Solution manual vector mechanics engineers dynamics 8th beer chapter 14

Symbols: dm mass flow ratedt =exhaust relative to the airplane u =speed of airplane v =drag force D =Principle of impulse and momentum... Apply the impulse - momentum principle to the mo

The masses are mA=mB =1350 kg and mC =5400 kg.

Letv vA, B, andvC be the sought after final velocities, positive to the left

(mB +m vC) BC =m vB( )B 0 +m vC( )C 0

6750vBC = +0 5400 2.2222 vBC =1.77778 m/sCar-truck strikes car

Conservation of linear momentum for block, cart, and bullet together

m vv

Consider block and bullet alone

Principle of impulse and momentum

Also, just after impact, the velocity of the cart is zero

Accelerations after impact

The block moves 0.424 to the left relative to the cart

(b) This places the block 1.000 0.424 0.576 m− = from the left end of the cart

0.576 m from left end of cart

Initial momentum of system: m vA( )A 0 +m vB( )B 0+m vF( )F 0 = 0.

There are no horizontal external forces acting during the time period under consideration Momentum is

conserved

0= m vA A +m vB B +m vF F124.2vA +114.9vB +1366.5vF = 0 (1) The relative velocities are given as

/ /

7 ft/s3.5 ft/s

The masses are A WA, B WB, and F WF.

Initial momentum of system: m vA( )A 0+m vB( )B 0+m vF( )F 0 = 0

There are no horizontal external forces acting during the time period under consideration Momentum is

1.02 7.65 6.63 ft/s1.02 7.5 6.48 ft/s

impact takes place before the load has time to acquire velocity

Momentum immediately after impact:

m v′+m +m v′= m +m v′ (2) Equating (1) and (2) and solving for v′ ,

3 0

m vv

The masses are m for the bullet and m and A m for the blocks B

(a) The bullet passes through block A and embeds in block B Momentum is conserved

m v m vm

v v

−

++

A A

mv m vv

(a) Woman dives first

v

v′ = ′ +

(a) Woman dives first

(1)(2)(3)( ) Solving for a v vA, B, andv C, v =A 1.333 m/s vA =(1.333 m/s)j

( ) ( ) ( ) (1.800 kg m /s2 ) (0.900 kg m /s2 ) (3.60 kg m /s2 )

The masses are mA = mB = mC =9 kg.

Position vectors (m): rA = 0.9 ,k rB = 0.6i +0.6j+0.9 ,k rC = 0.3i+1.2j Coordinates of mass center G expressed in m

( )(9 0.9 ) ( )(9 0.6 0.6 0.9 ) ( )(9 0.3 1.2 )

270.3 0.6 0.6

Position vectors expressed in ft

(7.83 lb s) (5.22 lb s)

m =v ⋅ j− ⋅ k
(c) Position vectors relative to the mass center G (ft)

3 6 2.667 1.333 2.6670.333 4.667 2.667

6 3 2.667 1.333 2.6672.667 4.667 0.333

3 3 2.667 1.333 2.6670.333 1.333 0.333

2.667 1.333 2.667

0 7.83 5.2227.826 13.913 20.870 27.827 13.913 20.870

Linear momentum of each particle expressed inkg m/s.⋅

Position vectors, (meters): rA =3j k+ , rB =3i +2.5 ,k rC = 4i+2j k +

The mass center moves as if the projectile had not exploded

2 2

There are no external forces The mass center moves as if the explosion had not occurred

Mass center at time of first collision

: 432440 3600= v tBj

(3600 3.6363432440)( ) 30.034

B

Mass center at time of first collision

( )1 ( )1

: 432440 12240= vB , vB =35.33 ft/s,j

48.1 mi/h

C

Projectile motion ax =0, ay = − = −g 9.81 m/s ,2 az = 0

( )vx 0 =165 m/s, ( )vy 0 = 0, ( )vz 0 = 0After the chain breaks the mass center continues the original projectile motion

Place the vertical y axis along the initial vertical path of the rocket Let the x axis be directed to the right (east) Motion of the mass center: ax = 0, vx = 0, x = 0

Velocities of pieces C and D after impact and fracture

2.1 3 m/s, 3tan30 m/s0.7

2.1 2.333 m/s, 2.3333tan m/s0.9

C

C D

o o

Solving simultaneously,

Weight of arrow: 2 oz 0.125 lb.

