Solution manual vector mechanics engineers dynamics 8th beer chapter 16

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... Kinemat

or μ=0.0848!

20.675 lb2.9282

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

(a) Four-wheel drive:

continued

(a) Sliding impends:

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:

D M

(a) + 60° Σ − ∑F t ( )F t eff:

(B y− A)cos30°− B xsin 30° +W cos30°=ma

Recalling equation (2), we have,

Bar AB

N 4 9.81 sin 30 43

x

21.5717 m/s

4 1.5717 0.15 11.772

20.841 N0.866 0.6

AB

or T AB=20.8 N!

Chapter 16, Solution 14

Bar AC

0

C M

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

B Bn

a =a =rω =

271.061 m/s

From Problem 16.15 a G =71.061 m/s2 60°

261.541 m/s

From symmetry B y =C y =194.0 N (Same as what we got for Prob 16.15)

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

max

M = Area under V – Diagram from B to G

1194.0 N 0.375 m2

36.4 N m

max

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: ( ) ( ) 2

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

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

( )2

1000.69813

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 2 2

T e T

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

4T −2.5663T =250 N T =174.38 NFrom (1): T2 =4 174.38( )−250 T2=447.51 N

Substituting for T1andT2into (2):

( )( )2 ( 2)

1

10 kg 0.225 m 0.25313 kg m2

2.2474 rad/s0.25313

Chapter 16, Solution 26

1

10 kg 0.225 m 16 rad/s 4.05 N m2

0.25 176.4 N

or F =176.4 N 44°!

or I = 20.3 lb ft s⋅ ⋅ !

Disk A 2

12

B =

Iα ⎛ ⎞ ⎛ ⎞ α

= =⎜ ⎟ ⎜ ⎟

412

A A

a = r α =⎛⎜ ⎞⎟α

⎝ ⎠

812

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

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

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

Substitute for αA from Eq (1), and 1 2

α =

G k B

2 k N B mr

g r

Chapter 16, Solution 37

21sin

φ

While slipping occurs: F = µk N = µk P

Substitute from Eqs (1) + (2): 0

m m

m (QED)

(a) Angular accelerations

While slipping occurs:

Based on the solution for P16.39

( )0 ( )0

20; 750 rpm 25 rpm

Chapter 16, Solution 41

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

α

Kinematics The acceleration of P is i

( ) ( ) ( )( ) ( ) 2

∆ a = ∆ a+ ∆ α r× − ∆The sum of the effective forces is

Since G is the mass center, Σ ∆ri′ m i = 0

Also, for each particle, ri′× =ri′ 0

Thus,

( i′ m i i) i′ ( i i′) m i

Σ r × ∆ a = Σr × α ×r ∆ Since α ⊥ ,ri′ we have ( ) 2

28.05 ft/s

2 wr

t Pg

2(1.200 rad/s )

(b) With all rockets except D:

2 eff

= −

eff( G) : 3

G

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

1 lb , 2.6833 ft/s32.2

Chapter 16, Solution 48

23.5603 rad/s

α=

22.6833 ft/s

1.5110 3.5603 0.42441(3.560 1(3.5603)

Chapter 16, Solution 50

2 2

2800 lb lb s

ft32.2 ft/s

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

μ

= +

See Problem 16.54

2

4 0.12 11 lb4

12 lb 3.2

ft1232.2 ft/s

k P mr

32.2 ft/s

12 lb32.2 ft/s

2.11 m/s2 3.3

11(0.3) = 3.3 11(0.3) = 3.3

1.81 m/s2,

3 m/s2

g

=

a !

B

g

=

a !

