# Solution manual vector mechanics engineers dynamics 8th beer chapter 14

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Chapter 14, Solution 1.

The masses are mA = mB = 1350 kg and mC = 5400 kg.
Let v A , vB , and vC be the sought after final velocities, positive to the left.

( v A )0 = ( vB )0 = 0, ( vC )0 = 8 km/h = 2.2222 m/s

Initial velocities:

First collision. Truck C strikes car B. Plastic impact: e = 0
Let ( vBC )0 be the common velocity of B and C after impact.

Conservation of momentum for B and C :

( mB + mC ) vBC

= mB ( vB )0 + mC ( vC )0

6750 vBC = 0 + ( 5400 )( 2.2222 )

vBC = 1.77778 m/s

Second collision. Car-truck BC strikes car A.
Elastic impact. e = 1
v A − vBC = − e ( v A )0 − ( vBC )0  = −1.77778 m/s

(1)

Conservation of momentum for A, B, and C.

mAv A + ( mB + mC )( vBC ) = mA ( v A )0 + ( mB + mC )( vBC )0

1350 v A + 6750 vBC = 0 + ( 6750 )(1.77778 )

(2)

Solving (1) and (2) simultaneously for v A and vBC ,

v A = 2.9630 m/s, vBC = 1.18519 m/s = vB = vC
v A = 10.67 km/h

v B = 4.27 km/h
vC = 4.27 km/h

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

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Chapter 14, Solution 2.

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

components

mB v0 = ( mA + mB + mC ) v f

:

vf =

( 0.028)( 550 ) = 1.7058 m/s
mBv0
=
mA + mB + mC 5 + 0.028 + 4
v f = 1.706 m/s

(a)
Consider block and bullet alone.

Principle of impulse and momentum.
components
components

:

N ( ∆t ) − mg ( ∆t )

N = mg

: mB v0 − µk N ( ∆t ) = ( mA + mB ) v'

Just after impact, ∆t is negligible. The velocity then is

v0' =

( 0.028)( 550 ) = 3.0628 m/s
mB v0
=
mA + mB
5 + 0.028

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

Accelerations after impact.

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

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∑ F = ma :

Block and bullet:

µ k ( mA + mB ) g = ( mA + mB ) a AB
a AB = µk g = ( 0.50 )( 9.81)
= 4.905 m/s 2

∑ F = maC :

Cart:

µ k ( mA + mB ) g = mC aC :
aC =

µ k ( mA + mB ) g
mC

=

( 0.50 )( 5.028)( 9.81)
4

= 6.1656 m/s 2
Acceleration of block relative to cart.

a AB/C = 4.905 − ( − 6.1656 ) = 11.0706 m/s 2
Motion of the block relative to the cart.

( vAB/C )2
2

( v' )2
2

= 2 ( a AB/C )( s AB/C )

In the final position, v AB/C = 0

s AB/C = −

( v' )2
2 a AB/C

2
3.0628 )
(
=−
= − 0.424 m
( 2 )(11.0706 )

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

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

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Chapter 14, Solution 3.

The masses are mA =

( 22 )( 2000 ) = 1366.5 slugs
4000
3700
= 124.2 slugs, mB =
= 114.9 slugs, and mF =
32.2
32.2
32.2

Let v A , vB , and vF be the sought after velocities in ft/s, positive to the right.

( v A )0

Initial values:

Initial momentum of system:

= ( vB )0 = ( vF )0 = 0.

mA ( v A )0 + mB ( vB )0 + mF ( vF )0 = 0.

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

0 = mAv A + mBvB + mF vF

(1)

124.2v A + 114.9vB + 1366.5vF = 0
The relative velocities are given as

v A/F = v A − vF = − 7 ft/s

(2)

vB/F = vB − vF = − 3.5 ft/s

(3)

Solving (1), (2), and (3) simultaneously,

v A = − 6.208 ft/s, vB = − 2.708 ft/s, vF = 0.7919 ft/s
v F = 0.792 ft/s

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

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Chapter 14, Solution 4.

