ − =

1 1 2 2

mg mg ma g− = =a

Kinematics:

Plane motion = Translation + Rotation

continued



(a) /A G A G= + =a a a a 2b α+

2Ag=a

2 2b g

b3 +

4

g= 14A g=a "

(b) /B G B G= + =a a a a 2b α+

12B g=a

3 2 2b g

b +

54

g= 54B g=a "




2

21

2 4

bI m mb

⎛ ⎞= =⎜ ⎟⎝ ⎠

Statics: 1 21 1

2 2T T W mg= = =

Kinetics: ( )eff:G GM MΣ = Σ

1 2

bT I α⎛ ⎞ =⎜ ⎟⎝ ⎠

21 1

2 2 4

bmg mb

⎛ ⎞ =⎜ ⎟⎝ ⎠

g

b=α

( )eff

:y yF F W T maΣ = Σ − =

1 1

2 2

mg mg ma g− = =a

Kinematics:


(a) /A G A G= + =a a a a 2

b α+ 1

2g=

2

b g

b⎛ ⎞+ ⎜ ⎟⎝ ⎠ ; 0A =a "

(b) /B G B G= + =a a a a 2

b α+ 1

2g=

2

b g

b⎛ ⎞+ ⎜ ⎟⎝ ⎠ ; B g=a "




2

21 1

2 2 8

bI m mb

⎛ ⎞= =⎜ ⎟⎝ ⎠

Statics: 1 21 1

2 2T T W mg= = =

Kinetics: ( )effG GM MΣ = Σ

1 2

bT I α=

21 1

2 2 8

bmg mb α⎛ ⎞ =⎜ ⎟

⎝ ⎠

2g

b=α


1 1

2 2

mg mg ma g− = =a

Kinematics:


(a) /A G A G= + =a a a a 2

b α+ 1

2g= 2

2

b g

b⎛ ⎞+ ⎜ ⎟⎝ ⎠ ;

1

2A g=a "

(b) /B G B G= + =a a a a 2

b α+ 1

2g= 2

2

b g

b⎛ ⎞+ ⎜ ⎟⎝ ⎠ ;

3

2B g=a "



Chapter 16, Solution 64. Kinetics: 2I mk=

( )eff :x xF F F maΣ = Σ =

k kmg ma gµ µ= =a

( )eff :G GM M Fr I αΣ = Σ =

( ) 2kmg r mkµ α=

2k grk

µ=α

Kinematics: 0v v at= −

0 kv v gtµ= −

For 0v = when 1t t=

00 1 10 ; k

k

vv gt tg

µµ

= − = (1)

0 tω ω α= −

0 2k gr tk

µω ω= −

For 0ω = when 1t t=

2

0 1 1 020 ; k

k

gr kt tgrk

µω ωµ

= − = (2)

continued



Set Eq. (1) = Eq. (2)

2

00 0 02;

k k

v k r vg gr k

ω ωµ µ

= = (3)

Distance traveled: 21 0 1 1

12

s v t at= −

( )2 2

0 0 01 0 1

1 ; 2 2k

k k k

v v vs v g sg g g

µµ µ µ

= − =

(4)

Radius of gyration of a sphere 2 225

k r=

(a) Eq. (3): 00 0

2

52 25

r vvrr

ω = = 00

52

vr

=ω "

(b) Eq. (1) 01

k

vtgµ

= "

(c) Eq. (4) 20

1 2 k

vsgµ

= "




Kinetics: ( )effx xF FΣ = Σ

kmg maµ =

k gµ=a

N mg= ( )eff :G GM M Fr I αΣ = Σ =

225

I mr= ( ) 225kmg r mrµ =

52

k gr

µ=α

Kinematics: When sphere rolls, instant center of rotation is at C and when 1,t t= v rω= (1)

0 0 kv v at v gtµ= − = − (2)

0 0 152 kt gtω ω α ω µ= − + = − +

When 1:t t=

Eq. (1): :v rω= 0 1 0 152

kk

gv gt t rr

µµ ω − = − +

10 1 0

52

kk

gtv gt rr

µµ ω− = − +

( )0 01

27 k

v rt

gω

µ+

= (3)

0 05 m/s, 9 rad/s, 100 mm 0.10 mv rω= = = =

(a) ( )( )( )1 2

5 m/s 0.01 m 9 rad/s2 1.7184 s7 0.1 9.81 m/s

t+

= =

1or 1.718 st = !

continued



(b) ( )( )21 0 5 m/s 0.1 9.81 m/s 1.7184 s 3.3142 skv v gtµ= − = − =

1or 3.31 m/s v = !

(c) ( )2 20.1 9.81 m/s 0.981 m/ska gµ= = =

( ) ( )( )22 21 0 1 1

1 15 m/s 1.7184 s 0.981 m/s 1.7184 s 7.1436 m2 2

s v t at= − = − =

1or 7.14 ms = !




Use equations derived in P16.64

Have 0 05 m/s, 18 rad/s, 0.10 mv rω= = =

(a) ( )

( )1 2

5 m/s 0.1 18 rad/s21.98049

7 0.1 9.81 m/st

⎡ ⎤+⎢ ⎥= =⎢ ⎥⎣ ⎦

1or 1.980 st = "

(b) ( )( )21 0 5 m/s 0.1 9.81 m/s 1.9805 skv v gtμ= − = −

3.0571 m/s=

1or 3.06 m/s=v "

(c) ( )0.1 9.81 m/s 0.981 m/ska gμ= = =

( )( ) ( )( )22 21 0 1 1

1 15 m/s 1.9805 s 0.981 m/s 1.9805 s 7.9785 m

2 2s v t at= − = − =

1or 7.98 ms = "




Kinetics: ( )eff:x xF F F maΣ = Σ =

kmg maμ =

k gμ=a

( )eff:G GM M Fr I αΣ = Σ =

( ) 22

5kmg r mrμ α=

5

2k g

r

μ=α

Kinematics: kv at gtμ= = (1)

5

2k g

t tr

μω α= = (2)

C = Point of contact with belt

5

2k

C kg

v v r gt t rr

μω μ ⎛ ⎞= + = + ⎜ ⎟⎝ ⎠

7

2C kv gtμ=

(a) When sphere starts rolling ( )1 ,t t= we have

1 1 17

; 2C kv v v gtμ= = 1

12

7 k

vt

gμ= "

(b) Velocities when 1t t=

Eq (1): 12

7 k

vv g

gμ

μ⎛ ⎞

= ⎜ ⎟⎝ ⎠

12

7v=v "

Eq (2): 15 2

2 7k

k

g v

r g

μωμ

⎛ ⎞⎛ ⎞= ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

15

7

v

r=ω "




0 : 0;yF N W N W mgΣ = − = = =

( )eff:x xF F F maΣ = Σ =

kmg maµ = µ= k ga (1)

( )eff :G GM M Fr I αΣ = Σ = (2)

making 2 in Equ. (2):I m k=

2k mgr m kµ α= (3)

Kinematics: 2k grttk

µω α= = (4)

kv at gtµ= = (5)

2k

A kgrtv v r gt r

kµω µ = + = +

2

21A krv g tk

µ

= +

(a) When the wheel starts rolling (t = t1), we have

1Av v= 1

2

121krg t vk

µ

+ =

continued



1

2 2

12kk rg t v

kµ + =

2

11 2 2

k

v ktg r kµ

=+

!!!!

(b) Substituting for t in (5) and (4) the value found for t1:

2

12 2 ,k

k

v kv gg r k

µµ

= +

2

1 2 2kv v

r k=

+ !!!!

2

12 2 2 ,k

k

gr v kgk r k

µωµ

= +

12 2v r

r k=

+ωωωω !!!!

Note: We check that 1 .v v rω= −




Kinetics: ( )eff:x xF F F maΣ = Σ =

kmg maµ =

k gµ=a

( )eff:G GM M Fr I αΣ = Σ =

( ) 225kmg r mrµ α=

52

k gr

µ=α

Kinematics: 0 0 kv v at v gtµ= − = − (1)

52

k gt tr

µω α= = (2)

C = Point of contact with belt

Cv v rω= − +

52

kC

gv v r tr

µ= − +

52k

Cgv v tµ= − + (3)

But, when 1,t t= 0v = and 1cv v=

Eq (3): 1 15 ;

2k gv tµ= 1

12

5 k

vtgµ

= !

Eq (1): 0 kv v gtµ= −

continued



When 1, 0,t t v= =

10

20 ;5k

k

vv gg

µµ

= −

0 1

25

v v= !

Distance when 1:t t=

20 1 1

12

S v t at= −

( )2

1 11

2 2 1 25 5 2 5k

k k

v vS v gg g

µµ µ

= −

21 4 2 ;

25 25k

vSgµ = −

212

25 k

vSgµ

= !




tOG r rα= =a

We first observe that the sum of the vectors is the same in both figures. To have the same sum of moments about G, we must have

( )( ):G G tM M I ma GPαΣ = Σ =

( )2mk mr GPα α=

2k

GPr

= (Q.E.D.) "

Note: The center of rotation and the center of percussion are interchangeable. Indeed, since ,OG r= we may write

2 2

or k k

GP GOGO GP

= =

Thus, if point P is selected as center of rotation, then point O is the center of percussion.




21 12

a r I mLα= =

( )eff :C CM MΣ = Σ

( )2LP r ma r I α + = +

( ) 2112

mr r mLα α= +

2 212 12LP r m r L α + = +

Substitute data:

( )210 lb

32.2 ft/sm =

3 ftL = 9 in. 0.75 ftr = =

(a) ( )2

22

10 lb 315 lb 0.75 ft 1.5 ft 0.751232.2 ft/s

α

+ = +

( )2 233.75 lb ft 4.0761 lb ft s ; 82.7999 rad/sα α⋅ = ⋅ ⋅ =

or 282.8 rad/sα = !

(b) ( )eff: 0y y yF F C WΣ = Σ − = or 10 lbyC = !

( )eff : 0 x x x xF F C F C P maΣ = Σ − = − = −

( )( )22

10 lb15 lb 0.75 ft 82.8 rad/s32.2 ft/sxC − = −

4.2857 lbxC = − or 4.29 lbxC = !




(a)

21

12a r I mLα= =


P ma=

( )P m rα=

P

mrα = (1)

( )eff:

2G GL

M M P I αΣ = Σ =

21

2 12

LP mL α=

21

2 12

L PP mL

mr⎛ ⎞= ⎜ ⎟⎝ ⎠

2 1 36 in.

; 2 12 6 6

L Lr L r

r= = =

or 6 in.r = "

(b)

Eq (1): 6

6

P P PLmr mLm

α = = =⎛ ⎞⎜ ⎟⎝ ⎠

( )( )

2

2

6 15 lb96.6 rad/s

10 lb3 ft

32.2 ft/s

α = =⎛ ⎞⎜ ⎟⎝ ⎠

or 296.6 rad/s=α "




21

2 12

La I mLα= =

( )effA AM MΣ = Σ

( )2

LPL ma I α= +

21

2 2 12

L Lm mLα α⎛ ⎞= +⎜ ⎟⎝ ⎠

21

3PL mL α=

(a) ( )

( )( )23 3.5 N3

11.667 rad/s1 kg 0.9 m

P

mLα = = =

211.67 rad/s=α "

(b) ( ) ( )( )2

eff: 0 1 kg 9.81 m/sy y y yF F A W AΣ = Σ − = ⇒ =

9.81 Ny =A "

( )eff:x x xF F A P maΣ = Σ − = −

( ) ( )20.9 m3.5 N 1 kg 11.667 rad/s

2 2xL

A P m α⎛ ⎞ ⎛ ⎞= − = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

1.750 N=

1.75 Nx =A "




(b)

21

2 12

La I mLα= =


P ma=

2

LP m α⎛ ⎞= ⎜ ⎟

⎝ ⎠

2P

mLα =

( )( )( )

22 3.5 N7.7778 rad/s

1 kg 0.9 mα = =

27.78 rad/s=α "

(a)

( )eff:G GM MΣ = Σ

:2

LP h I α⎛ ⎞− =⎜ ⎟⎝ ⎠

21 2

2 12 6

L P PLP h mL

mL⎛ ⎞ ⎛ ⎞− = =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

( )2 2; 900 mm

2 6 2 6 3 3

L L L Lh h L⎛ ⎞− = = + = =⎜ ⎟⎝ ⎠

or 600 mmh = "




Determination of m a

Since 20, ω= = = −na a rαααα

( )2 r zω= − +a i k

But 2 21 = − − = − −

rx r rπ π

( )10 ft 0.3633812 = −

0.30282 ft= −

And 2 2 10 ft 0.53052 ft12

rzπ π

= = =

180 rpm 18.85 rad/sω = =

Thus: ( ) ( )218.85 0.30282 0.53052= − − +a i k

( ) 2107.60 188.51 ft/s= + −i k

And ( )5 16.708 29.271 lb32.2

= = −ma a i k (1)

Equivalence of external and effective forces



External Forces:

( ) ( ) ( ) ( )5 lb at 0.30282 ft 0.5 ft 3.53052 ft= − = − + +GrW j i j k

( )2 at 1 ft= + =x AA AA i k r j

+ at 0= + =x y z BB B BB i j k r

Effective Forces:

From Equ. (1): ( ) ( )16.708 lb 29.271 lb at= − Gma i k r

( )eff :∑ = ∑ × + × = ×B B G A G mM M r W r A r a

0.30282 0.5 0.53052 0 1 00 5 0 0

− +− x zA A

i j k i j k

0.30282 0.5 5.305216.708 0 29.271

= −− −

i j k

Equating the coefficients of k:

( )( ) ( ) ( )( )0.30282 5 1 16.708 0.5xA− = −

9.868 lbxA = + ( )9.87 lb=xA i !