A B

WW

=Conservation of momentum: Let v be velocity immediately after impact

xz

A B C

(−600i +750j−800k)+2400j+2400j =vA Aλλλλ +4vB Bλλλλ +4vC Cλλλλ Resolving into components,

: 600 0.66667 1.69160: 5550 0.66667 3.58828 3.81792: 800 0.33333 0.51260 1.19312

19.51

271

explosition have the directions of the unit vectors

C

19.51

271

rvt

/ 27 1500 ft/s0.018

B D B B

rvt

C/ 22.5 1875 ft/s0.012

D C C

rvt

(2)(3)Since mass is conserved, W =WA +WB +WC =6 lb (4) Solving equations (1), (2), and (4) simultaneously,

i i i i

This condition is satisfied if,

( ) 0 Point has zero velocity

or ( ) Point coincides with the mass center

or ( ) is parallel to Velocity is directed along line

This condition is satisfied if

( ) 0 The frame is newtonian

or ( ) Point coincides with the mass center

or ( ) is parallel to Acceleration is directed along line

The masses are m for the bullet and m and A m for the blocks B

The bullet passes through block A and embeds in block B Momentum is conserved

m v m vm

v v

−

++

A A

mv m vv

Data and results from Prob 14.1

2.9630 m/s,

A

v = vB =vc =1.18519 m/sInitial kinetic energy: 0 1 ( )20 1 ( )20 1 ( )20

2 1 10.67 kJ

T =T =

Initial momentum of system: m vA( )A 0 +m vB( )B 0 +m vF( )F 0 = 0.

There are no horizontal external forces acting during the time period under consideration Momentum is

conserved

0= m vA A +m vB B +m vF F124.2vA +114.9vB +1366.5vF = 0 (1) The relative velocities are given as

/ /

7 ft/s3.5 ft/s

From the solution to Prob 14.27,

Initial velocity of 6-lb shell: v0=1464 ft/s

Velocity of mass center: (mA +mB)v =mA Av +mB Bv

−+

mE

S

mE

(a) A strikes B and C simultaneously

During the impact, the contact impulses make 30° angles with the velocity v 0

component: 0 0 sin 30 sin30

23

Second impact: A strikes C

During the impact, the contact impulse makes a 30o angle with the velocity v 0

Thus, vC =vC cos30° −i sin30°j

(a) Velocity of B at maximum elevation At maximum elevation ball B is at rest relative to cart A vB = v AUse impulse-momentum principle

components:

x m vA 0 + =0 m vA A +m vB B = (mA+m vB) B

0 A B

m vv

=+
(b) Conservation of energy:

Velocity vectors: v0 =v0(cos30° −i sin30°j ) v0 =15 ft/s

vv

i: v0cos30° = vBsin 30° +vCcos30° j: −v0sin30° = − −vA vBcos30° +vCsin30° Solving for vB and ,v C

A

3 15 7.5 6.4952 ft/s2

B

1 15 7.5 11.25 ft/s2

C

Velocity vectors: v0 =v0(cos45° +i sin 45°j ) v0 =15 ft/s

vv

i: v0cos 45° =vBsin 60° +vCcos60° j: v0sin 45° = vA −vBcos60° +vCsin 60° Solving for vB and ,v C

1sin ,

3

θ = cos 8,

3

θ = θ =19.471°Velocity vectors v0 = −v0j

Divide by m and resolve into components

i: 0 2 cos= vA θ −uBsinθ

−j: v0 = +2 sinvA θ +uBcosθ −vCSolving for vA and ,u B 1( 0 ) 0.94281( 0 )

1 8sin , cos , 19.471

v′ = vCord becomes taut

B =  v °

v +[0.94281v0 19.471° ] [0.95743 0

T TT

Fraction of energy lost = 0.0556

(a) Use part (a) of sample Problem 14.4 with mA=mB=m.

or v = with B 0 vA=v0
(c) Consider positions when θ θ= max and θ θ= min

Since vB A/ = at these states 0

v =vConservation of momentum

mv =mv +mv

0

12

v =v = v

Both A and B keep moving to the right with A and B stopping intermittently

Relative velocity and acceleration

B/A B/A

ta

There are no external forces Momentum is conserved

−

There are no external forces Momentum is conserved

(a) Moments about D : 0.9m vA 0 =1.8m vC C +0.9m vB B

( )( ) ( )( )

0

0.9 0.9 0.5 12 0.5 3.5 4.251.8 A 1.8 B

−

(a) Linear and angular momentum.

B= v

v i

Let the system consist of spheres A and B

State 1 Instant cord DC breaks

( ) ( )HG 2 = HG 1

0

32

The system is spheres A and B and the ring D

Initial velocities: vA =v0(−cos30° −i sin 30°j )

0 cos30 sin 300

B D

Let m be the mass of one ball

Conservation of linear momentum: (Σmv) (= Σmv )0

( )vC y =( ) tan 30vC x ° =2.25tan30° =1.2990 ft/s

From (1), ( )vB y = −1.2990 ft/s

( )vB y= −1.299 ft/s
(2.25 ft/s) (1.299 ft/s)

Let m be the mass of one ball

Conservation of linear momentum: (Σmv) (= Σmv )0

( )vC y=( ) tan 45vC x ° = 2tan 45° =2 ft/s

From (1), ( )vB y= −2 ft/s

( )vB y= −2.00 ft/s
(2.00 ft/s) (2.00 ft/s)

Conservation of linear momentum: WA WB 0 WA A WB B

Let ω be the spin rate

Velocities in m/s Lengths in meters Assume masses are 1.0 for each ball.