The masses are mA =

WA
W
W
, mB = B , and mF = F .
g
g
g

Let the final velocities be v A , vB , and vF = 0.34 ft/s, positive to the right.

( v A )0

Initial values:

= ( vB ) 0 = ( vF ) 0 = 0

mA ( v A )0 + mB ( vB )0 + mF ( vF )0 = 0

Initial momentum of system:

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

0 = mAv A + mB vB + mF vF =

WA
W
W
v A + B vB + F vF
g
g
g

WF = −

Solving for WF ,

WAv A + WBvB
vF

(1)

From the given relative velocities,

v A = vF + v A/F = 1.02 − 7.65 = − 6.63 ft/s
vB = vF + vB/F = 1.02 − 7.5 = − 6.48 ft/s
Substituting these values in (1),

WF = −

( 4000 )( − 6.63) + ( 3700 )( − 6.48)
1.02

= 49506 lb
WF = 24.8 tons

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

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Chapter 14, Solution 5.

(

)

(

)

The masses are the engine mA = 80 × 103 kg , the load mB = 30 × 103 kg , and the flat car

(m

C

)

= 20 × 103 kg .

( v A )0

Initial velocities:

= 6.5 km/h = 1.80556 m/s,

( vB ) 0

= ( vC )0 = 0.

No horizontal external forces act on the system during the impact and while the load is sliding relative to the flat
car. Momentum is conserved.

mA ( v A )0 + mB ( 0 ) + mC ( 0 ) = mA ( v A )0

Initial momentum:

(1)

(a) Let v′ be the common velocity of the engine and flat car immediately after impact. Assume that the
impact takes place before the load has time to acquire velocity.
Momentum immediately after impact:

mAv′ + mB ( 0 ) + mC v′ = ( mA + mC ) v′

(2)

Equating (1) and (2) and solving for v′,

v′ =

m A ( v A )0

mA + mC

(80 × 10 ) (1.80556) = 1.44444 m/s
=
(100 × 10 )
3

3

v′ = 5.20 km/h
(b) Let v f be the common velocity of all three masses after the load has slid to a stop relative to the car.
Corresponding momentum:

mAv f + mBv f + mC v f = ( mA + mB + mC ) v f

(3)

Equating (1) and (3) and solving for v f ,

vf =

m A ( v A )0

mA + mB + mC

(80 × 10 ) (1.80556) = 1.11111 m/s
=
(130 × 10 )
3

3

v f = 4.00 km/h

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

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Chapter 14, Solution 6.

The masses are m for the bullet and mA and mB for the blocks.
(a) The bullet passes through block A and embeds in block B. Momentum is conserved.

Initial momentum:

mv0 + mA ( 0 ) + mB ( 0 ) = mv0

Final momentum:

mvB + m Av A + mB vB

Equating,

mv0 = mvB + mAv A + mBvB

m=

( 3)( 3) + ( 2.5)( 5) = 43.434 × 10−3 kg
mAv A + mB vB
=
v0 − vB
500 − 5
m = 43.4 g

(b) The bullet passes through block A. Momentum is conserved.

Initial momentum:

mv0 + mA ( 0 ) = mv0

Final momentum:

mv1 + m Av A

Equating,

mv0 = mv1 + mAv A

(

)

43.434 × 10−3 ( 500 ) − ( 3)( 3)
mv0 − mAv A
=
= 292.79 m/s
v1 =
m
43.434 × 10−3
v1 = 293 m/s

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

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Chapter 14, Solution 7.

(a) Woman dives first.
Conservation of momentum:
120
300 + 180
v1 = 0
(16 − v1 ) −
g
g
v1 =

(120 )(16 )
600

= 3.20 ft/s

Man dives next. Conservation of momentum:

300 + 180
300
180
v1 = −
v2 +
(16 − v2 )
g
g
g

v2 =

480v1 + (180 )(16 )
480

= 9.20 ft/s

v 2 = 9.20 ft/s

(b) Man dives first.
Conservation of momentum:

180
300 + 120
v1′
(16 − v1′ ) −
g
g
v1′ =

(180 )(16 )
600

= 4.80 ft/s

Woman dives next. Conservation of momentum:

300 + 120
300
120
v1′ = −
v′2 +
(16 − v′2 )
g
g
g

v′2 =

420v1′ + (120 )(16 )
420

= 9.37 ft/s

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

v′2 = 9.37 ft/s

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Chapter 14, Solution 8.