Equating the coefficients of i:

( )( ) ( ) ( )( )0.53052 5 1 0.5 29.271+ = −zA

17.288 lbzA = − ( )17.29 lb= −zA k !




(mass) 2 3

2

mz z gr L

L Lω ⎛ ⎞⎛ ⎞⎛ ⎞= −⎜ ⎟⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠⎝ ⎠

2

3

2ymgz mgz z

F T LL L

⎛ ⎞Σ = − = −⎜ ⎟⎝ ⎠

31 3

2

mgz zT

L L⎛ ⎞= + −⎜ ⎟⎝ ⎠

or

(a) 3

42

mgz zT

L L⎛ ⎞= −⎜ ⎟⎝ ⎠

!

(b) maxT at ,z L=

max 2.5AT T mg= =

( )( )2.5 2 lb 5 lb= =

or 5 lbAT = !




Determination of mass center of disk

We determine the centroid of the composite area:

1 1 2 2 1 2 1 1 2 2or ( )xA x A x A x A A x A x A= − − = −

2 2

1 2 22 2 2 2

1 2

0 (160) (60) (160)(60)6.667 mm

(300) (60) (300) (60)

xA x Ax

A A

ππ π

− −= = = − = −− − −

Mass center G coincides with centroid C

480 rpm 50.265 rad/sω = =

2 3 2 2(6.667 10 m)(50.265 rad/s) 16.844 m/sa rω −= = × =

2eff( ) : (30 kg)(16.844 m/s )Σ = Σ = =F F mA a

505 N=A !




(a)

1 0.35 kg/m(0.2 m) 0.07 kgm = =

( )( )2 0.35 kg/m 2 0.2 mm π=

0.43982 kg=

4 rad/sω π=

( )( )( )2

0.07 kg 0.2 m 4 rad/s: 1.1054 N

2x xF Aπ−

Σ = = −

1 2: (0.07)(9.81) (0.43982)(9.81) 5.0013 Ny yF A m g m gΣ = + = + =

1 (0.07)(9.81)(0.2): 0.06867 N m

2 2Am gr

M MΣ = − = = ⋅

or 0.0687 N m= ⋅M !

and 2 2( 1.1054) (5.0013) 5.122 NR = − + =

or 5.12 N=R 71.5° !

(b)

: 0x xF AΣ =

1 2 1.1054y yF A m g m gΣ = − − =

6.1067 NyA =

:AMΣ or 0=M !

and 6.11 N=R !

Ay Ax

m1g

m2g

=

21 2

rm ω




0 2

Laω α= =

( )eff:

2 2A AL L

M M W I maαΣ = Σ = +

21

2 12 2 2

L L Lmg mL mα α⎛ ⎞= + ⎜ ⎟

⎝ ⎠

21 3

2 3 2

L gmg mL

Lα= α =

( )eff

:2y yL

F F A mg ma m αΣ = Σ − = − = −

3

2 2

L gA mg m

L⎛ ⎞⎛ ⎞− = − ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

3

4A mg mg− = −

1

;4

A mg= 1

4mg=A !

/ 0B A B A Lα= + = +a a a

3 3

2 2Bg

L gL

⎛ ⎞= =⎜ ⎟⎝ ⎠

a ; 3

2B g=a !




Let , 2LS b a Sα= − =

( )eff :C CM M WS I maSαΣ = Σ = +

( )2112

mgS mL m S Sα= +

2 2112

mgS m L S α = +

22

12

SgL S

α =+

(1)

For rotation about C: 2ALa S= −

2

2 22 2

2 2

12 12

A

L LS Sg S Sa g

L LS S

− − = =

+ +

( )2

2 22 6

12ALS Sa gL S

−=+

Differentiate with respect to S.

( )( ) ( )( )

( )( )

2 2 2

22 2

12 4 2 246

12A

L S L S LS S Sda gdS L S

+ − − −=

+



Set numerator equal to zero 3 2 2 3 2 34 12 48 24 48 0L SL S L S S L S− + − − + =

3 2 24 12 0L SL S L− − =

( )2 24 12 0L L SL S− − =

( )( )6 2 0L L S L S− + =

and 2 6L LS S= − =

(a) For , 2LS b L= − = and support was at B, impossible

For , 6 3L LS b= = this results in max Aa

3Lb = "

(b) Eq. 1 with 6LS =

22

136 6 1 2

912 6

L g g gL LL L

α α= = = +

3 16 2 2AL ga S g

Lα = = =

1Max:

2Aa g= "

( )eff:y yF F C mg maΣ = Σ − = −

C mg mSα− = −

36 2L gC mg m

L − = −

14

C mg mg− = −

34

mg=C "




(a) 2

2 21 119

4 3 12ABr

I mr mL m= + =

219

2 12zL mr

M mg mgr α⎛ ⎞Σ = = =⎜ ⎟⎝ ⎠

12

19

g

rα =

( )12 242

19 19Dg g

a rr

= = 24

19Dg−=a j!

(b)

4 12 4 16

3 19 3 19

r g gαπ π π

⎛ ⎞ = =⎜ ⎟⎝ ⎠

4 4

319G

g

π⎡ ⎤⎛ ⎞= − +⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

a i j !

or:




/: = +B A B AKinematics a a a

aB = aA 2

Lα =

2

3

Lα 2

Lα

aG =

3

2 3x AE

mLF T

α⎛ ⎞Σ = =⎜ ⎟⎜ ⎟

⎝ ⎠

1

2 2y AE BFmL

F T T mgα⎛ ⎞Σ = + − = −⎜ ⎟

⎝ ⎠

21 1

2 2 2 12G AE BFL L

M T T mL α⎛ ⎞⎛ ⎞ ⎛ ⎞Σ = − + =⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠

Solve for, (a) 3

4

g

Lα = !

3( ) ,

2 8AE BFmg mg

b T T= = !

unknowns: , ,AE BFT T α

30°

3

Lα

2

Lα




We represent the effective forces of the disk by a couple DI αααα and the effective forces of the rod by vectors ( )R R tm a and ( )R R nm a attached at C and a couple .RI αααα We have,

2 3 21 (3 kg)(0.120 m) 21.6 10 kg m2DI −= = × ⋅

2 3 214 kg, (4 kg)(0.48 m) 76.8 10 kg m12

−= = = × ⋅R Rm I

( ) 0.12α=R ta

2( ) 0.12ω=R na

eff( ) :A AM MΣ = Σ

(0.12) 4 (0.12) ( ) (0.12)α α− = + +D R R R tP g I I m a

3 30.12( 4 ) 21.6 10 76.8 10 4(0.12 )(0.12)α α α− −− = × + × +P g

4 1.300α− =P g (1)

eff( ) : ( )x x x R R nF F A m aΣ = Σ = −

24(0.12 )xA ω= − 20.48xA ω= − (2)

eff( ) : 4 3 ( )Σ = Σ − − − =y y y R R tF F A P g g m a

7 4(0.12 )α= + +yA P g 7 0.48α= + +yA P g (3)



(a) making 236 rad/s in (1):α =

4(9.81) 1.3(36) 86.04 NP = + = 86.0 N=P "

(b) making 212 rad/s in (2) and 86.04, 36 rad/s in (3) :Pω α= = =

20.48(12) 69.12 NxA = − = − 69.1 Nx =A "

86.04 7(9.81) 0.48(36)yA = + + 172.0 N=yA "




(4 lb)(0.5 ft) (4 lb)(1.25 ft) 6 lb ftAMΣ = + − ⋅ 2 2

2 21 4 lb 4 lb2 (1 ft) (0.5 ft)

12 32.2 ft/s 32.2 ft/sα α

= +

2 2

2 24 lb 4 lb(1.25 ft) [0.5 ft(0.866)]

32.2 ft/s 32.2 ft/sα α

+ +

(7 6) lb ft (0.26915)α− ⋅ = 23.7154 rad/sα =

or 23.72 rad/s=αααα !

(a)

(b)



(4 lb)(0.25 ft) (1 ft)(sin 30 )Σ = − °BM T

22 2

1 4 lb 4 lb(1 ft) (1.25 ft)(0.25 ft)12 32.2 ft/s 32.2 ft/s

α α

= +

24 lb (0.5 ft)(cos.30 )(0.433 ft)

32.2 ft/sα

+ °

0.07246 (0.07246)(3.7154) 1 (0.5) 0.26922Tα= = = − =

1.462 lbT =

or 1.462 lbT = !




From P16.84

ABC, 7 0.26915AM M αΣ = − =

BC, 1 0.5 0.07246BM T αΣ = − =

1 (0.5)(0.45) 0.07246 α− =

(a) 210.695 rad/sα =

or 210.70 rad/sα = !

(b) 7 0.2695 4.12 ft lbM α= − = ⋅

or 4.12 ft lbM = ⋅ !




eff( ) :2 2

α Σ = Σ − = +

! !!A A AB ABM M mg B I ma

21 12 2 2 2

α α − = +

! !! ! ! AB ABmg B m m

21 12 3

α− =! ! ! ABmg B m (1)

Rod CD ( 0)CDω =

eff( ) :2 2 2

α Σ = Σ + = +

! ! !D D CD CDM M mg B I ma

21 1 12 2 2 2 2

α α + = +

! !! ! ! CD CDmg B m m

Multiply by 2: 22

3α+ =! ! ! CDmg B m (2)

Adding (1) and (2): 23 1 22 3 3

α α = +

! ! AB CDmg mg (3)

continued



We rewrite eq. (3) as follows:

922

α α + = !AB CD

g (3')

Kinematics

we must have ( Ba ) ( B′= a )

2AB CDα α= !!

2α α= !AB CD (4)

substituting for ABα from (4) into (3'):

1 922 2

α α + = !CD CD

g

5 92 2

α = !CD

g 1.8α =!CDg (5)

(a) Acceleration of C:

(1.8)C CDga α = =

! !

! 1.8=C ga "

(b) Force on knob B:

substituting for CDα from (5) into (2):

22 (1.8)3

+ =

! ! !!gmg B m

1.2B mg mg= − 0.2= mgB "




( ) ( )eff:C C CM M M ma r I αΣ = Σ Σ = +

( )mr r Iα α= +

( )2CM mr I αΣ = +

But, we know that 2CI mr I= +

Thus: C CM I αΣ = (Q.E.D.) !




Kinematics:

/C G C= +a a a

( )/ /C G C G C= + × + × ×a α r ω ω r

or, since ω ⊥ /G Cr

2/ /C G C G Cω r= + × −a a α r

(1)

Kinetics:

( ) /eff:C C C G CM M I mΣ = Σ Σ = + ×M α r a

Recall Eq. (1): ( )2/ / /C G C C G C G CI m ω rαΣ = + × + × −M r a α r

( ) 2/ / / / /C G C C G C G C G C G CI r ma mr r mωαΣ = + × + × × − ×M α r r

But: / / 0G C G C× =r r and α ⊥ /G Cr

( ) 2/ / /G C G C G Cr m mr× × =α r α

Thus: ( )2/ /C G C G C CI mr mΣ = + + ×M α r a

Since 2/C G CI I mr= +

/C C G C CI mΣ = + ×M α r a (2)

Eq. (2) reduces to C CM I αΣ = when / 0;G C Cm× =r a that is, when /G Cr and Ca are collinear.

Referring to the first diagram, we note that this will occur only when points G, O, and C lie in a straight line.

(Q.E.D.) !




Kinematics: 20

1

2S v t at= +

( )215 m 0 40 s

2a= +

20.00625 m/s=a

Since 30 mm 0.03 mr = =

( )2; 0.00625 m/s 0.03 ma rα α= =

20.20833 rad/sα =

Kinetics:

( )eff:C CM MΣ = Σ

( ) ( )sin15mg r I ma rα° = +

( ) ( )2sin15mg r mk mr rα α° = +

( )2 2sin15gr k r α° = +

( )( ) ( ) ( )22 2 29.81 m/s 0.03 m sin15 0.03 m 0.20833 rad/sk⎡ ⎤° = +⎢ ⎥⎣ ⎦

2 20.36472 m 0.60392 mk k= ⇒ =

or 0.604 mk = !




