Solution manual vector mechanics engineers dynamics 8th beer chapter 12

Let y be positive downward position for all blocks... Let y be positive downward for both blocks... Let a be the acceleration of the plywood, P a be the acceleration of the T truck, and

Chapter 12, Solution 3

Assume g =32.2 ft/s2

W m g

W m g

750 10

59.046 lb s /ft12.7019 10

mv m v

v v dv = a dx

2 0

1

2v a x f

( ) ( )( )

f

v x

a

= − = −

+

Chapter 12, Solution 10

Kinematics: Uniformly accelerated motion (x0 = 0, 0v0 = )

( )( ) ( )

Chapter 12, Solution 12

Let the positive directions of x A and x B be down the incline

Constraint of the cable: x A +3x B = constant

For block A: Σ =F ma: m g A sin 30° −T =m a A A (1)

For block B: Σ =F ma: m g B sin 30° −3T =m a B B = −m a B A (2)

Eliminating T and solving for a A,

Chapter 12, Solution 13

Let the positive directions of x and A x be down the incline B

Constraint of the cable: x A +3x B =constant

Chapter 12, Solution 14

2

55000 lb 1708.1 lb s / ft32.2 ft/s

A

2 2

44000 lb

1366.5 lb s / ft32.2 ft/s

2 2

44000 lb

1366.5 lb s / ft32.2 ft/s

b x

F a

3110 lb compression

c

3.88 lb 10 lb0.20 sin 25 cos 25

Chapter 12, Solution 18

Assume aB>aA so that the boxes separate Boxes are slipping

k

µ µ= 0:ΣF y= N mg− cos15° = 0

Chapter 12, Solution 19

Let y be positive downward position for all blocks

Constraint of cable attached to mass A: y A +3y B = constant

Chapter 12, Solution 20

Let y be positive downward for both blocks

Constraint of cable: y A+y B=constant

v t a

continued

(a) Acceleration of block A

Chapter 12, Solution 21

(a) Maximum acceleration The cable secures the upper beam; only the lower beam can move

For the upper beam, ΣF y =0: N1−W =0

(b) Maximum deceleration of trailer

Case 1: Assume that only the top beam slips As in Part (a) N1=mg

: 0.25

20.25 8.05 ft/s

continued

Case 2: Assume that both beams slip As before N2=2W.

Chapter 12, Solution 22

Since both blocks move together, they have a common acceleration Use

blocks A and B together as a free body

f B

F N

Chapter 12, Solution 23

(a) Kinematics of the belt v o = 0

1 Acceleration phase with 2

(b) Motion of the package

1 Acceleration phase Assume no slip ( ) 2

Since 3.2 m/s2 < 3.43 m/s ,2 the package does not slip

F a

Chapter 12, Solution 25

Let a be the acceleration of the plywood, P a be the acceleration of the T

truck, and aP T/ be the acceleration of the plywood relative to the truck

(a) Find the value of a so that the relative motion of the plywood with T

respect to the truck is impending a P = a Tand

12.5 m/s ,0.4

P T

P T

x a

Chapter 12, Solution 26

At maximum speed a = 0 F0 = kv02 = 0 0

2 0

F k v

=When the propellers are reversed, F is reversed 0

2 0

0.347m v

x

F

t t

= −+

0

0

0 2

h v

dy

v k

= −+

0

0 2

1ln

= −

−

0 0

f v

1ln2

f v

f

kh k

e mg

0 0

Chapter 12, Solution 30

Let y A , y B , and y C be the position coordinates of blocks A, B, and C respectively measured downward from the

upper support Then the corresponding velocities and accelerations are positive downward

Constraint of cable: y A− y B + y A+2y B + y C =constant

Assume that block B slides downward relative to block A Then the friction

force F is directed as shown Its magnitude is 1

Assume that block A slides downward relative to the fixed plane The

friction force F is directed as shown Its magnitude is 2

Chapter 12, Solution 32

Let the positive direction for position coordinates, velocities, and accelerations be to the right Let the origin

lie at the fixed anchor

Constraint of cable: 3(x C − x A) (+ x C − x B) (+ −x B) = constant

A

From (3), ( )(2 6.0605 32.2)( ) 2

39.0 ft/s10

B

41.5 ft/s20

Substitute (2), (3) and (5) into (1).