(b) Acceleration of A:

/

30.866 + 0.5

20.866

Chapter 16, Solution 61

112

2

16

g b

Chapter 16, Solution 62

2

21

⎛ ⎞

+ ⎜ ⎟

⎝ ⎠ ; aB = g "

B = g

a "

2 0

0

52

v r

µ

= "

(c) Eq (4)

2 0

v s

g

µ

= "

k g r

k k

k k

k g r

7 k

v t

v r

=

ω "

1 2

k k

µµ

2 1

k k

g

µω

Note: We check that v = −v1 ωr

k g r

k C

k C

1 0

2

5

k k

v

g

µµ

25

2

25 k

v S

g

µ

Chapter 16, Solution 70

t

OG =r a =rα

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

( )( ):

( )2

mk α = mrα GP

2

k GP r

1

Chapter 16, Solution 72

(a)

21

2 12

L

P = mLα

21

6 15 lb

96.6 rad/s

10 lb

3 ft32.2 ft/s

21

L

PL = ma +Iα

21

211.67 rad/s

Chapter 16, Solution 74

(b)

21

= −  0.30282 ft= −And 2 2 10 ft 0.53052 ft

12

r z

Chapter 16, Solution 77

Determination of mass center of disk

We determine the centroid of the composite area:

xA x A x

Axm1g

m2g

=

2 12

r

Chapter 16, Solution 79

0

2

L a

A−mg = − mg

1

;4

112

mgS =m L +S α

2 2

Differentiate with respect to S

Set numerator equal to zero

g r

Chapter 16, Solution 82

/: B = A+ B A

g L

We represent the effective forces of the disk by a couple I Dαα and the effective forces of the rod by vectors m a R( )R t and m a R( )R n attached at C and a couple I Rα We have, α.

1(3 kg)(0.120 m) 21.6 10 kg m2

(a) making α =36 rad/s in (1):2

(4 lb)(0.5 ft) (4 lb)(1.25 ft) 6 lb ft

A M

92

Eq (2) reduces to ΣM C = I Cα when rG C/ ×maC =0; that is, when rG C/ and aC 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.) !

I = mk

m =rα( )eff: ( sin ) ( )

s

r g

k

β = μ +

2tan s 1 r

80 lb 80

lb s /ft32.2

s F N

( )

8

ft 0.66667 ft12

a = rα =⎛⎜ ⎞⎟α = α

2 2

2

10 lb 6

ft1232.2 ft/s

20.07764 lb ft s

10 lb 83.3333 lb ft ft 0.07764 lb ft s

α = or α =15.46 rad/s2 !

ft 15.4558 rad/s 10.3039 ft/s12

Chapter 16, Solution 94

8

ft 0.66667 ft12

a =rα =⎛⎜ ⎞⎟α = α

2 2

2

10 lb 6

ft1232.2 ft/s

I = mk =⎛⎜ ⎞⎛⎟⎜⎝ ⎞⎟⎠

20.07764 lb ft s

( )eff ( )

12: 5 lb ft

s

F N

( )μs min =0.020!

ft 0.6666712

a =rα =⎛⎜ ⎞⎟α = α

2 2

2

10 lb 6

ft1232.2 ft/s

I =mk =⎛⎜ ⎞⎛⎟⎜⎝ ⎞⎟⎠

20.07764 lb ft s

( )eff ( )

4: 5 lb ft

α =

or α = 7.73 rad/s2 !

ft 7.7280 rad/s 5.1520 ft/s12

( )minμs =0.340!

Chapter 16, Solution 96

8

ft 0.66667 ft12

a = rα =⎛⎜ ⎞⎟α = α

2 2

2

10 lb 6

ft1232.2 ft/s

I = mk =⎛⎜ ⎞⎛⎟⎜⎝ ⎞⎟⎠

20.07764 lb ft s

α =

or α =7.73 rad/s2 !

ft 7.7280 rad/s 5.15201 ft/s12

Assume disk rolls: a =rα =(0.16 m)α

( )( )2

2 5 kg 0.12 m

I = mk =

20.072 kg m

(0.16 m 24 rad/s) ( 2) 3.84 m/s2

23.84 m/s

Assume disk rolls: a = rα =(0.16 m)α

( )( )2 2

5 kg 0.12 m

I = mk =

20.072 kg m

Chapter 16, Solution 101

(a)

10 832.2 12

Chapter 16, Solution 102

From 16.101, For BE, equation (1)

20.7071−0.6667R =0.01380α (1)