2I mk=

m rα=

( ) ( ) ( )eff: sinC CM M mg r I ma rβ αΣ = Σ = +

2 2sinmg r mk mrβ α α= +

2 2

singr

r kα β=

+ (1)

eff : sinF F F mg maβΣ = Σ − = −

sinF mg mrβ α− = −

sinF mg mrβ α= −

eff : cos 0F F N mg βΣ = Σ − =

cosN mg β=

If slipping impends ors sF

F NN

μ μ= =

sinsin

cos coss

rgF mg mr

N mg

β αβ αμ

β β

−−= = =

Substitute for α from Eq (1)

2 2sin sin

coss

r gr

g r kβ β

μβ

− ⋅+=

2 2

2 2 2 2tan 1 tansr k

r k r kμ β β

⎡ ⎤ ⎡ ⎤= − =⎢ ⎥ ⎢ ⎥

+ +⎣ ⎦ ⎣ ⎦

2 2

2tan sr k

kβ μ +=

2

tan 1sr

kβ μ

⎡ ⎤⎛ ⎞= +⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦

!




General case: 2I mk a rα= =

( )effC CM MΣ = Σ

( )sinw r I marβ α= +

2 2sinmg r mk mrβ α α= +

2 2sinrg

r kβα =

+

(1)

For pipe: 2

2 21sin sin2P

rk r a g gr r

β β= = =+

For cylinder: 2

2 22

2

1 2sin sin2 3

2

Crk r a g g

rrβ β= = =

+

For sphere: 2

2 2

2 2

2 5sin sin25 75

Srk r a g g

r rβ β= = =

+

(a) Between pipe and cylinder

/2 1 1sin sin3 2 6C P C Pa a a g gβ β = − = − =

2 2/ /

1 1 1 sin2 2 6C P C Px a t g tβ = =

US units: ( ) ( )22/

1 1 32.2 ft/s sin 6 s 20.08427 ft2 6C Px β = =

/ 20.1 ftC Px = !

SI units: ( ) ( )22/

1 1 9.81 m/s sin 6 s 611884 m2 6C Px β = =

/ 6.12 mC Px = !

continued



(b) Between sphere and cylinder

/5 2 1sin sin7 3 21S C S Ca a a g gβ β = − = − =

2 2/ /

1 1 1 sin2 2 21S C S Cx a t g tβ = =

US units: ( ) ( )22/

1 1 32.2 ft/s sin12 6 s 5.7384 ft2 21S Cx = ° =

or / 5.74 ftS Cx = !

SI units: ( ) ( )22/

1 1 9.81 m/s sin12 6 s 1.7482 m2 21S Cx = ° =

or / 1.748 mS Cx = !




Kinematics:

(a)

12 in. = 1 ftr =

22

80 lb 80lb s /ft

32.232.2 ft/sm = = ⋅

80

50 80 sin 15 (1)32.2x

F F mrα α⎛ ⎞∑ = − ° − = = ⎜ ⎟⎝ ⎠

80 cos 15 0, 80 cos 15 77.274 lbyF N N∑ = − °= = °=

2 21 3

50(1) (80 sin 15 )(1) ( )2 2

α α α∑ = − ° = + =CM mr mr r mr

27.8607 rad/s, = 7.86 ft/sGaα = 15° !

(b)

80

50 80 sin 15 (1)(7.8607) 9.7648 lb32.2

F = − °− =

9.7648

0.1263777.274s

F

Nμ = = = 0.1264sμ = !




( )8ft 0.66667 ft

12a rα α α⎛ ⎞= = =⎜ ⎟

⎝ ⎠

2

22

10 lb 6ft

1232.2 ft/sI mk

⎛ ⎞= = ⎜ ⎟⎝ ⎠

20.07764 lb ft s= ⋅ ⋅

( ) ( ) ( )eff

8: 5 lb ft

12C CM M ma r I α⎛ ⎞Σ = Σ = +⎜ ⎟⎝ ⎠

( )2

22

10 lb 83.3333 lb ft ft 0.07764 lb ft s

1232.2 ft/sα α

⎛ ⎞⎛ ⎞⋅ = + ⋅ ⋅⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

215.4558 rad/sα = or 215.46 rad/sα = !

(a) ( )2 28ft 15.4558 rad/s 10.3039 ft/s

12a rα ⎛ ⎞= = =⎜ ⎟

⎝ ⎠

or 210.30 ft/s=a !

(b) 0 10 lb 0 10 lbyF N NΣ = − = =

( ) ( )22eff

10 lb: 5 lb 10.3039 ft/s

32.2 ft/sx xF F F ma

⎛ ⎞Σ = Σ − = = ⎜ ⎟

⎝ ⎠

1.800 lbF =

Now ( )min

1.800 lb0.180

10 lbsF

Nμ = = =

or ( )min0.18sμ = !




( )8ft 0.66667 ft

12a rα α α⎛ ⎞= = =⎜ ⎟

⎝ ⎠

2

22

10 lb 6ft

1232.2 ft/sI mk

⎛ ⎞⎛ ⎞= = ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

20.07764 lb ft s= ⋅ ⋅

( ) ( )eff

12: 5 lb ft

12C CM M mar I α⎛ ⎞Σ = Σ = +⎜ ⎟⎝ ⎠

( )( ) ( )22

10 lb 85 lb ft 0.66667 ft ft 0.07764 lb ft s

1232.2 ft/sα α

⎛ ⎞ ⎛ ⎞⋅ = + ⋅ ⋅⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

( )2 25 lb ft 0.21567 ft s , 23.1839 rad/sα α⋅ = ⋅ =

or 223.2 rad/s=α !

(a) ( )2 28ft 23.1839 rad/s 15.4559 ft/s

12a rα ⎛ ⎞= = =⎜ ⎟

⎝ ⎠

or 215.46 ft/s=a !

(b) 0: 10 lb 0 10 lbyF N NΣ = − = =

( ) ( )22eff

10 lb: 5 lb 15.4559 ft/s


⎛ ⎞Σ = Σ − = = ⎜ ⎟

⎝ ⎠

0.20 lbF =

( )min

0.200.020

10sF

Nμ = = =

( )min0.020sμ = !




8ft 0.66667

12a rα α α⎛ ⎞= = =⎜ ⎟

⎝ ⎠

2

22

10 lb 6ft

1232.2 ft/sI mk

⎛ ⎞⎛ ⎞= = ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

20.07764 lb ft s= ⋅ ⋅

( ) ( )eff

4: 5 lb ft

12C CM M mar I α⎛ ⎞Σ = Σ = +⎜ ⎟⎝ ⎠

( )21.6667 lb ft 0.21567 ft s α⋅ = ⋅

27.7280 rad/sα =

or 27.73 rad/sα = !

(a) ( )2 28ft 7.7280 rad/s 5.1520 ft/s

12a rα ⎛ ⎞= = =⎜ ⎟

⎝ ⎠

or 25.15 ft/s=a !

(b) 0: 10 lb 0, 10 lbyF N NΣ = − = =

( ) ( )22eff

10 lb: 5 lb 5.1520 ft/s


⎛ ⎞Σ = Σ − = = ⎜ ⎟

⎝ ⎠

3.3999 lbF =

( )min

3.3999 lb0.39999

10 lbsF

Nμ = = =

( )min0.340sμ = !




( )8ft 0.66667 ft

12a rα α α⎛ ⎞= = =⎜ ⎟

⎝ ⎠

2

22

10 lb 6ft

1232.2 ft/sI mk

⎛ ⎞⎛ ⎞= = ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

20.07764 lb ft s= ⋅ ⋅

( ) ( )eff

4: 5 lb ft

12C CM M mar I α⎛ ⎞Σ = Σ = +⎜ ⎟⎝ ⎠

( )21.6667 lb ft 0.21567 ft s α⋅ = ⋅

27.7280 rad/sα =

or 27.73 rad/s=α !

(a) ( )2 28ft 7.7280 rad/s 5.15201 ft/s

12a rα ⎛ ⎞= = =⎜ ⎟

⎝ ⎠

or 25.15 ft/s=a !

(b) 0: 5 10 lb 0 5 lbyF N N= + − = =

( ) ( )22eff

10 lb: 5.15201 ft/s


⎛ ⎞Σ = Σ = = ⎜ ⎟

⎝ ⎠

1.600 lbF =

( )min

1.600 lb0.32

5 lbsF

Nμ = = =

or ( )min0.32sμ = !




Assume disk rolls: ( )0.16 ma rα α= =

( )( )22 5 kg 0.12 mI mk= =

20.072 kg m= ⋅

( ) ( )( ) ( )eff: 20 N 0.16 mC CM M ma r I αΣ = Σ = +

( )( ) ( )2 23.2 N m 5 kg 0.16 m 0.072 kg mα α⋅ = + ⋅

216 rad/sα =

or 216 rad/s=α !

( )( )2 20.16 m 16 rad/s 2.56 m/sa rα= = =

or 22.56 m/s=a !

( )eff: 20 Nx xF F F maΣ = Σ − + =

( )( )220 N 5 kg 2.56 m/sF− + =

7.2 NF =

( ) ( )( )2

eff: 0 5 kg 9.81 m/sy yF F N mg NΣ = Σ − = =

49.05 N=

( )0.25 49.05 N 12.2625 Nm sF Nμ= = =

Since ,mF F< disk rolls with no sliding !




Assume disk rolls:

( )0.16a r mα α= =

( )( )22 25 kg 0.12 m 0.072 kg mI mk= = = ⋅


( )( ) ( )2 24.8 N m 5 kg 0.16 m 0.072 kg mα α⋅ = + ⋅

224.0 rad/sα = or 224.0 rad/sα = !

( )( )2 20.16 m 24 rad/s 3.84 m/sa = =

23.84 m/s=a !

( )eff: 20 Nx xF F F maΣ = Σ − + =

( )220 N 5 kg 3.84 m/s , 0.8 NF F− + = =

( )20: 0 5 kg 9.81 m/s 49.05 NyF N mg NΣ = − = = =

( )0.25 49.05 12.2625 Nm sF Nμ= = =

Since ,mF F< disk rolls with no sliding !




(a)

Assume disk rolls:

( )0.16 ma rα α= =

( )( )22 5 kg 0.12 mI mk= =

20.072 kg m= ⋅

( ) ( )( ) ( )2eff : 20 N 0.08 mC CM M I mr αΣ = Σ = +

( )( )221.6 N m 0.072 kg m 5 kg 0.16 m α ⋅ = ⋅ +

28 rad/sα =

( )( )2 20.16 m 8 rad/s 1.28 m/sa rα= = =

( )eff : 20 Nx xF F F maΣ = Σ − =

( )( )220 N 5 kg 1.28 m/s ; 13.6 NF F− = =

( )( )20: 0; 5 kg 9.81 m/s 49.05 NyF N mg NΣ = − = = =

( )0.25 49.05 N 12.263 Nm sF Nµ= = =

,mF F> Disk slides !

(b) Since disk slides, ( )0.20 0.20 49.05 N 9.81 Ns mFµ = ⇒ = =

( ) ( ) ( )( )eff : 0.16 m 20 N 0.08 mG GM M F I αΣ = Σ − =

( )( ) ( )( ) ( )29.81 N 0.16 m 20 N 0.08 m 0.072 kg m α− = ⋅

20.42222 rad/sα = −

or 20.422 rad/sα = !

continued



( )eff : 20 N 9.81 Nx xF F ma∴ Σ = Σ − =

( ) 210.19 N 5 kg 2.038 m/sa a= =

or 22.04 m/s=a !




Assume disk rolls: ( )0.16 ma rα α= =

( )( )22 5 kg 0.12 mI mk= =

20.072 kg m= ⋅


( )( ) ( )2 21.6 N m 5 kg 0.16 m 0.072 kg mα α⋅ = + ⋅

28 rad/sα = or 28 rad/sα = !

( )( )2 20.16 m 8 rad/s 1.28 m/sa rα= = =

or 21.28 m/s=a !

( ) ( )( )2eff

: 5 kg 1.28 m/s 6.40 Nx xF F F maΣ = Σ = = =

( )( )20 20 N 0, 20 N 5 kg 9.81 m/s 0yF N mg NΣ = + − = + − =

29.05 NN =

( )0.25 29.05 N 7.2625 Nm sF Nμ= = =

Since ,F Fm< disk rolls without sliding !




(a)

10 832.2 12

ma α =

where α is angular acceleration of bar AD and bar BE about fixed axes of rotation. The plate is translating so A Ba a=

Plate: 2 30;∑ = = −GM R R

Bar AD: 2

24 8 1 3 8(3 lb) ft (cos 45 ) ft ft

12 12 3 32.2 12DM R α ∑ = ° − =

20.7071 0.6667 0.01380R α− = (1)

Note for bar BE EM∑ would yield

40.7071 0.6667 0.01380R α− =

Therefore 4 2 1 3;= =R R R R

210 82 (10 lb) (cos 45 ) = ft

32.2 12F R α Σ = + °

22 7.0711 0.20704R + = (2)

Solve eq (1) and (2) for α and 2R

20.6667 0.01380 0.70711R α+ = (3)

22 (0.20704) 7.0711R α− = − (4)

continued



From equations (3) and (4):

237.00 rad/sα =

2 28 8ft ft (37.00 rad/s ) 24.667 ft/s10 12

a α = = =

or 224.7 ft/sa = 45° !