A

0.20 32.2 29.37 ft/s0.31056

Since a A, a B, and a are to the right, the friction forces C

, F F A B, and F are to the left as assumed C

(b) Tension in the cable T = 5.56 lb W

Chapter 12, Solution 34

Let the positive direction of x and y be those shown in the sketch, and let

the origin lie at the cable anchor

Constraint of cable: x A + y B A/ = constant or a A + a B A/ = 0,where the positive directions of a A and a B A/ are respectively the x and the y

Chapter 12, Solution 35

Motion of B relative to A Particle B is constrained to move on a circular path with its center at point A (aB A t/ ) is the component of aB A/ lying along the circle, say to the left in the diagram and (aB A n/ ) is directed

toward point A Initially, aB A/ = 0, since the system starts from rest

(a) aB = aA +aB A/ = (a A 25° +) (a B A/ ) Crate B: ΣF x = Σma x: 0 = m a B B A/ − m a B Acos 25°

/ cos 25 0.4cos 25 0.363

2 / 0.363 m/s

4 sin 25 sin 50

3.0642 ftsin 25

Chapter 12, Solution 39

(a) 0:ΣF y = Tsinθ −W = 0

16sin sin 60

W T

g

ρθ

°

Chapter 12, Solution 40

2

, 1 tan or 452

OB OC

C n

By Eq (2), v2 = (0.10261 5.2198)( ) (+ 0.15 0)( ) = 0.5356 m / s2 2

For 0 ≤ T BA, T BC, T DA, T DE ≤ 75 N, 0.732 m/s ≤ v ≤ 4.34 m/s W

By (2), v2 = (0.115702 116)( ) (+ 0.046587 41.216)( ) =15.34 m / s2 2

Try T CA = 0 By 1 , ( )a T CB = −50.78 N unacceptable( )Try T CB = 0 By 1 , ( )b T CA = 64.03 N acceptable( )

By (2), v2 = (0.115702 64.030)( ) +0 = 7.408 m / s2 2

2.72 m/s

v =For 0 ≤ T CA, T CB ≤116 N, 2.72 m/s ≤ v ≤ 3.92 m/s W

v t a

Just before point B v =509.2 ft/s, ρ =(4)(5280) = 21120 ft

( )2 2

2

509.2

12.277 ft/s21120

n

v a

5.124 506.2

21120

Chapter 12, Solution 48

(a) v =160 km/h = 44.44 m/sWheels do not touch the road

v g

A n

C n

Effective forces at B: 2 120 222.2 103 230 lb

32.2 3600

B n

W v ma

(88.889) cos51.875 (200) (9.81)sin 51.875

=(88.889) sin 51.875 (200)(9.81) cos51.875

(c) Minimum speed F = −µN

2 2

θ φρ

θ φρ

Then, θ φ+ =sin−1u =14.90°

Chapter 12, Solution 56

If the collar is not sliding, it moves at constant speed on a circle of radius ρ =rsin θ v=ρω

Normal acceleration a n v2 ρ ω2 2 ( sin )r θ ω2

Since F >µs N, the collar slides

Since the collar is sliding, F =µk N

n F

Form the ratio F

N, and set it equal to µs for impending slip

s s

644× −6

=

This is the friction force available to cause the trunk to slide

The normal force N is calculated from equilibrium of forces in the

n n

4.0041, the ratio becomes

4.0041 sin0.7

B

N v

θµ

d d

(a) Minimum value of µs for no slip ( )µs min = 0.258 W

(b) Corresponding values of θ θ =14.5 and 165.5° ° W

n n

For impending motion to the right: F tan s 0.35

Chapter 12, Solution 64

Consider the motion of one electron For the horizontal motion, let x = 0

at the left edge of the plate and x = l at the right edge of the plate At the screen,

2

x = l + L

Horizontal motion: There are no horizontal forces acting on the electron so that a x = 0

Let t1 = when the electron passes the left edge of the plate, 0 t = when t1

it passes the right edge, and t = when it impacts on the screen For t2

uniform horizontal motion,

Vertical motion: The gravity force acting on the electron is neglected since

we are interested in the deflection produced by the electric force While the electron is between plates (0 ≤ ≤t t1), the vertical force on the electron is

Chapter 12, Solution 65

Consider the motion of one electron For the horizontal motion, let x = 0

at the left edge of the plate and x = l at the right edge of the plate At the screen,

2

x = l + LHorizontal motion: There are no horizontal forces acting on the electron so that a =x 0

Let t = when the electron passes the left edge of the plate, 1 0 t = when t1

it passes the right edge, and t = when it impacts on the screen For t2uniform horizontal motion,

Vertical motion: The gravity force acting on the electron is neglected since

we are interested in the deflection produced by the electric force While the electron is between the plates (0 ≤ ≤t t1), the vertical force on the electron is Fy = eV d/ After it passes the plates (t1 ≤ ≤t t2),

0.4252

eV

dmdvl <

(a) Acceleration of B relative to the rod

m r&&−rθ& = r&&= rθ& = =

2 / rod 258 ft/s

Chapter 12, Solution 74

Let r and θ be polar coordinates of block A as shown, and let y be the B

position coordinate (positive downward, origin at the pulley) for the rectilinear motion of block B