Before impacts: ( )vA 0 = v0i = 4 , i ( )vB 0 =( )vC 0 = 0

After impacts: vA = −1.92 , j vB =( )vB xi+( )vB yj, vC =vCi

Conservation of linear momentum: v0 = vA+vB + v C

i: 4 0= +( )vB x +vC ( )vB x = −4 vCj: 0 = −1.92+( )vB y +0 ( )vB y =1.92

Velocities in m/s Lengths in meters Assume masses are 1.0 for each ball.

Before impacts: ( )vA 0 = v0i = 5 , i ( )vB 0 =( )vC 0 =0

After impacts: vA = −vAj, vB =( )vB xi +( )vB yj, vC =3.2i

Conservation of linear momentum: v0 = vA+vB + v C

i: 5 0= +( )vB x +3.2 ( )vB x =1.8j: 0 = − +vA ( )vB y +0 ( )vB y = vA

0 = v0

v i The initial velocities in this system are ( ) ( )v′A 0, v′B 0 and ( )vC′ 0, each having a magnitude of lω They are directed 120° apart Thus,

( ) ( )v′A 0+ v′B 0+( )v′C 0 = 0(a) Conservation of linear momentum:

( )A 0 ( )B 0 ( )C 0 ( A 0) ( B 0) ( C 0)

m v′ +m v′ +m v′ =m v −v +m v −v +m v −v

0= vAj−vi + −vBj−vi + vCi−v i Resolve into components

(c) 3 m/s

0.1501

ll

ω

Use a frame of reference that is translating with the mass center G of the system Let v be its velocity 0

0 =v0

v i The initial velocities in this system are ( ) ( )v′A 0, ,v′B 0 and ( )v′C 0, each having a magnitude of lω They are directed 120° apart Thus,

( ) ( )v′A 0+ v′B 0+( )v′C 0 = 0Conservation of linear momentum:

( )A 0 ( )B 0 ( )C 0 ( A 0) ( B 0) ( C 0)

m v ′+m v′ +m v′ =m v −v +m v −v +m v −v

0= vAj−vi + −vBj−vi + vCi−v i Resolve into components

i: vC −3v0 = 0 vC =3v0 =( )( )3 0.4 =1.2 m/sj: vA−vB = 0 vB =vA

Initial kinetic energy: 1 3 1 02 3 1 2 2

Conservation of angular momentum about G

Mass flow rate As the fluid moves from section 1 to section 2 in time ∆t, the mass ∆m moved is

Consider velocities measured with respect to the plate, which is moving with velocity V The velocity of the stream relative to the plate is

(1000)(600 10 )

Let F be the force that the wedge exerts on the stream Assume that the fluid speed is constant v = 60 ft/s.Volumetric flow rate: Q =475 gal/min = 1.0583ft /s3

slug/ft 1.0584 ft /s 2.051 slug/s32.2

mtdmvdt

(48 ft/s )

Volumetric flow rate: Q =500 gal/min = 1.1141ft /s3

slug/ft 1.1141 ft /s 2.1590 slug/s32.2

dm

Q

Use a frame of reference that is moving with the wedge to the left at 12 ft/s In this frame of reference the

upstream velocity vector is

mtdmudt

Let F be the force exerted on the chips Apply the impulse-momentum principle to the chips Assume that the feed velocity is negligible

16.89 lb

x

D

4040 N

=

The line of action of F passes through point O

Components : ( )∆m uA+ F t( )∆ sinα = ∆( )m uBcosθ

θ

α =Thus point C lies at the midpoint of arc AB

( )800 ( )( )

13.333 L/s 1.000 kg/L 13.333 L/s 13.333 kg/s60

∆

∆93.6 N

Mass flow rate: 62.4 lb/ft 40 ft /min2 1.29193 lb s/ft

60 s/min32.2 ft/s

moments aboutD: ( )( sin 60 ) 15 ( )( cos60 ) 23 ( ) 15

Volumetric flow rate: Q =300 gal/min= 66840 ft /s3

slug/ft 0.6684 ft /s 1.2954 slug/s32.2

Moments about C: ( )∆m v a WA − P( )∆t lcosθ = ∆( )m v bB

Volumetric flow rate: Q =300 gal/min= 66840 ft /s3

slug/ft 0.6684 ft /s 1.2954 slug/s32.2

moments about C: ( )∆m v a WA − p( )∆t l cosθ = ∆( )m v bB

Symbols: dm mass flow rate

dt =exhaust relative to the airplane

u =speed of airplane

v =drag force

D =Principle of impulse and momentum

Let F be the force exerted on the slipstream of one engine Then, the force exerted on the airplane is −2F as shown