(b)

21 ( 7.0711 lb) (0.20704)(37.00) 0.2946 lb2

R = − + =

Now 2 3cos 45 ( ) cos 45 2(0.2946) (0.707)= °+ − °=BF R R

0.417 lb= 0.417 lbBF = !




From 16.101, For BE, equation (1)

20.7071 0.6667 0.01380R α− = (1)

30 0GM R∑ = ⇒ =

210 8

(10)(cos 45 )32.2 12

F R α⎛ ⎞ ⎛ ⎞∑ = + ° = ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

(2)

Solving equations (1) and (2)

20.6667 (0.01380) 0.7071R α+ = (3)

2 0.20704 7.0711R α− = − (4)

Take 1.5 ∗ (3) – (4) and solve for α:

28(35.706) 23.804 ft/s

12α ⎛ ⎞= =⎜ ⎟

⎝ ⎠

or 223.8 ft/sa = 45° !

And

2 7.0711 (0.20704) (35.706) 0.321 lbR = − + =

or 2 0.321 lbR = 45° !




Kinematics:

Since gear D is fixed, we have for point E of gear C:

( ) = 0E ta But /E B E B= +a a a

/( ) ( ) ( )E t B t E B ta a a= +

0 0.2 0.1α α= −AB C 12

α α=AB C (1)

Kinetics—gear C:

eff( ) :B BM M∑ = ∑

(0.1 m) α= C CQ I

2(5 kg)(0.075m) α= C

0.28125α= CQ (2)



Bar AB and gear C

212AB ABI m L=

21 (3 kg)(0.2 m)12

=

(a) eff( ) :A AM M∑ = ∑

(0.1) (0.1) (0.2) ( ) 0.1 ( )0.2α α− + = + + +AB C AB AB AB C B C CW Q W m a I m a I

21(3) (0.1) (0.1) (5 ) (0.2) (3)(0.1 ) (0.1) (3)(0.2)12AB ABg Q g α α− + = +

25(0.2 )0.2 5(0.075)α α+ +AB C

(1.3) 0.1 0.24 0.028125ABg Q α α− = −

Substituting for ABα and Q from (2) and (1):

11.3 0.1(0.28125 ) 0.24 0.0281252

α α α − = +

C C Cg

1.3 0.17625 α= Cg 7.3759(9.81)α =C

72.36α =C 72.4 rad/sC =α "

(b) 10.2 0.2 0.1 0.1(72.36)2

α α α = = = =

B AB C Ca

27.24 m/s=Ba "

Note: The same numerical values are obtained for Cα and Ba in Prob. 16.104




Kinematics:

Since gear D is fixed, we have for point E of gear C:

( ) = 0E ta But /E B E B= +a a a

/+ ( ) ( ) ( )E t B t E B ta a= + a

0 0.2 0.1α α= −AB C

12AB Cα α= (1)

Gear C

eff( ) :B BM M∑ = ∑

(0.1m) α= C CQ I

2(5 kg) (0.075 m) Cα=

0.28125 α= CQ (2)



Bar AB and gear C

(a) eff( ) :A AM MΣ = Σ

(0.1) (0.2) (0.3) ( ) 0.1 ( )0.2α α+ − = + + −AB C AB AB AB C B C CW W Q m a I m a I

21(3) (0.1) (5) (0.2) (0.3) 3(0.1 )0.1 (3)(0.2)12AB ABg g Q α α+ − = +

25(0.2 )0.2 5(0.075)α α+ −AB C

(1.3) 0.3 0.24 0.028125α α− = −AB Cg Q

Substituting for ABα and Q from (2) and (1):

11.3 0.3(0.28125 ) 0.24 0.0281252

α α α − = −

C C Cg

1.3 0.17625 α= Cg 7.3759(9.81)α =C

72.36Cα = 272.4 rad/s=Cαααα !

(b) 10.2 0.2 0.1 0.1(72.36)2B AB C Ca α α α = = = =

27.24 m/s=Ba !

Note: The same numerical values were obtained for cα and Ba in Prob. 16.103




Let 2

2 23 2 83,

8 5 320= − = − =r mr

OG r I mr mr

2sin ( ) ( )ρ θ α αΣ = − = + −M mg r r I m r r

2

( ) sin

( )

mg r r

I m r r

θα −=+ −

22

5sin

8

83 5320 8

rmg

mr rm

θα

⎛ ⎞⎜ ⎟⎝ ⎠=

⎛ ⎞+ ⎜ ⎟⎝ ⎠

25 sin

26

g

r

θα = !




Let 2 24 1,

3 2π= = = −r

OG r I mr mr

( )2

2 1 169 32

2 9 18

mrmr π

π π2

2 2⎛ ⎞= − = −⎜ ⎟⎝ ⎠

From Prob 16.105:

( )

( )2

sinmg r r

I m r r

θα

−=

+ −

( )22

4sin

3

49 32

318

rmg r

mr rm r

θπα

πππ

22

⎛ ⎞⎛ ⎞− ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠=

⎛ ⎞⎛ ⎞− + − ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

41 sin

33 82 3

g

r

θπα

π

⎛ ⎞⎛ ⎞− ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠=

⎛ ⎞−⎜ ⎟⎝ ⎠

( )

( )2 3 sin

9 16

g

r

π θα

π− 4

=−

!




2

22

2

2 41G

rI mr m mr

π π⎛ ⎞ ⎛ ⎞= − = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

2

1xF F mrαπ

⎛ ⎞Σ = = −⎜ ⎟⎝ ⎠

2 2

2

4 2 21 1 1

2Dr

M P mr mrα απ ππ

⎛ ⎞ ⎛ ⎞⎛ ⎞Σ = = − + − −⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠

2 2 2 2 22 2

4 2 42mr mr mr mr mrα α

ππ π⎛ ⎞ ⎛ ⎞= − + − +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

2 2 22 2mr mr α

π⎛ ⎞= −⎜ ⎟⎝ ⎠

(a) 2

4 1

P

mrα

π

=⎛ ⎞−⎜ ⎟⎝ ⎠

or 0.688P

mrα = !

(b)

21

2 44 1

PP

F mrmr

π

π

⎛ ⎞−⎜ ⎟⎝ ⎠= =⎛ ⎞−⎜ ⎟⎝ ⎠

0:yF N mg PΣ = = +

and ( )4sF P

N mg Pμ = =

+ or ( )

0.25s

P

mg Pμ =

+ !




22

41GI mr

π⎛ ⎞= −⎜ ⎟⎝ ⎠

2

1xF F P mrαπ

⎛ ⎞Σ = − + = −⎜ ⎟⎝ ⎠

22 2

2

3 4 21 1

2Dr

M P r mr mrα αππ

⎛ ⎞ ⎛ ⎞ ⎛ ⎞Σ = − = − + −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠ ⎝ ⎠⎝ ⎠

3 4 2

1 2 2 12

P mr mrα απ π

⎛ ⎞ ⎛ ⎞ ⎛ ⎞− = − = −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠ ⎝ ⎠⎝ ⎠

(a) 2 3

84

P

mrα

π

⎛ ⎞⎜ ⎟−= ⎜ ⎟⎜ ⎟−⎜ ⎟⎝ ⎠

!

0.1843

or P

mrα⎛ ⎞=⎜ ⎟

⎝ ⎠

(b) 2 2 3

18

4

PF P mr

mrππ

⎛ ⎞⎜ ⎟−⎛ ⎞= − − ⎜ ⎟⎜ ⎟

⎝ ⎠ ⎜ ⎟−⎜ ⎟⎝ ⎠

3

1 , or 0.9332 2

PF P

⎛ ⎞= + =⎜ ⎟⎜ ⎟

⎝ ⎠

0, yF N mgΣ = =

Minimum

31

2

2s

PF

N mgμ

⎛ ⎞+⎜ ⎟⎜ ⎟

⎝ ⎠= = s0.933

or P

mgμ = !



Chapter 16, Solution 109. Kinematics: ω = 0 / 0 [ α= + = +O A O A ra a a ]

/= = +G O G Oa a a a

[ α= r ] + [ αx ]

Thus x rα=a , y xα=a (1)

Kinetics:

( ) ( ) ( )eff : αΣ = Σ = + +A A x yM M Wx ma r ma x I

( ) ( ) 2mg x mr r mx x mkα α α= + +

2 2 2gx

r x kα =

+ + (2)



( )eff : ;x x xF F F ma F mrαΣ = Σ = =

( )eff: ;y y yF F N W ma N mg mxαΣ = Σ − = − = −

minF mrN mg x

αµα

= =−

minr

g xαµ

α=

− (3)

For a hemisphere

38

x OG r= = 2

2 22 35 8

= − = −

OI I mx mr m r

2 22 95 64

I mr mr= − 2 22 95 64

Ik rm

= = −

Substituting into (2)

2 2 2

3 3158 8

79 2 9 56564 5 64

g rg gr rr r r

α

= = = + + −

(a) Substituting into (3)

min

150.2678656

3 15 0.8995518 56

µ = = −

min 0.298µ = "

(b) ( ) ( ) 3015256 56B

gga r rr

α = = 2 = 0.536=B ga "

Note: In this Problem and the next we cannot use the equation ,A AM I α∑ = since points A, O, and G are not aligned.




Kinematics: Choose positive Bv and Ba to left

Trans. with B + Rotation Abt. B = Rolling motion

2Ba rω = + a B

r aR +

Kinetics:

( )eff :C CM MΣ = Σ

( ) ( )y xPR Wr ma r ma R I α− = + +

( )2 2 BB B

r aPR mgr m a r m a r R mkR R

ω − = + + +

22 2

BB

r k vma R mr RR R r

= + + +

2 2

22 21B B

r r k rP mg ma m vR R R

+ = + + + (1)



Substitute: 23.5 lb3.5 lb,

32.2 ft/smg m= =

0.9 ft 0.075 ft12

r = =

3 ft 0.25 ft12

R = =

2.2 ft 0.18333 ft12

k = =

( )20.075 ft 3.5 lb3.5 lb0.25 ft 32.2 ft/s BP a = +

( ) ( )( ) ( )

2 22

2 2 20.075 ft 0.18333 ft 3.5 lb 0.75 ft 1

32.2 ft/s0.25 ft 0.25 ftBv

+ × + +

( ) ( )2 2 2 21.05 lb 1.7693 lb/ft s 0.13043 lb/ft sB BP a v= + ⋅ + ⋅

Substitute: 1.05 ft/sBv = ; 1.05 ft/sBv = +

23.6 ft/sBa = ; 23.6 ft/sBa = +

( )( ) ( )( )22 2 2 21.05 lb 0.17693 lb/ft s 3.6 ft/s 0.13043 lb/ft s 1.05 ft/sP = + ⋅ + ⋅

1.05 lb 0.63695 lb 0.14380 lb 1.83075 lb= + + =

1.831 lbP = "




Kinematics:

= aA + Lα, 0.7071Aa Lα=

aG = aA + 2

Lα = 0.7071 Lα + 2

Lα =

PM m+ ∑ = ( )0.70712

Lg

1

12m= 2L mα +

2 2

L Lα ⎛ ⎞⎜ ⎟⎝ ⎠

3

0.70712

g

Lα ⎛ ⎞= ⎜ ⎟⎝ ⎠

232.2 ft/s , 3 ft,g L= = (a) 211.38 rad/s=α !

( )0.70712x

m LmLF T

α∑ = = = ( )0.5 3

2

g

L

3

8mg

⎛ ⎞=⎜ ⎟⎝ ⎠

(b) 5.63 lb=T !

3 5

,8 8yF N mg mg N mg∑ = − = − =

(c) 9.38 lb=N !

aB

2

Lα 45°




y components

16

sin 30 sin 60Ga α°= °12

( ) 16 30.5

2Ga α=12

From Kinematics ⇒ 16

3Ga α=12

L 16

Note: ft;2 12

= 32

ft;12

PA = 16

3 ft12

PG =

( )2

2

16 1 16 lb 3216 lb ft cos30 ft

12 12 1232.2 ft/sPM α

⎛ ⎞⎛ ⎞ ⎛ ⎞∑ = ° = ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠

( ) ( )2

2

16 lb 16ft 3 3

1232.2 ft/sα

⎛ ⎞⎛ ⎞+ ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

( )18.4752 0.29446 2.6501 α= +

(a) 26.2744 rad/sα =

26.27 rad/s=α !

(b) ( )2

22

16 1 16 lb 32ft cos30 ft 6.2744 rad/s

12 2 1232.2 ft/sG AM N⎛ ⎞⎛ ⎞ ⎛ ⎞∑ = ° =⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠

1.600 lbAN =

or 1.600 lbAN = !