(a) Woman dives first.
Conservation of momentum:

120
300 + 180
v1 = 0
(16 − v1 ) +
g
g
v1 =

(120 )(16 )
600

= 3.20 ft/s

Man dives next. Conservation of momentum:

300 + 180
300
180
v1 = −
v2 +
(16 − v2 )
g
g
g
v2 =

− 480v1 + (180 )(16 )
= 2.80 ft/s
480

v 2 = 2.80 ft/s

(b) Man dives first.
Conservation of momentum:

180
300 + 120
v1′ = 0
(16 − v1′ ) −
g
g
v1′ =

(180 )(16 )
600

= 4.80 ft/s

Woman dives next. Conservation of momentum:

300 + 120
300
120
v1′ =
v′2 +
(16 − v′2 )
g
g
g
v′2 =

−420v1′ + (120 )(16 )
= −0.229 ft/s
420
v′2 = 0.229 ft/s

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

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Chapter 14, Solution 9.
The masses are mA = mB = mC = 9 kg.

Position vectors (m):

rA = 0.9k ,

rB = 0.6i + 0.6 j + 0.9k ,

rC = 0.3i + 1.2 j

HO = rA × ( mA v A ) + rB × ( mB v B ) + rC × ( mC vC )

In units of kg ⋅ m 2 /s,

i

j

= 0
0

k

0
0.9 +
( 9vA ) 0

i

j

k

i

j

0.6 0.6 0.9 + 0.3 1.2
0 0
( 9vB ) 0 0

k
0
( 9vC )

= ( − 8.1v Ai ) + ( 8.1vB j − 5.4vBk ) + (10.8vC i − 2.7vC j)
= ( − 8.1v A + 10.8vC ) i + ( 8.1vB − 2.7vC ) j + ( − 5.4vB ) k

(

)

But, H O is given as − 1.8 kg ⋅ m 2 /s k
Equating the two expressions for H O and resolving into components,

i:

− 8.1v A + 10.8vC = 0

(1)

j: 8.1vB − 2.7vC = 0

(2)

k: − 5.4vB = −1.8

(3)

(a) Solving for v A , vB , and vC ,

v A = (1.333 m/s ) j

v A = 1.333 m/s

v B = ( 0.333 m/s ) i

vB = 0.333 m/s

vC = (1.000 m/s ) k

vC = 1.000 m/s

Coordinates of mass center G in m.

r =
=

mArA + mBrB + mC rC
mA + mB + mC

( 9 )( 0.9k ) + ( 9 )( 0.6i + 0.6 j + 0.9k ) + ( 9 )( 0.3i + 1.2 j)
27

= 0.3i + 0.6 j + 0.6k
Position vectors relative to the mass center in m.

rA′ = rA − r = ( 0.9k ) − ( 0.3i + 0.6 j + 0.6k ) = −0.3i − 0.6 j + 0.3k
rB′ = rB − r = ( 0.6i + 0.6 j + 0.9k ) − ( 0.3i + 0.6 j + 0.6k ) = 0.3i + 0.3k
rC′ = rC − r = ( 0.3i + 1.2 j) − ( 0.3i + 0.6 j + 0.6k ) = 0.6 j − 0.6k
mA v A = ( 9 )(1.333j) = (12 kg ⋅ m/s ) j
mB v B = ( 9 )( 0.333i ) = ( 3 kg ⋅ m/s ) i
mC vC = ( 9 )(1k ) = ( 9 kg ⋅ m/s ) k

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

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(b)

HG = rA′ × ( mA v A ) + rB′ × ( mB v B ) + rC′ × ( mC vC )
= ( − 0.3i − 0.6 j + 0.3k ) × (12 j) + ( 0.3i + 0.3k ) × ( 3i ) + ( 0.6 j − 0.6k ) × ( 9k )
= ( − 3.6i − 3.6k ) + ( 0.9 j) + ( 5.4i ) = 1.8i + 0.9 j − 3.6k

(

) (

) (

)

HG = 1.800 kg ⋅ m 2 /s i + 0.900 kg ⋅ m 2/s j − 3.60 kg ⋅ m 2/s k

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

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Chapter 14, Solution 10.