Let F be the force exerted on the slipstream of one engine

Use a frame of reference moving with the plane Apply the impulse-momentum principle Let F be the force that

the plane exerts on the air

The thrust on the fluid is dm( B A)

720 = 22.36 slug/s, 1860 ft/s, 560 mi/h 821.33 ft/s32.2

dm

(22.36 1860 821.33)( − )−D1 =0 D1 = 23225 lbDrag force factor:

After control surface malfunction: k2 =1.2k1 =0.04131 lb s /ft⋅ 2 2

When the new cruising speed is attained,

Apply the impulse - momentum principle to the moving air Use a frame of reference that is moving with the airplane Let F be the force on the air

270 km/h 75 m/s

600 m/s

vu

Symbols: n = number engines operating

dm

dt = mass flow rate for one engine

u = discharge velocity relative to jetliner = 800 m/s

2

(250) 1 37.879 m/s3(800 250)

dm

k dt

−(a) One engine lost: n = 2

2

37.8792(800 )

u be the velocity of the stream relative to the velocity of the blade u=(v V− )

t ρ

∆ =

∆Principle of impulse and momentum

(∆m u F t) − t( ) (∆ = ∆m u) cosθ(1 cos ) ( )(1 cos )

The force Ft on the fluid is directed to the left as shown By Newton’s law of action and reaction, the

tangential force on the blade is Ft to the right

Output power: Pout=FVt =ρAv vA( A−V V) (1 cos )− θ

(a) V for maximum power output

A

v = V
(b) Maximum power

P = ρAv − θ
Input power = rate of supply of kinetic energy of the stream

( ) (1 cos )1

×

×

mechanical efficiency 0.551=

∆ =

∆Input power 1 3

input power 10,000 W, 36 km/h 10 m/sdT

dtv

ρ

4

Mass flow rate:

mass density volume

density area length

Q = b gd d d +d

Mass flow rate:

mass density volume

density area length

Q =b gd d d +dData: g =9.81 m/s, b = 3 m, d1 =1.25 m, d2 =1.5 m

1

3 9.81 1.25 1.5 1.25 1.5 15.09 m /s2

From hydrostatics, the pressure at section 1 is p1= rh = ρgh.

The pressure at section 2 is p2 = 0

Calculate the mass flow rate using section 2

mass =density×volume= density×area×length

62.4 lb/ft

Q 11.1408 10 ft /s32.2 ft/s

23.413 ,

β = α =60o −β = 36.587o Moments about O:

sin21.590 10 0.72648 60 sin 36.587 0.726480.56093 0.011395 lb ft

ωω

Consider the conservation of the horizontal component of momentum of the railroad car of mass m and the 0sand mass qt.

v

dt = = m qt

+Integrating, using x =0 0 and x = L when t =tL,

0 qL m vL

v = v e−

Apply the impulse-momentum principle

Letρ be the mass per unit length of chain Apply the impulse - momentum to the entire chain Assume that the reaction from the floor it equal to the weight of chain still in contact with the floor

Calculate the floor reaction

(a) Letρ be the mass per unit length of chain The force P supports the weight of chain still off the floor

Letρbe the mass per unit length of chain Consider the impulse-momentum applied to the link being brought to

forces contributing to moments about O in the diagram

Moments about O:

( )∆m v+[ρgh C t− ]∆ = ∆( )m v

C = ρghEquating the two expressions for C,

=

Letρbe the mass per unit length of chain Assume that the weight of any chain above the hole is supported by the floor It and the corresponding upward reaction of the floor are not shown in the diagrams

Case 1 Apply the impulse-momentum principle to the entire chain

750 lb/s,dW

Thrust of one engine:

As all the fuel is consumed: m1=1500 1200− =300 kg

1 15 9.81 0.05 9.81300

u

Using (1) and (2) for a and 1 a and substituting into (3), 0

Pqu

Apply the principle of impulse and momentum to the satellite plus the fuel expelled in time ∆t

Data from Problem 14.97: m0 =5000 kg, u = 4200 m/s

Apply conservation of momentum to the rocket plus the fuel

Apply conservation of momentum to the rocket plus the fuel

0 1 fuel 1 1200 lb

W =W +W =W +

1 1

1200 exp 450 1.08690

5400

WW

1

1200 0.08690

See sample Problem 14.8 for derivation of

0 0

ln1ln

2

z z

260 8.0745 lb s/ft32.2

wqg

= & = = ⋅

2 fuel

fuel 4000 124.22 lb s /ft

32.2

Wm

g

fuel 124.22 15.385 s8.0745

mtq