2

1 2b PG rπ

⎛ ⎞= = −⎜ ⎟⎝ ⎠

2

2 22

rI mr mα α

π⎡ ⎤⎛ ⎞= −⎢ ⎥⎜ ⎟⎝ ⎠⎢ ⎥⎣ ⎦

2 81mr α

π2⎛ ⎞= −⎜ ⎟⎝ ⎠

PM m∑ = g r2

22

2 8 21 1 2 1mr α

π ππ

⎡ ⎤⎛ ⎞ ⎛ ⎞ ⎛ ⎞− = − + −⎢ ⎥⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎢ ⎥⎣ ⎦

m= 2r8

3απ

⎛ ⎞−⎜ ⎟⎝ ⎠

(a)

21

,8

3

g

r

πα

π

⎛ ⎞−⎜ ⎟⎝ ⎠=⎛ ⎞−⎜ ⎟⎝ ⎠

231.4 rad/s=α !

(b)

22

12

18

3x B

mgF N mb mr

α παπ

π

⎛ ⎞−⎜ ⎟⎝ ⎠⎛ ⎞∑ = = = − =⎜ ⎟⎝ ⎠ ⎛ ⎞2 −⎜ ⎟⎝ ⎠

5.71 NBN = !




From 16.113:

22 8

1 3PM mg r Fr mr απ π

⎛ ⎞ ⎛ ⎞∑ = − − = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

227.5 rad/s=α !




Kinematics: Assume αααα ( )0ω =

/= +A B A Ba a a

aA 25° = aB + 1.5 α 35°

Law of sines:

1.5sin 60 sin 65

Ba α=° °

1.4333Ba α=

/ 1.4333 α= = + =G B G Ba a a a 0.75 α+ 35°

0.75 cos 35α= °xa 0.61436 α=

1.4333 α= −ya 0.75 sin 35α+ ° 1.0031 α=

Kinetics:

Law of sines: ( )sin 60, 1.5 m 1.4333 msin 60 sin 65 sin 65

°= = =° ° °

EB AB EB



( )1.4333 m 0.75 m cos55 1.0031 mED EB DB= − = − °=

0.75 sin 55 0.61436 mDG = °=

(a) ( ) ( ) ( ) ( )eff : α∑ = ∑ = + +E E x yM M W ED I ma DG ma ED

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )216 9.81 1.0031 6 1.5 6 0.61436 0.6143612

α α= +

( ) ( ) ( )6 1.0031 1.0031α+

( )59.04 1.125 2.2646 6.3778 α= + +

59.04 9.427 6.263,α α= = + 26.26 rad/s=α "

( ) 20.61436 6.263 3.848 m/sxa = =

( ) 21.0031 6.263 6.282 m/sya = =

(b) ( )eff :A AM M∑ = ∑

( ) ( )0.75 cos55 1.5 sin 55W B° − °

( ) ( )0.75 cos55 0.75 sin 55y xI ma maα= + ° − °

( ) ( ) ( )6 9.81 0.75 cos55 1.5 sin 55B° − °

( ) ( ) ( ) ( ) ( ) ( ) ( )21 6 1.5 6.263 6 6.282 0.75 cos55 6 3.848 0.75 sin 5512

= + ° − °

25.321 1.2287 7.046 16.214 14.184− = + −B

16.245 1.2287B= 13.22 N=B "



Chapter 16, Solution 116. Kinematics: 24 m/sB =a ( )0ω =

/A B A B= +a a a

aA 25° = 4 + 1.5 α 35°

Law of sines:

1.5 4sin 65 sin 60

α =° °

22.7907 rad/s=αααα

/ 4G B A B= = + =a a a a ( )0.75 2.7907+ 35°

( )0.75 2.7907 cos35x = °a 21.7145 m/s=

4y =a ( )0.75 2.7907 sin 35+ ° 22.7995 m/s=

Kinetics: ( )( )22 21 1 6 kg 1.5 m 1.125 kg m12 12

I m= = = ⋅!

Law of sines:

( )sin 60 1.5, 1.4333 m

sin 60 sin 65 sin 65EB AB EB

°= = =

° ° °



1.4333 0.75 cos55 1.0031 mED EB DB= − = − °=

0.75 sin 55 0.61436 mDG = °=

( ) ( ) ( ) ( ) ( )eff :E E x yM M P EB W ED I ma DG ma EDα∑ = ∑ − = + +

( ) ( ) ( ) ( ) ( ) ( ) ( )1.4333 6 9.81 1.0031 1.125 2.7907 6 1.7145 0.61436P − = +

( ) ( ) ( )6 2.7995 1.0031+

1.4333 59.0425 3.1395 6.3199 16.8491P − = + +

1.4333 85.531P =

59.5 N=P "




Kinematics: Crank AB:

12 rad/sAB =ωωωω , 280 rad/sABα =

( ) ( )22 23 ft 12 rad/s 36.0 ft/s12B na rω = = =

( ) ( )2 23 ft 80 rad/s 20.0 ft/s12B ta rα = = =

Rod BD Velocities:

( ) ( )3 12 3 ft/s12B ABAB= = =v ωωωω

3 ft/s0.5 ft

BBD

vBC

= =ωωωω

6 rad/sBD =ωωωω

Accelerations:



Translation with B and rotation about D = Planar motion

/D B D B= +a a a

Da = 20 + 36 ( )/D B na+ ( )/D B t

a+ (1)

Where ( ) ( ) ( )22 2/

10 6 30 ft/s12D B BDn

a BD ω= = =

And ( ) ( )/1012D B BD BDt

a BD α α= =

x components:

( )4 3 1020 30 05 5 12 BDα − − + =

1 20 24 44.02 BDα = + =

288 rad/sBDα =

Since ( )/ is directedB D ta :

288.0 rad/sBD =αααα

/ /12G B G B B D B= = + = +a a a a a a

20=a + 36 ( )1 302

+ ( )1 10 882 12 +

( ) ( )4 320 15 36.667 10.005 5xa = − − + = −

210.00 ft/sx =a

( ) ( ) 23 436 15 36.667 74.33, 74.33 ft/s5 5y ya = + + = =a

continued



Kinetics: 21 3 10 0.17361

12 12I

g g = =

( )eff6 4 4 3: ft ft

12 12 12 12B B y xM M D W I ma maα ∑ = ∑ − = + −

( ) ( ) ( )1 0.17361 3 3 110.5 3 88 74.33 103 3 4

Dg g g

− = + −

( )10.5 1 82.108 , 7.10 lb32.2

D D− = = 7.10 lb=D "




Kinematics: 12 rad/sAB =ωωωω , 280 rad/sAB =αααα Same analysis as for Prob. 16.117 except that ABαααα ,

Resulting in ( )B ta Eq. (1) must be replaced by,

Da = 20 + 36 ( )/D B n+ a ( )/D B n

+ a (1)

Where we have again ( ) 2/ 30 ft/sD B n

a =

and ( )/ /1012D B B Dt

a α=

x components:

( )4 3 1020 30 05 5 12 BDα − + =

1 20 24 4.002 BDα = − + =

28.00 rad/sBDα =

Since ( )/ is directedD B ta :

28.00 rad/sBD =αααα

/ /12G B G B B D B= = + = +a a a a a a

20=a + 36 ( )1 302

+ ( )1 10 8.002 12 +

( ) ( ) 24 320 15 3.333 10.00 ft/s5 5xa = − + = +

( ) ( ) 23 436 15 3.333 47.67 ft/s5 5ya = + + = +

Thus: 210.00 ft/sx =a , 247.67 ft/sy =a

continued



Kinetics:

21 3 10 0.17361

12 12I

g g = + =

( )eff6 4 3 4: ft ft ft ft

12 12 12 12B B y yM M D W I ma maα ∑ = ∑ + = + +

( ) ( ) ( )1 0.17361 3 3 110.5 3 8.00 10 4.7673 4 3

Dg g g

− = + +

( )10.5 1 56.556 , 5.513 lb32.2

D D− = =

5.51 lb=D "




Kinematics: Since ( ) 0,D n =v

( ) ( )cos60 cos30D Dx yv v° = °

( ) ( ) ( ) ( )( )0.75 m 6 rad/sD B BAx xv v AB ω= = =

( ) ( ) cos60cos30D Dy xv v °=

°

( )( ) ( )1

0.75 620.75 633

2

= =

But ( ) ( )D BDyv BDω=

( ) ( )0.75 6 6 rad/s

0.75 3 3D y

BD

v

BDω = = =

( ) ( )22 20.75 m 6 rad/s 27 m/sB BAAB ω= = =a

/D B D Ba a a= + 2136 1 rad/s3BDα = +

256.7846 rad/s=

continued



( )eff :B BM M∑ = ∑ ( )( ) ( )39.24 N 0.375 m cos30 0.75 mBM∑ = − + °D

( )( ) ( )( ) 10.1875 56.7846 54 0.375 1 N m3

= − − ⋅

( )0.64952 14.1715 10.6471 8.5587 N m= + − ⋅D

25.87 N=D 60°

or 25.9 N=D 60° "




Kinematics:

Velocity

( )sin 30B AB ABAB CDω= = °v ω

( )( )0.45 m sin 30 5 rad/s= °

1.125 m/s=

sin 30CD BCB = °vω

( )1.125 m/s sin 30

0.45 mCD

°=ω

1.25 rad/sCD =ω

( )( )1.125 m/s cos30 0.97428r = ° =&

Acceleration

( ) ( )( ) ( )( )2 20.45 2 13.5 m/s sin 30 5.625 m/s cos30CD CDr+ = ° + °α ω

220.4127 rad/sCD =α

Kinetics: C C CDM I∑ = α

( ) ( ) ( ) ( )39.24 N cos30 0.45 m 0.45 mCM N∑ = ° +

( ) ( )214 kg 0.9 m

3 CDα=

( ) ( )0.45 m 15.292 22.0457 N mN = − + ⋅

15.01 NN = or 15.01 N=N 60° !




Kinematics: From Prob. 16.120

21.25 rad/sCDω = (1.125 m/s)(0.866) 0.97428 m/sr = =&

Acceleration

2 2(0.45 m) 2 (13.5 m/s )(sin 30 ) (5.625 m/s )(cos30 )CD CDrα ω+ = ° − °&

2(0.45 m)( ) (1.125 m/s)(0.866)(1.25 rad/s) 1.8786 m/sCDα − =

29.5873 rad/sCDα = Kinetics:

C C CDM I α=∑

21

( 39.24 N)(cos30 )(0.45 m) (0.45 m) (4 kg)(0.9 m)3C CDM N αΣ = − ° + =

Substituting 29.5873 rad/sCDα =

57.0 NN = !




2 ft/s2 rad/s

1 ftω = = Ga / :A G A= +a a

15

12rα α=

221 1 8.05 30

(3)12 12 32.2 12

m α ⎛ ⎞⎛ ⎞= ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

l

0.390625 lb ft= ⋅

8.056.25 1.5625 lb

32.2Gma⎛ ⎞= =⎜ ⎟⎝ ⎠

15 15

(0.6) (0.8) 0.39062512 12G AM N P

⎛ ⎞+Σ = − =⎜ ⎟⎝ ⎠

(a) 4.48 lb=P !

(b) 6.49 lbA =N !

0

23 rad/sα = 26.25 ft/sGa =

8.05 lb 1.5625 lby AF N+Σ = − = −

6.4875 lbAN =




Crank BC:

( ) ( ) ( )2 25 ft 15 rad/s 6.25 ft/s12B ta BC α = = =

( ) ( ) ( )2 2 25 ft 6 rad/s 15 ft/s12B na BC ω = = =

Rod ABD: 1 1 0.1 msin sin 300.2 m

BCAB

β − −= = = °

/A G A Ga a a= +

[ Aa ] [6.25= 15+ ] 1012

α+ β

100 15 cos3012

α= − °

( )2

215 ft/s

20.7846 rad/s10 ft cos3012

α = = °



Kinetics:

( ) ( ) 2eff

10 1: ft cos3012 12G GM M A I mLα α Σ = Σ ° = =

( ) ( )2

22

1 10 lb 200.72168 ft ft 20.7846 rad/s12 1232.2 ft/s

A =

2.07041 lbA =

2.07 lb=A !




Kinematics: Aa=a

where ( )/G A trα=a θ

Kinetics:

( ) ( ) ( )eff : 0 cosA A AM M I mr r ma rα α θΣ = Σ = + −

2 2 cosAmk mr ma rα α θ+ =

2 2 cosAa rk r

α θ=+

Setting ,ddωα ωθ

= and using 0.44 m, 0.25 mr k= =

( )( ) ( )2 2

0.44 mcos 1.7181 cos

0.44 m 0.25 mA

Aad a

dωω θ θθ

= =+

20 01.7181 cosf

Ad a dπω ω ω θ θ=∫ ∫

22 20

0

1 1.7181 sin 3.43622

f

A f Aa aπω

ω θ ω= ⇒ = (1)



Given 22 m/sAa =

( )2 3.4362 2 6.8724 2.6215 rad/sf fω ω= = ⇒ =

or 2.62 rad/sfω = !