(35.706) 23.804 ft/s12

Bar AB and gear C

2

12

2

1(3 kg)(0.2 m)12

Bar AB and gear C

r mg

m

θα

g r

θ

9 32

318

θπα

2 2

Chapter 16, Solution 108

2

2

41

P mr

3 15 0.899551

r

Note: In this Problem and the next we cannot use the equation ∑M A=I Aα, since points A, O, and G

are not aligned

Kinematics: Choose positive v and B a to left B

Trans with B + Rotation Abt B = Rolling motion

ft 0.25 ft12

2.2

ft 0.18333 ft12

( )2

0.075 ft 3.5 lb3.5 lb

g L

α= ⎛⎜⎝ ⎞⎟⎠

232.2 ft/s , 3 ft,

(0.7071)2

x

m L mL

2

g L

Chapter 16, Solution 112

y components

16sin 30 sin 60

G

12

( ) 16 30.5

=

2 2

g r

πα

=

(b)

2212

1

83

=

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

sin 60 sin 65 sin 65

25.321 1.2287− B=7.046 16.214 14.184+ −

59.5 N

=

P "

Kinematics: Crank AB:

B BD

v BC

Translation with B and rotation about D = Planar motion

1012

2 12

 +   

Chapter 16, Solution 118

Kinematics: ωAB=12 rad/s , αAB=80 rad/s2

Same analysis as for Prob 16.117 except that αα , ABResulting in ( )aB t Eq (1) must be replaced by,

( )( ) ( )

10.75 62

0.75 6

33

D y BD

v BD

=

continued

°

=ω

1.25 rad/s

CD =ω

Kinematics: From Prob 16.120

21.25 rad/s

29.5873 rad/s

rα = α

2 2

Crank BC:

( ) ( ) 5 ( 2) 2

ft 15 rad/s 6.25 ft/s12

0.44 m

cos 1.7181 cos0.44 m 0.25 m

A

A a

d

a d

Given data: ωf = 2 rad/s

2 =3.4362a A ⇒ a A =1.1641 m/s

or a A =1.164 m/s2 !

3

=

ft12

B BD

v BD

ft12

B BD

v BD

Chapter 16, Solution 135

Kinematics: We resolve the acceleration of G into the acceleration of A

and the acceleration of G relative to A:

Substitute for a A from (3) into (2):

=

(b) Substitute data into equation (5):

2

4(9.81)sin 20 13.4209

9.93451.35093

1( ) : (0.25 m) = (6 kg) (0.25 m)

2

Σ = Σ

Bar: x-Dir :− +B1 (19.62 N) cos 45°=(2 kg) (0.25 m)α +(cos 45 ) (2 kg) (0.375 m)° αBD

y-Dir :B2+(19.62 N) sin 45 = sin 45 (2 kg) (0.375 m)° ° αBD

Six equations and six unknowns:

continued

a r

P a

A

P m

P−B =m L α + Lα 

continued

P mL

α = − − 

 

307

BC

P mL

Data: L =25 in.= 2.0833 ft

2 2

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

⋅

65

Pb mL

P L mL

 

611

P mL

2 2

6 lb

0.18634 lb s /ft32.2 ft/s

Chapter 16, Solution 141

Since ωAB=ωBC=0: ( )aA x=( )aB x=( )aB x=(aBC x) =a C

1( )

g L

External forces: A W x, ,AJ axial force F shear J, V and bending moment J, M J

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

From answers to Prob 16.79:

External forces : Reaction A, distributed load per unit length mg/L, shear V bending moment J, M J

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

( )eff: ( )cos 60 cos30 cos 60

Lever ABC: Static Equilibrium (Friction Force )

θ =

1786.77094 125.22 rev

3 12.075 ft/s

6 lb32.2 ft/s

0.75 lb

4.025 ft/s

6 lb32.2 ft/s

g r

0.90 m

g L

2

( ) ( ) ( ) ( 2) 2 / 0.15 m 46.266 rad/s 6.9399 m/s

Kinematics: Bar AC Rotation about C