The masses are

mA = mB = mC = 9 kg.

rA = 0.9k ,

Position vectors (m):

rB = 0.6i + 0.6 j + 0.9k ,

rC = 0.3i + 1.2 j

Coordinates of mass center G expressed in m.

r =
=

mArA + mBrB + mC rC
mA + mB + mC

( 9 )( 0.9k ) + ( 9 )( 0.6i + 0.6 j + 0.9k ) + ( 9 )( 0.3i + 1.2j)
27

= 0.3i + 0.6 j + 0.6k
Position vectors relative to the mass center expressed in m.

rA′ = rA − r = ( 0.9k ) − ( 0.3i + 0.6 j + 0.6k ) = −0.3i − 0.6 j + 0.3k
rB′ = rB − r = ( 0.6i + 0.6 j + 0.9k ) − ( 0.3i + 0.6 j + 0.6k ) = 0.3i + 0.3k
rC′ = rC − r = ( 0.3i + 1.2 j) − ( 0.3i + 0.6 j + 0.6k ) = 0.6 j − 0.6k
Angular momenta.

HO = rA × ( mA v A ) + rB × ( mB v B ) + rC × ( mC vC )
HG = rA′ × ( mA v A ) + rB′ × ( mB v B ) + rC′ × ( mC vC )
Subtracting,

HO − HG = ( rA − rA′ ) × ( mA v A ) + ( rB − rB′ ) × mB v B + ( rC − rC′ ) × mC vC
0 = r × ( mA v A ) + r × ( mB v B ) + r × ( mC vC )
= r × ( mA v A + mB v B + mC vC ) = r × L

L = λr

L is parallel to r.

λ2 =

L ⋅ L = λ 2r ⋅ r

2

( 45) = 502 ,
L⋅L
=
r⋅r
( 0.9 )2

λ = ± 50 N ⋅ s/m

mA v A + mB v B + mC vC = λ r

( 9 )( vA j) + ( 9 )( vBi ) + ( 9 ) ( vC k ) = ± 50 ( 0.3i + 0.6 j + 0.6k )

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

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(a) Resolve into components and solve for v A , vB , and vC .
v A = 3.333 m/s

v A = ( 3.33 m/s ) j

vB = 1.6667 m/s

v B = (1.667 m/s ) i

vC = 3.333 m/s

vC = ( 3.33 m/s ) k

(b) Angular momentum about O expressed in kg ⋅ m 2 /s.

HO = rA × ( mA v A ) + rB × ( mB v B ) + rC × ( mC vC )
i
= 0
0

j

k

i

0
0.9 +
( 9vA ) 0

j

k

i

j

0.6 0.6 0.9 + 0.3 1.2
0 0
( 9vB ) 0 0

k
0
( 9vC )

= ( − 8.1v Ai ) + ( 8.1vB j − 5.4vBk ) + (10.8vC i − 2.7vC j)
= ( −8.1v A + 10.8vC ) i + ( 8.1vB − 2.7vC ) j + ( −5.4vB ) k
= 9i + 4.5j − 9k

(

) (

) (

)

HO = 9.00 kg ⋅ m 2 /s i + 4.50 kg ⋅ m 2 /s j − 9.00 kg ⋅ m 2 /s k

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

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Chapter 14, Solution 11.

Position vectors expressed in ft.

rA = 3i + 6 j,

rB = 6 j + 3k ,

rC = 3i + 3k

Momentum of each particle expressed in lb ⋅ s.
WA v A
4
1
= ( 42i + 63j) = (168i + 252 j)
g
g
g
WB v B
4
1
= ( − 42i + 63j) = ( −168i + 252 j)
g
g
g
WC vC
28
1
=
( − 9 j − 6k ) = ( − 252 j − 168k )
g
g
g

Angular momentum of the system about O expressed in ft ⋅ lb ⋅ s.