Given data: 2 rad/sfω =

( )2 22 3.4362 1.1641 m/sA Aa a= ⇒ =

or 21.164 m/sAa = !




Kinematics:

Kinetics:

/B AB B Aa rα= =

Da 9= ←⎯⎯ + ( )0.9 , 0.9 0.866 4.5BD BDα α =

210rad/s

3BDα =

9Ga = ←⎯⎯ +

100.45

3=

4 kgm =

( )( )2 21 104 kg 0.9 m rad/s

12 3Iα

⎛ ⎞= ⎜ ⎟⎝ ⎠

2.7

N m3

= ⋅

Bar AB

( ) ( )( ) ( )2 210.9 m 4 kg 0.9 m 10 rad/s

3A xM BΣ = =

12.0 NxB =

Bar BD

( ) ( )0.939.24 N 0.45 m

sin 30EM F⎛ ⎞Σ = + −⎜ ⎟°⎝ ⎠

0.9

(12.0) N mtan 30

⎛ ⎞− ⋅⎜ ⎟°⎝ ⎠

( )( ) ( )4 4.52.7 0.9

4(9) 0.45 N mtan 303 3

⎡ ⎤⎛ ⎞= + + ⋅⎢ ⎥⎜ ⎟°⎝ ⎠⎣ ⎦

35.2 N=P !

4.5 3 9

9 M/S2




Kinematics:

Kinematics: 2 2/ 22.5 m/sB AB B Aa rω= =

4.5 5

rad/s rad/s0.9 3 3

BDω = =

Da = 22.5 + 20.9 7.550.93

BDα⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

+ ←⎯⎯⎯⎯⎯ =

2125 1 rad/s

3 3BDα ⎛ ⎞= −⎜ ⎟

⎝ ⎠

Ga = 22.5 + 1

(6.45)(25) 13 3

⎛ ⎞− +⎜ ⎟⎝ ⎠

=

2 21 1(4 kg)(0.9 m) 25 1 rad/s

12 3 3I α ⎛ ⎞= −⎜ ⎟

⎝ ⎠

1

4.6875 1 N m 5.45096 N m3 3

⎛ ⎞= − ⋅ = ⋅⎜ ⎟⎝ ⎠

Kinetics: Bar AB

0 since 0, 0A AB xM BαΣ = = =

Bar BD

(1.8 m) (39.24 N)(0.45 m) 0PM F= + +

3.75

5.45096 4(3.75)(0.9) 3 4 11.25 (0.45) N m3

⎡ ⎤⎛ ⎞= − + + + ⋅⎢ ⎥⎜ ⎟⎝ ⎠⎣ ⎦

13.567 NF = 13.57 N=F !

37.5

3.75

3.7511.25

3

⎛ ⎞+⎜ ⎟⎝ ⎠




Kinematics: 0BFω =

Therefore 2 ,AB AB BD BE AB BDl l l lω α= = 2 2225 rad/sBE ABα ω= =

Bar BE

aG = + (10/12) (15)2 =

Kinetics: :AB 105 ft lb ft12A xM B Σ = ⋅ +

2

21 4 lb 10 ft3 1232.2 ft/s ABα =

BE: 28 lb 10 ft

1232.2 ft/sy y ABF B α Σ = − =

(a) 25 24.84 rad/s0.1725 0.02875ABα = =

+

224.8 rad/sAB =αααα !!!!

2

22

1 4 lb 10 ft (225 rad/s )12 1232.2 ft/sDM =

4 lbD N= +

BE: 10 10ft (8 lb) ft12 12BM N Σ = −

22

21 8 lb 20 (225 rad/s )

12 1232.2 ft/sDM − =

25.469 NN = (b) 25.469 (4 lb)D = +

= 29.46 lb 29.5 lb=D !!!!

Radial acceleration of B/A is equal to the tangential acceleration of B/D,

( )10 /12 ABB α

10/12 (15)2

(10/12) αAB




From Prob. 16.128

2 20, 900 rad/sBE BE ABω α ω= = =

210 10ft ( ) ft (96 rad/s )

12 12G ABa α⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

280 ft/s=

AB: 10

ft12A xM M B⎛ ⎞Σ = + ⎜ ⎟⎝ ⎠

2

22

1 4 lb 10ft (96 rad/s )

3 1232.2 ft/s

⎛ ⎞⎛ ⎞= ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

BE: 2

2

8 lb(80 ft/s )

32.2 ft/sx xF B

⎛ ⎞Σ = − = ⎜ ⎟

⎝ ⎠

19.876 lbxB = −

(a) 2.761 16.563 19.324 ft lbM = + = ⋅

19.3 ft lbM = ⋅ !

Sleeve: 2

22

1 4 lb 10ft (900 rad/s )

12 1232.2 ft/sDM

⎛ ⎞⎛ ⎞= ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

6.470 ft lb, 4 lbDM D N= ⋅ = +

BE: 10 10

ft (8 lb) ft12 12BM N⎛ ⎞ ⎛ ⎞Σ = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

2

22

1 8 lb 20ft (900 rad/s )

12 1232.2 ft/sDM

⎛ ⎞⎛ ⎞− = ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

77.876 lbN =

(b) (77.876 4) lb 81.876 lbD = + =

or 81.9 lb=D !




Kinematics:

Velocity

0ABω =

( )( )0.06 m 6 rad/sB Av v= =

( )0.18 m BCω=

2 rad/sBCω =

Acceleration

( )( )2 20.06 m 6 rad/s 2.16 m/sA = =a

( )( )2 20.18 m 2 rad/s 0.72 m/sB = =a

( )( )2 20.09 m 2 rad/s 0.36 m/sBC = =a

( ) ( ) 2 21 1 2.16 0.72 m/s 1.44 m/s2 2AB A Ba a a= + = + =

0.12A B ABa a α= +

( )2 22.16 m/s 0.72 m/s 0.12 m ABα= +

21.2 rad/sABα =

continued



Kinetics: ( )( )22 22 kg 0.12 m1 0.0024 kg m

12 12ABI mL= = = ⋅

Rod BC:

Since 0, 0BC aα = =

0CMΣ = yields 0xB =

Rod AB:

( )eff : 0x x xF F AΣ = Σ =

( ) ( ) ( ) ( )eff : 0.12 m 0.06 m 0.06 mA A y AB AB AB ABM M B mg I m aαΣ = Σ − = −

( ) ( )( )( )20.12 m 2 kg 9.81 m/s 0.06 myB −

( )( ) ( )( )( )2 2 20.0024 kg m 12 rad/s 2 kg 1.44 m/s 0.06 m= ⋅ −

8.61 NyB =

or 8.61 N=yB !

( )eff:y y y y AB ABF F A B mg m aΣ = Σ + − = −

( )( ) ( )( )2 28.61 N 2 kg 9.81 m/s 2 kg 1.44 m/syA = − + −

8.13 NyA =

or 8.13 N=yA !




Kinematics: Velocity of all elements C= Acceleration:

( )( )2 20.06 m 18 rad/s 108 m/sB A= = =a a

26 rad/s0.18 m

BBC BC

aα α= ⇒ =

( )( )2 20.09 m 6 rad/s 0.54 m/sBCa = =

21.08 m/sAB A Ba a a= = =

Kinetics: ( ) ( )( )2 2 21 1 3 kg 0.18 m 0.0081 kg m12 12BC BCI m BC= = = ⋅

Rod BC:

( ) ( ) ( )eff : 0.18 m 0.09 mA A x BC BC BC BCM M B I m aαΣ = Σ = +

( ) ( )( ) ( )( )( )2 2 20.18 m 0.0081 kg m 6 rad/s 3 kg 0.54 m/s 0.09 mxB = ⋅ +

1.08 NxB =

on AB, 1.08 NxB =

continued



Rod AB:

( )eff :x x x x AB ABF F A B m aΣ = Σ − =

( )( )21.08 N 2 kg 1.08 m/s , 3.24 Nx xA A− = =

( ) ( )0: 0.12 m 0.06 m 0A yM B mgΣ = − =

( )( )2 0.06 m2 kg 9.81 m/s 9.81 N0.12 myB

= =

0: 0y y yF A B mgΣ = + − =

( )( )22 kg 9.81 m/s 9.81 N 9.81 NyA = − =

"





2600 rpm 62.832 rad/s60ABπω = =

( ) ( )22 24 ft 62.832 rad/s 1315.9 ft/s12B ABa AB ω = = =

Also: ( ) ( )4 ft 62.832 rad/s 20.944 ft/s12B ABv AB ω = = =

Connecting rod BD: Velocity Instant, center at D.

2

220.944 ft/s 25.133 rad/s10 ft12

BBD

vBD

ω = = =

Acceleration:

/D B D B Ba= + = a a a ( ) 2BDBD ω +

21315.9 ft/sD = a ( )210 ft 25.133 rad/s12

+ 2789.53 ft/s =

( ) [1 1 1315.92 2BD B Da a= + =a 789.53+ ] 21052.7 ft/s=

Kinetics of piston:

( )22

4.5 lb 789.53 ft/s 110.34 lb32.2 ft/sD = =a

110.3 lbD =

Force exerted on connecting rod at D is: 110.3 lbD =

continued



Kinetics of connecting rod (neglect weight)

( )eff :x xF FΣ = Σ

( )22

3 lb110.34 lb 1052.7 ft/s32.2 ft/s

B − =

208.42 lbB =

or 208 lb=B "

110.3 lb=D "





2600 rpm 62.832 rad/s60ABπω = =

( ) ( )22 24 ft 62.832 rad/s 1315.9 rad/s12B ABa AB ω = = =

Also: ( ) ( )4 ft 62.832 rad/s 20.944 ft/s12B ABv AB ω = = =

Connecting rod BD:

Velocity Instant center at D:

20.944 ft/s 25.133 rad/s10 ft12

BBD

vBD

ω = = =

Acceleration

/D B D B Ba a a= + = a ( ) 2BDBD ω +

21315.9 ft/sD = a ( )210 ft 25.133 rad/s12

+ 21842.3 ft/s =

( ) (1 1 1315.92 2BD B Da a= + =a 1842.3+ ) 21579.1 ft/s=

Kinetics of piston:

( )22

4.5 lb 1842.3 ft/s 257.46 lb32.2 ft/sD = =a

257.46 lbD =

Force exerted on connecting rod at D is: 257.46 lbD =

continued



Kinetics of connecting rod (neglect weight)

( )eff :x xF FΣ = Σ

BD BDB D m a− =

( )22

3 kg257.46 lb 1579.1 ft/s32.2 ft/s

B − =

404.581 lbB =

or 405 lb=B "

257 lb=D "




Kinematics:

( 520 mm)BD = aD =

0.48Da α∴ = and 0.20Ba α=

so that DEα α= and ABα α= G = Midpoint of BD: 0.2Ga α= + 0.26α

aG =

210.2 (0.2)

3A x AM B I mα αΣ = = =

0.2 0.06673xB m mα α= =

(0.48) (0.24)E y EM D mg I αΣ = + =

21 (0.48)

3m α=

0.16 0.5yD m mgα= −

continued

0.1 α0.24

Kinetic analysis

aB+ Where BDα α=



0.0667PM mΣ = − (0.2) [0.16 mα − 0.5 mα − ](0.48)g

m+ 1(0.24)12

g m= 2(0.52) 0.1mα + (0.1) 0.24 mα + (0.24)α

0.48 0.180273g α=

2 29.81 m/s , 26.1 rad/sg α= =

226.1 rad/sA =αααα !

226.1 rad/sBD =αααα !

226.1 rad/sDE =αααα !




Kinematics: We resolve the acceleration of G into the acceleration of A and the acceleration of G relative to A:

/G A G A= = +a a a a

/A G A= +a a a

where /12G Aa Lα= (1)

Kinetics: eff( ) :A AM MΣ = Σ

/0 ( )2G ALI maα = +

( cos )2ALma β −

But, from (1)

/12G Aa Lα=

and 2112

I mL=

Thus: 21 1 ( cos ) 012 2 2 2A

L LmL m L maα α β + − =

2cos3Aa Lβ α= (2)

eff( ) :x xF FΣ = Σ

/sin ( )cosA G AW ma maβ β= −

1sin cos2Amg ma m Lβ α β = −

1sin cos2Aa g Lβ α β= + (3)

continued



Substitute for Aa from (3) into (2):

21 2sin cos cos2 3

g L Lβ β α β α+ =

21sin cos (4 3cos )6

g Lβ β α β= −

26 sin cos

4 3cosg

Lβ βα

β=

− (4)

Substitute for α into (2) and solving for :Aa

2 22 6 sin cos 4 sin,

3cos 4 3cos 4 3cosA AL g ga

Lβ β β

β β β= =

− −a β

(5)

We note that the senses of αααα and aA are always as indicated, since 24 3cos 0.β− >

eff /( ) : cos siny y G AF F A W maβ βΣ = Σ − = −

1cos sin2

A mg m Lβ α β = −

Substitute for α the expression obtained:

2sin 6 sin coscos2 4 3cos

L gA mg mL

β β βββ

= −−

2 2 2

2 23sin 4 3cos 3sincos 1 cos

4 3cos 4 3cosmg mgβ β ββ β

β β − −= − = − −

2cos

4 3cosmg β

β=

−A β (6)

Given data: 24 kg, 1.5 m, 20 , 9.81 m/sm L gβ= = = ° =

(a) Substitute into equation (4):

26(9.81) sin 20 cos 20 0.3213939.24 9.3354

1.5 1.350934 3cos 20α ° °= = =

− °

29.34 rad/s=αααα !