Constraint of cable: r + y B =constant,

W g a

Chapter 12, Solution 75

Let r and θ be polar coordinates of block A as shown, and let y be the B

position coordinate (positive downward, origin at the pulley) for the rectilinear motion of block B

Radial and transverse components of v A.Use either the scalar product of vectors or the triangle construction shown, being careful to note the positive directions of the components

cos306cos30 5.19615 ft/s

v r

Substituting Eq (6) into Eq (4) and solving for a A,

v r

F r

Since the particle moves under a central force, aθ = 0

continued

Magnitude of acceleration:

2

2 0

But

2 0

r

θ =     Hence,

2 0

n

v a r

0

r r

ρ = W

θ

°

2 0 0

8

t

mv F r

 

 Using r = 3R as given leads to

398.06 10 86.1624 10

42.145 10 m4

14.077 10 86.1624 10

138.334 10 ft4

(b) In SI units:

42.145 10

GM v

GM v

πτ

Solving for r,

1/ 3

2 2 2

Chapter 12, Solution 83

Let M be the mass of the sun and m the mass of Venus

For the circular orbit of Venus,

GM M

GM g R

126.98 10 617.76 10

1.071 10 m4

2 126.98 10

8.209 10 m/s1.4420 10

398.06 10 7200

8.055 10 m4

6.37 10

8.055 10

R r

×37.74

θ = °

(75.48 7200)( )2

2 6.7584 10

8.3923 10 s173.032 10

π

2 2 constant4

44

GM M

G

ππ

4 34.068 10

39.1 1034.4 10

7.33 10 10.4 10

5.196 10 m/s14.67 10

A A B

B

r v v

14.67 10

B B

×45

φ = °

3

5.196 10 7.349 10 m/ssin 45 sin 45

B B

4.6849 10

97.70 10 ft/s490.8 10

E

E

GM v

A A B B

r v v r

×For circular orbit of Mars, r M =141.5 10 mi = 747.12 10 ft× 6 × 9

3 sun

9

4.6849 10

79.187 10 ft/s747.12 10

M

M

GM v

34.4 10 5.03 10

4.8382 10 ft/s7.392 10

A v

2

34.4 10 5.03 10

5.0208 10 ft/s6.864 10

B v

3 1

7.392 10 4.7522 10

5.1178 10 ft/s6.864 10

B

B

r v v

9.81 6.37 10

7.55173 10 m/s6.98 10

A

A

gR v

circ

9.81 6.37 10

7.73102 10 m/s6.66 10

B

B

gR v

1

6.66 10 7.81602 10

7.54416 10 m/s6.90 10

B B C

C

r v v

r

×For path CD

6

6.90 10 7.62316 10

7.53579 10 m/s6.98 10

D

D

r v v

Let y be the position coordinate of B, positive upward with origin at O

Constraint of the cord: r− y =constant or y=r

(a) Kinematics: ( )a B y = y = r and ( )a A r = − r rθ2

Collar B: ΣF y = m a B B: T −W B = m y B=m r B (1) Collar A: ΣF r = m a A( )A r: −T = m r A(− rθ2) (2)

Adding (1) and (2) to eliminate T,

0.08075 0.6 10 2.60.08075 0.08075

v r

Adding (1) and (2) to eliminate T,

0.08075 1.2 3.00 2.6

10.70 ft/s0.08075 0.08075

A

F a

r

θθ

B rod

r r

r r

θ

(2)(2.9432)(0.750)0.600

Chapter 12, Solution 96

2 2

dθ + = = mh u by Eq (12.37 )

2mh u 2mh F

dθ + = − = mh u by Eq (12.37 )

5mh u 5mh F

r The minus sign

indicates that the force is repulsive, as shown in Fig P12.97

2 2

0 0

0

0 2

0

0 0

+

2 2

u r

12

GM

r

θθ

=Data: r0=350 10 km× 3 =350 10 m× 6

GM v

×

10.599 10 ft/s

A A

h v r

GM v

r

×

×Elliptic orbit

Using Eq (12.39), 1 2 cos A and 1 2 cos B

162.74 10108.27 10 m /s

9

3 6

108.27 10

6.614 10 m/s16.370 10

A A

h v r

108.27 10146.37 10

B B

h v r

For elliptic orbit AB, r A =325 10 m, × 9 r B =148 10 m× 9

Using Eq (12.39), 1 2 cos A and 1 2 cos B

473 105.1907 10 m /s

A B AB

GMr r h

5.1907 10

15.971 10

325 10

AB A

A

h v r

( ) ( 18)( 9)( 9)

9

15 2

2 132.474 10 325 10 137.6 102

462.6 105.0609 10 m /s

A B AB

GMr r h

5.0609 10

15.572 10 m/s

325 10

AB A

A

h v

5.0609 10

36.780 10 m/s137.6 10

AB B

B

h v

9

15 2

2 132.474 10 137.6 10 264.7 102

402.3 104.8977 10 m /s

B A

B A

GMr r h

4.8977 10

35.594 10 m/s137.6 10

B A B

B

h v

For Venus, GM =0.82 GMearth =326.41 10× 12 m /s3 2

For a parabolic trajectory with r A =15×10 m6

GMr r h

continued

( ) 9 3

6 2

96.571 10

6.4381 10 m/s

AB A

A

h v

B

h v

GMr r h

B

h v

BC C

C

h v

r

×

×Final circular orbit r C = ×9 10 m6

For Venus, GM =0.82GMearth =326.409 10 m /s× 12 3 2

Transfer orbit AB: v A=6500 m/s,r A =15 10 m× 6

B

h v

r

×

×

Second transfer orbit BC r C = ×9 10 m6

507.56 10

B C BC

GMr r h

B

h v

75.968 10

8.4409 10 m/s

9 10

BC C

C

h v

r

×

×Circular orbit with r C = ×9 10 m.6

6 2

2 1.51188 102

10.0901 10 ft/s29.7 10

A

A

GM v

r

×

×

First elliptic transfer orbit AB

Using Eq (12.39), 1 2 cos A and 1 2 cos B

623.7 10

A B AB

GMr r h

π

τ = τ =

3 9

311.85 10 132.82 10

444.95 10 s292.45 10

AB

π

GM v

ππ

×For elliptic orbit of spacecraft it is given that

3 0

3 8.5038 10 s2

2 29.543 10 28.703 102

626.54 10 ft /s8.5038 10

626.54 10

27.789 10 ft/s22.546 10

A A

h v r

0 3/ 2

0

3/ 2 3

3/ 2

3/ 2 0

0 3/ 2

R GM

r R

πτ

2 15.458 102

3.753 10 s823.78 10

BC

A t

h

×

×3

For the moon, GM =(0.01230 14.077 10) ( × 15) =173.149 10 ft /s× 12 3 2

For elliptic orbit AB, r A =1110 mi = 5.861 10 ft, × 6 r B = 2240 mi =11.827 10 ft× 6

Using Eq (12.39), 1 2 cos A and 1 2 cos B

36.839 10 3.1148 10 ft/s11.827 10

AB B A

h v r

Dividing Eq (2) by Eq (1),

33.726 10

2.8516 10 ft/s11.827 10

BC B

B

h v

r

×

×( ) ( )2 1 263.2 ft/s

19.536 10

0.9343 rad 53.53420.909 10

AB s R

×For an elliptic trajectory, 1 GM2 (1 cos )

C A

r r

εε

−1.2348 1

0.68661.2348 cos153.233

337.5 10

13.07 10 ft/s25.819 10

C C

h v r

×

14.0771 10

25.3159 10 ft/s21.9648 10

A

A

GM v

15 2 9

14.0771 10

5.997512 cos 50522.6946 10

B A

εε

Chapter 12, Solution 122

For an ellipse 2a =r A+r B and b = r r A B

Using Eq (12.39), 1 2 cos A and 1 2 cos B

ττ

=

1

h a

=

Using (2) and (3) to eliminate N AB and solve for a A,

(a) Acceleration of block A aA = 8.63 ft/s2 W Substituting for a into (1) and solving for A a B A/ ,

(b) Acceleration of B relative to A aB A/ = 32.2 ft/s2 25° W

Chapter 12, Solution 127

2

1805.59 lb s /ft32.2

W m g

6.18 ft/s ≤ ≤v 13.05 ft/s W

1 sin

with 0cos

b b

r b

(b) Forces P and Q exerted on the pin by the arm OA and the wall of the

slot DE, respectively

4 r

M G

πτ

=Data: r = 384.5 10 m× 6

4 384.5 10

6.04 1066.73 10 2.3604 10

Chapter 12, Solution 134

Let r and θ be polar coordinates with the origin lying at the shaft

Constraint of rod: θB =θA +π radians; θ&B =θ&A =θ θ& &&; B =θ&&A =θ&& (a) Components of acceleration

Sketch the free body diagrams of the balls showing the radial and transverse components of the forces acting on them Owing to frictionless sliding of B along the rod, ( )FB r = 0

Radial component of acceleration of B

( ) :

Transverse components of acceleration

( )aA θ =rAθ&&+2r&Aθ& =raθ&&

( )aB θ =rBθ&&+2r& (1) Bθ&

Since the rod is massless, it must be in equilibrium Draw its free body diagram, applying Newton’s 3rd Law

m r θ&+m r θ& = m r θ& + m r θ&

Applying to the final state with ball B moved to the stop at C,

( )2

0 0

GM v

r

×

×For the descent trajectory,