(b) Substitute data into equation (5):

24(9.81)sin 20 13.4209 9.9345

1.350934 3cos 20Aa °= = =− °

29.93 m/sA =a 20° ! (c) Substitute data into equation (6):

2cos 20 36.875(4)(9.81) 27.295

1.350934 3cos 20A °= = =

− °

27.3 N=A 70° !




Kinematics:

Kinetics:

Disk: eff( ) : (58.86 N) (cos 45 ) (6 kg) (0.25 m)x xF F B F αΣ = Σ − + ° −

eff 2( ) : (58.86 N)(sin 45 ) = 0y yF F N BΣ = Σ + − °

2

eff1

( ) : (0.25 m) = (6 kg) (0.25 m)2G GM M F αΣ = Σ

Bar: x-Dir 1: (19.62 N) cos 45 (2 kg) (0.25 m) (cos 45 ) (2 kg) (0.375m) BDB α α− + °= + °

y-Dir 2: (19.62 N) sin 45 = sin 45 (2 kg) (0.375 m) BDB α+ ° °

Six equations and six unknowns: 13.1967 N, = 48.3878 NF N=

(a)

50.2 N= 60.3° !

(b) 13.1967

0.27348.3878

F

N= = !

13.1967 48.3878

45° = 24.884

43.547 = N + F



Chapter 16, Solution 137. Kinematics: Assume Aα and Bα 0A Bω ω= =

D Arα=a

E D Aa rα= =a

/B E B Ea a= +a

( )A Br rα α= +

( )B A Br α α= +a

Kinetics: Disk A:

( )eff :A AM MΣ = Σ

ATr I α=

212 ATr mr α=

2A

Tmr

α = (1)

Disk B: ( )eff :B BM MΣ = Σ

BTr I α=

212 BTr mr α=

2B

Tmr

α = (2)

continued



From (1) and (2) we note that A Bα α=

( ) ( )eff :E E B BM M Wr I ma rαΣ = Σ = +

( )212 B A BWr mr mr rα α α= + +

:A Bα α= 252 AWr mr α= (a) 2

5Agr

=α "

25B

gr

=α "

Substitute for Aα into (1):

2 25

g Tr mr

= (b) 15

T mg= "

( ) ( ) 22 25B A B A

ga r r rr

α α α = + = =

(c) 45B g=a "



Chapter 16, Solution 138. Kinematics: Cylinder: 0ω =

Rolling without sliding ( ) 0C xa = ( ) /A C A Cx= +a a a

0 Arα= +

A Arα=a

AA

ar

α =

Rod AB: 2A AB ABLa aα= −

Kinetics: Cylinder: ( )eff :C CM MΣ = Σ

x A AA r ma r I α= +

212

Ax A

aA r ma r mrr

= +

32x AA ma=

32 2x AB AB

LA m aα = −

(1)

continued



Rod AB: ( )eff :A AM MΣ = Σ

2AB ABLPL ma I α= +

2

2 12AB ABL mPL ma L α= + (2)

( )eff :x x x ABF F P A maΣ = Σ + =

Substitute from (1): 32 2 AB AB AB

LP m a maα + − =

5 32 4AB ABP ma mLα= − (3)

Multiply by :9L 21 5 1

9 18 12AB ABLPL ma mL α= − (4)

(4) + (2): 10 1 5 79 2 18 9AB ABPL mLa mLa = + =

107AB

Pam

= (5)

(5) (3) 5 10 32 7 4 AB

PP m mLm

α = −

25 37 4 ABP P mLα= −

18 37 4 ABP mLα− = 24

7ABP

mL=α

24 102 2 7 7A AB ABL L P Pa a

mL mα = − = −

12 10 ;7 7A

Pam

= −

27A

Pm

=a "

24 102 2 7 7B AB ABL L P Pa a

mL mα = + = +

12 10 ;7 7B

Pam

= +

227B

Pm

=a "




Kinematics: Assume ABα BCα and 0AB BCω ω= =

2AB ABL α=a

B ABLα=a

12 2BC B BC AB BCLa L Lα α α= + = +a

Kinetics: Bar BC ( )eff :B BM MΣ = Σ

( )2BC BCLPL I maα= +

2

12 2 2BC AB BCm L LL m Lα α α = + +

1 12 3AB BCP mL mLα α= + (1)

( )eff :x x x BCF F P B maΣ = Σ − =

12x AB BCP B m L Lα α − = +

(2)

continued



Bar AB: ( )eff :A AM MΣ = Σ

( )2x AB ABLB L I maα= +

2

12 2 2AB ABm L LL mα α = +

13x ABB mLα= (3)

Add (2) and (3): 4 13 2AB BCP mL mLα α= + (4)

Subtract (1) from (4)

5 106 6AB BCmL mLα α= +

5BC ABα α= − (5)

Substitute for BCα in (1): ( )1 1 752 3 6AB AB ABP mL mL mLα α α= + − = −

67AB

PmL

α = − (6)

Eq. (5) 657BC

PmL

α = − −

307BC

PmL

α = (7)

Data: 25 in. 2.0833 ftL = =

22

6 lb 0.18634 lb s /ft32.2 ft/s

m = = ⋅

5 lbP =

( )( )26 6 5 lb7 7 0.18634 lb s /ft 2.0833 ft

ABP

mLα = − = −

⋅

211.040 rad/s= −

211.04 rad/sAB =αααα "

( )( )230 30 5 lb7 7 0.18634 lb s /ft 2.0833 ft

BCP

mLα = =

⋅

255.201 rad/s=

255.2 rad/sBC =αααα "




Kinematics: We choose AB BCα α α= =

2ABL α=a

B Lα=a

32BC Lα=a

Kinetics: Bars AB and BC (Acting as rigid body)

2ABCm m=

( )( )21 2 212

I m L=

223

I mL=

( ) ( )eff :A A ABC ABC BM M P L b I m a LαΣ = Σ + = +

( ) ( )( )22 23

P L b mL m L Lα α+ = +

( ) 283

P L b mL α+ = (1)

continued



Bar BC: ( )eff :B BM MΣ = Σ

( )2BC BCLPb I maα= +

2 312 2 2m LL m Lα α = +

256

Pb mL α=

265

PbmL

α = (2)

Substitute for α into (1):

( ) 22

8 63 5

PbP L b mLmL

+ =

16 16 11; 15 5 5

PL Pb Pb L b b + = = − =

511

b L=

Eq. (2) 26 55 11

P LmL

α =

611

PmL

α =

Data: 25 in. 2.0833 ft,L = =

22

6 lb 0.18634 lb s /ft32.2 ft/s

m = = ⋅

5 lbP =

(a) ( )5 5 2.0833 ft 0.94695 ft11 11

b L= = =

0.947 ftb = "

(b) ( )( )( )2

5 lb6 611 11 0.18634 lb s /ft 2.0833 ft

PmL

α = =⋅

27.0254 rad/s=

27.03 rad/s=α "




Kinematics: Assume ABαααα , BCαααα , Ca , 0AB BCω ω= =

Since 0: ( ) ( ) ( ) ( )AB BC A x B x B x BC x Cω ω= = = = = =a a a a a

1( )2BC y BCLα=a ( )B y BCLα=a

1 1( ) ( )2 2AB y B y AB BC ABL L Lα α α = + = +

a a

( ) ( ) ( )A y B y AB BC ABL L Lα α α= + = +a a

Kinetics—entire system:

eff( ) : 0 ( ) ( )x x AB x BC xF F m a m aΣ = Σ = +

0 2 Cm a=

Thus: ( ) ( ) ( ) ( ) 0A x AB x B x BC x C= = = = =a a a a a

continued



Rod AB

eff1 1( ) : ( )2 2B B AB ABM M mg L I ma Lα Σ = Σ = +

21 1 1 1

2 12 2 2AB BC ABmgL mL m L L Lα α α = + +

1 1 12 3 2AB BCg α α= +

23 AB BCL gα α+ = (1)

eff( ) :y y ABF F B w m a+ Σ = Σ − = −

12BC ABB mg m L Lα α = − +

(2)

Rod BC

eff1 1( ) : ( )2 2C C BC BCM M BL W L I m a Lα Σ = Σ + = +

21 1 1 1

2 12 2 2BC BCBL mgL mL m L Lα α + = +

1 12 3 BCB mg mLα= − + (3)



Subtract (3) from (2)

3 4 102 3 2BC ABmg m L Lα α = − +

3 98 8AB BCL L gα α+ = (4)

Subtract (1) from (4): 3 2 18 3 8ABL gα − =

7 1 ,24 8ABL gα− = 3

7ABgL

α −=

Substitute in (1): 2 3 ,3 7 BC

gL L gL

α− + =

97BC

gL

α =

(a) 9 3 67 7 7A BC ABa L L g g gα α= + = − = 6

7A g=a "

(b) 97B BCa L gα= = 9

7B g=a "




Rod AB: 2La α=

( )effA AM MΣ = Σ

( )2LPL ma I α= +

212 2 12L Lm mLα α = +

3PmL

α = (1)

( )eff :x x xF F A P maΣ = Σ − = −

3 1; 2 2 2 2x xL L P PA P m P m A P

mLα = − = − = − =

Portion AJ of rod:

External forces: , ,x AJA W axial force ,JF shear ,JV and bending moment JM

Effective forces: Since acceleration at any point is proportional to distance from A, effective forces are linearly distributed. Since mass per unit length is m/L, at point J we find

( )Jm ma xL L

α =

Using (1): 3J

m m PaL L mL

=

23

Jm PxaL L

=

( ) 2eff1 3 2:2 3J J J x

Px xM M M A x xL

Σ = Σ − = −

32

1 12 2J

PM Px xL

= − (2)



For 2max 2

3: 02 2

JdM P PM xdx L

= − =

3

Lx = (3)

Substituting into (2)

( )3

2max1 1 1 22 2 2 33 3 3J

PL P L PLML = − =

( )max 3 3JPLM = (4)

Note: Eqs. (3) and (4) are independent of W

Data: 0.9 m, 3.5 NL P= =

Eq. (3): 0.9 m 0.51962 m

3 3Lx = = =

0.520 mx =

Eq. (4): ( ) ( )( )max

3.5 N 0.9 m0.60622 N m

3 3jM = = ⋅

max 0.606 N m, 520 mm below M A= ⋅ "



Chapter 16, Solution 143. From answers to Prob. 16.79:

3 1 2 4Ba g A mg= =

We now find

32

Ba gL L

α = =

32J

ga x xL

α= =

Portion AJ of rod:

External forces: Reaction A, distributed load per unit length mg/L, shear ,JV bending moment .JM

Effective forces: Since : ,a x the effective forces are linearly distributed. The effective force per unit length at J is:

23 32 2J

m m g mga x xL L L L

= ⋅ =

( ) 2eff

1 3:4 2 2y y J

mg mg mgF F x V x xL L

Σ = Σ − + =

22

34 4J

mg mg mgV x xL L

= − +

( ) 2eff1 3:

2 4 2 32J J Jmg x mg mg xM M x x M x xL L

Σ = Σ − + =

2 32

1 14 2 4J

mg mg mgM x x xL L

= − +



Find min:V 23 20; 2 3

JdV mg mg x x Ldx L L

= − + = =

2

min min22 3 2 ;

4 3 4 3 12mg mg mg mgV L L V

L L = − + = −

Find maxM where 230: 04 4J J

mg mg mgV V x xL L

= = − + =

2 23 4 0x Lx L− + =

( )( )3 0 and 3Lx L x L x x L− − = = =

2 3

max1 1

4 3 2 3 4 3 27mg L mg L mg L mgLM

L L = − + =

2 3min

1 1 04 2 4

mg mg mgM L L LL L

= − + =

max 4mgV = "

max at from 27 3

mgL LM A= "




Links:

( ) ( )22 21.25 ft 6 rad/s 45 rad/sna rω= = =

( )( )2 21.25 ft 12 rad/s 15 rad/sta rα= = =

Bar AD is in translation B Ca a a= =

( ) ( ) ( )eff : sin 60 15 in. sin 60 15 in.G G CF BEM M F FΣ = Σ ° − °

CE BEF F=


( ) ( )( ) ( ) ( )sin 60 30 in. 15 lb 15 in. sin 30 15 in. sin 60 15 in.CF t nF ma ma° − = ° + °

( ) ( )( )2 22

15 lb25.981in. 225 lb in. 15 in. 15 rad/s sin 30 45 rad/s sin 6032.2 ft/sCFF

− ⋅ = ° + °

21.1586 lbCFF =

or 21.2 lb TCF BEF F= = !



( ) ( )eff : cos60 cos30 cos60x x BE CF t nF F F F P ma maΣ = Σ + ° + = − ° + °

22 2

15 lb 15 rad21.1586 lb cos30 45 rad/s cos 6032.2 ft/s s

P + = − ° + °

21.1586 lb 4.4299 lbP+ =

16.7287 lbP = − or 16.73 lb=P !




Lever ABC: Static Equilibrium (Friction Force )

0.35 NkF Nμ= =

( ) ( ) ( )0: 0.20 m 0.04 m 300 N 0.18 m 0AM N FΣ = − − =

( )0.20 0.04 0.35 N 54 NN − =

290.32 NN =

( )0.35 290.32 N 101.61 NkF Nμ= = =

Drum:

0.160 mr =

02

360 rpm60

πω ⎛ ⎞= ⎜ ⎟⎝ ⎠

0 12 rad/sω π=

( )eff:D DM MΣ = Σ

( )( ) ( )2; 101.61 N 0.16 m 18 kg mFr I α α= = ⋅

20.9032 rad/s (deceleration)α =

2 20 2 ;ω ω αθ= +

( ) ( )20 12 2 0.9032π θ= + −

786.77094 radθ =

1

786.77094 125.22 rev.2

θπ

⎛ ⎞= =⎜ ⎟⎝ ⎠

125.2 rev. θ = !




(a) While slipping occurs, a friction force F is applied to disk A, and F to disk B.

Disk A:

212A A AI m r=

( )( )21 8 kg 0.11 m2

=

20.0484 kg m= ⋅

: 30 NF N PΣ = =

( )0.15 30 4.5 NF Nµ= = =

( )eff :A A A A AM M Fr I αΣ = Σ =

( )( ) ( )24.5 N 0.11 m 0.0484 kg m Aα= ⋅

10.227Aα = 210.23 rad/sA =αααα !

Disk B:

212B B BI m r=

( )( )21 4 kg 0.08 m2

=

20.0128 kg m= ⋅

( )eff :B B B B BM M Fr I αΣ = Σ =

( )( ) ( )24.5 N 0.08 m 0.0128 kg m Bα= ⋅

228.125 rad/sBα = 228.1 rad/sB =αααα !

continued



(b)

( ) 2480 rpm 16 rad/s60A Bπω π = =

Sliding stops when C C′=V V or A A B Br rω ω=

( )0A A A B Br t r tω α α − =

( ) ( ) ( )( )2 20.11 m 16 rad/s 10.227 rad/s 0.08 m 28.125 rad/st tπ − =

1.6383 st =

( ) ( )( )20 16 rad/s 10.227 rad/s 1.6383 s 33.511 rad/sA A Atω ω α π= − = − =

( ) 633.511 rad/s 320.00 rpm2A

aωπ

= =

or 320 rpmA =ωωωω !

( )228.125 rad/s 1.6383 s 46.077 rad/sB Btω α= = =

( ) 6046.077 rad/s 440.226 rpm2Bωπ

= =

or 440 rpmB =ωωωω !




Disk: 21

2I mr=


P ma=

P

m=a

( )effG GM MΣ = Σ

Pr I α=

21

2Pr mr α=

2P

mrα =

Since ; w P

m a gg w

= = 2P

gwr

α =

(a) 2

1 3AP P

r g r gw r w

α ⎡ ⎤⎛ ⎞= + = + =⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

a a

( )

20.75 lb3 12.075 ft/s

6 lb32.2 ft/s

= = 2or 12.08 ft/sA =a !

(b) 2

1BP P

r g r gw r w

α ⎡ ⎤⎛ ⎞= − = − = −⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

a a

( )2

2

0.75 lb4.025 ft/s

6 lb

32.2 ft/s

= − = − 2or 4.03 ft/sB =a !




Kinetics:

( ) ( ) ( )2 22 2 22

1 1 400 lb 6.6 ft 3.6 ft 58.509 lb ft s12 12 32.2 ft/s

I m a b = + = + = ⋅ ⋅

( )eff

400: 400y y A BF F T T ag

Σ = Σ + − =

400 1A BaT Tg

+ = +

(1)

( ) ( )( ) ( )2eff : 3.3 ft 58.509 lb ft sG G A BM M T T I α αΣ = Σ − = = ⋅ ⋅

( ) ( )217.73 lb sA BT T α− = ⋅ (2)

Kinematics:

( ) 3.3A ya a α= +

3 3.3a α= + (3)

( ) 3.3B ya a α= −

1 3.3a α= − (4)

Solving (3); (4) 22 ft/s ,a = 20.30303 rad/sα =



From (1); (2) ( )2

22 ft/s400 lb 1 424.84 lb

32.2 ft/sA BT T

+ = + =

( )( )2 217.73 lb s 0.30303 rad/s 5.3727 lbA BT T− = ⋅ =

Solve 215.10636 lbAT = or 215 lbAT = !

209.73366 lbBT = or 210 lbBT = !




Statics: 1 21 12 2

T T W mg= = =

(a)

( )eff :2G GLM M T I α Σ = Σ =

21 12 2 12

Lmg mL α =

3gL

α = 3gL

=αααα !


12

mg mg ma− =

1 1 2 2

a g g= =a


/A G A G= +a a a

12 2A

Lg α= −a

1 32 2

L ggL

= −

;Aa g= − A g=a !



PROBLEM 16.149 CONTINUED (c) Acceleration of B:

/B G B G= +a a a

1 3 2 ;

2 2 2BL L ga a g g

Lα = + = + = +

2B g=a !




( ) ( )2

20.3 0.312C

mLM mg L m Lα αΣ = = +

0.3 451.730770.1733 26

g g gL L L

α = = =

(a) 360.826B

ga Lα= =

or 1813B

g=a !

(b) 0xR = 450.326y y

gF R mg mLL

Σ = − = −

135 25260 52y mg mg mg= − =R !




Statics: 1 21 12 2

T T W mg= = =

(a)

( )eff :2G GLM M T I α Σ = Σ =

21 12 2 12

Lmg mL α =

3gL

α = 3gL

=αααα !


12

mg mg ma− =

1 1 2 2

a g g= =a


/A G A G= +a a a

12 2A

Lg α= −a

1 32 2

L ggL

= −

;Aa g= − A g=a !



(c) Acceleration of B:

/B G B G= +a a a

1 3 2 ;

2 2 2BL L ga a g g

Lα = + = + = +

2B g=a !




( ) ( )2

20.3 0.3

12CmL

M mg L m Lα αΣ = = +

0.3 45

1.730770.1733 26

g g g

L L Lα = = =

(a) 36

0.826B

ga Lα= =

or 18

13Bg=a !

(b) 0xR = 45

0.326y y

gF R mg mL

L⎛ ⎞Σ = − = − ⎜ ⎟⎝ ⎠

135 25

260 52y mg mg mg= − =R !




2I mkα α=

ma mrα=

( ) ( ) ( )eff: sinC CM M W r ma r Iβ αΣ = Σ = +

( ) ( ) 2sinmg r mr r mkβ α α= +

( )2 2sinrg r kβ α= +

2 2

sinrg

r k

βα =+

2 2

sinrga r r

r k

βα= =+

2

2 2 sinr

a gr k

β=+

!




Kinematics: A rα=a , /B Aa rα=

B A=a a / B Aa+

B rα=a rα+

( )B xrα=a ( )B y

rα=a

Kinetics: 5h Bm m=

2 25h BI m r m r= =

(a) ( )eff:C CM MΣ = Σ

( ) ( )B h A B B B Bx yW r I m a r m a r m a rα= + + +

2 2 2 25 5B B B B Bm gr m r m r m r m rα α α α= + + +

212gr r α= 1

12

g

r=α !

(b) ( ) 1

12B xr gα= =a , ( ) 1

12B yr gα= =a !




(a) Kinematics:

/B A B A= +a a a

[ Ba ] [ Aa= ] [60 Lα° + ]20°

Law of Sines

sin 70 sin 50 sin 60A Ba a Lα= =

° ° °

1.0851A Lα=a 60°

0.88455B Lα=a

/ 2G ALa α= 20°

[/ 1.0851G A G Aa a a a Lα= = + = ]602L α° +

20 °

( ) ( )1.0851 cos60 0.5 sin 20xa L Lα α= ° + °

0.54254 0.17101 ; 0.71355xL L Lα α α= + =a

( ) ( )1.0851 sin 60 0.5 cos 20ya L Lα α= ° − °

0.93972 0.46985 ; 0.46985yL L Lα α α= − =a

continued



We have: 0.71355x Lα=a

0.46985y Lα=a

Kinetics:

Triangle ABE: 70ABE∠ = °

70 , 50ABE BAE∠ = ° ∠ = °

Law of Sines AB L=

; 0.88455sin 50 sin 60

BE L BE L= =° °

( )effE EM MΣ = Σ

( ) ( ) ( )0.46985 0.71355 0.46985x ymg L I ma L ma Lα= + +

( )( ) ( )( )210.46985 0.71355 0.71355 0.46985 0.4698512

mgL mL m L L m L Lα α α= + +

( )0.46985 0.81325mgL mL α2=

2

29.81 m/s0.57775 0.57775 6.2975 rad/s0.90 m

gL

α = = =

26.30 rad/s=αααα !

(b) ( ) ( )eff : sin 60 0.71355x x xF F A ma m LαΣ = Σ ° = =

( )( )( )( )2sin 60 8 kg 0.71355 0.90 m 6.2975 rad/s 32.354A ° = =

37.3589 NA = or 37.4 N=A 30° !




Crank AB:

60 mmAB =

2300 rpm 10 rad/s60ABπω π = =

( ) ( )( )0.06 m 10 rad/s 1.8850 m/sB ABAB ω π= = =v

( ) ( )( )22 20.06 m 10 rad/s 59.2176 m/sB ABAB ω π= = =a

Rod BD:

Velocity: Instant CTR at C

( )B BDv BC ω=

( )1.8850 m/s 0.24 m BDω=

7.854 rad/sBDω =

Acceleration:

( ) ( ) ( )( )/ 0.30 mD B BD BDta BD α α= =

( ) ( ) ( )( )22 2/ 0.30 m 7.854 rad/s 18.507 m/sD B BDn

a BD ω= = =

[/ :D B D B Da= +a a a ] 259.2176 m/s= ( )0.30 ft α + 218.507 m/s +

( ) ( )24 30 0 0.30 ft 18.507 m/s5 5BDα= + −

246.266 rad/sBDα =

continued



( ) ( ) ( )( )2 2/ 0.15 m 46.266 rad/s 6.9399 m/sG B BDt

a BD α= = =

( ) ( ) ( )( )22 2 2/ 0.15 m 7.854 rad/s 9.2533 m/sG B BDn

a BG ω= = =

( ) ( )/ / /B G B B G B G Bt na a a= + = + +a a a

259.218 m/sa = 26.9399 m/s + 29.2533 m/s +

2 23 459.217 6.9399 9.2533 m/s 70.785 m/s5 5xa = + + =

24 36.9399 9.2533 m/s 0

5 5ya = + =

Kinetics:

( )( )22 23 kg 0.3 m1 0.0225 kg m

12 12I mL= = = ⋅


( ) ( ) ( )0.24 m 0.12 m 0.09 mxD mg I maα− = +

( ) ( )( )( ) ( )( ) ( )( )( )2 2 2 20.24 m 3 kg 9.81 m/s 0.12 m 0.0225 kg m 46.266 rad/s 3 kg 70.785 m/s 0.09 mD = + ⋅ +

( )( )

3.5316 1.04099 19.11195 N m98.6856 N

0.24 mD

+ + ⋅= =

or 98.7 N=D !




Kinematics: Bar AC Rotation about C

( ) ( )0.3 ma BC α α= =

Bar BC:

/D B Lα=a must be zero since Da

0 andBD BDa aα∴ = =

Kinetics: Bar BD

( ) ( )eff

: 0.3 my y yF F B mg m αΣ = Σ − = −

( )( ) 23 kg 9.81 0.3 m/syB α= − (1)

( ) ( ) ( ) ( )( )( )eff : 0.51962 m 0.15 m 0.3 m 0.15 mB BM M D mg m αΣ = Σ − = −

( ) 20.86603 kg 9.81 0.3 m/sD α= − (2)

continued



Bar AC:

( )( )2223 kg 0.6 m

0.09 kg m12 12

mLI = = = ⋅

( ) ( )( ) ( )( )eff : 0.3 0.3 0.3C C yM M mg B I mα αΣ = Σ + = +

Substituting by from Eq. (1)

( ) ( ) ( )0.09 3 0.33 9.81 3 9.81 0.3

0.3α α

α + + − =

228.029 rad/sα =

Eq. (2): ( )( )2 20.86603 kg 9.81 m/s 0.3 m 28.029 rad/sD = −

1.21356 ND =

or 1.214 N=D !

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