HO = rA ×

WA v A
W v
W v
+ rB × B B + rC × C C
g
g
g

 i
j k
i
j k
i
j
k 
1

=  3
6 0 + 0
6 3 + 3 0
3 
g
−168 252 0
0 −252 −168 
 168 252 0
1
= {( −252k ) + ( −756i − 504 j + 1008k ) + ( 756i + 504 j − 756k )}
g
=

1
( 0i + 0 j + 0k )
g
H O = zero

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

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Chapter 14, Solution 12.

W = WA + WB + WC = 4 + 4 + 28 = 36 lb

(a)

mArA + mBrB + mC rC
W r + WBrB + mC rC
= A A
m
W
1
=
{( 4 )( 3i + 6j) + ( 4 )( 6j + 3k ) + ( 28)( 3i + 3k )}
36
= 2.667i + 1.333j + 2.667k

r =

r = ( 2.67 ft ) i + (1.333 ft ) j + ( 2.67 ft ) k

(b) Linear momentum

mv = mA v A + mB v B + mC vC =

1
(WA v A + WB v B + WC vC )
g

1
( 4 )( 42i + 63j) + ( 4 )( − 42i + 63j) + ( 28 )( − 9 j − 6k ) 
g
1
=
( 252 j − 168k )
32.2
mv = ( 7.83 lb ⋅ s ) j − ( 5.22 lb ⋅ s ) k
(c) Position vectors relative to the mass center G (ft).

=

rA′ = rA − r = ( 3i + 6 j) − ( 2.667i + 1.333j + 2.667k )
= 0.333i + 4.667 j − 2.667k
rB′ = rB − r = ( 6 j + 3k ) − ( 2.667i + 1.333j + 2.667k )
= −2.667i + 4.667 j + 0.333k
rC′ = rC − r = ( 3i + 3k ) − ( 2.667i + 1.333j + 2.667k )
= 0.333i − 1.333j + 0.333k
Angular momentum about the mass center.

W v 
W v 
W v 
HG = rA′ ×  A A  + rB′ ×  B B  + rC′ ×  C C 
 g 
 g 
 g 

i
j
k
i
j
k
i
1
=  0.333
4.667 −2.667 + −2.667
4.667 0.333 + 0.333
g
0
0
( 4 )( −42 ) ( 4 )( 63) 0
 ( 4 )( 42 ) ( 4 )( 63)
1
= {( 672i − 448j − 700k ) + ( −84i − 56 j + 112k ) + ( 308i + 56 j − 84k )}
g

1
(896i − 448j − 672k ) = 27.827i − 13.913j − 20.870k
32.2
= ( 27.8 ft ⋅ lb ⋅ s ) i − (13.91 ft ⋅ lb ⋅ s ) j − ( 20.9 ft ⋅ lb ⋅ s ) k
=

HG

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

j
−1.333
( 28)( −9 )

k
0.333
( 28)( −6 )

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From Problem 14.28,

HO = r × mv + H G

HO

i
j
k
= 2.667 1.333 2.667
0
7.83 −5.22

= − 27.826i + 13.913j + 20.870k + 27.827i − 13.913j − 20.870k
= 0i + 0 j + 0k
From Prob 14.11,

HO = rA × ( mA v A ) + rB × ( mB v B ) + rC × ( mC vC )

=

1
{rA × (WAv A ) + rB × (WB v B ) + rC × (WC vC )}
g

 i
j k
i
j k
i
j
k 
1

=  3
6 0 + 0
6 3 + 3 0
3 
g
−168 252 0
0 −252 −168 
 168 252 0
1
= {( − 252k ) + ( − 756i − 504 j + 1008k ) + ( 756i + 504 j − 756k )}
g

=

1
( 0i + 0 j + 0k )
g

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

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Chapter 14, Solution 13.

Linear momentum of each particle expressed in kg ⋅ m/s.

mA v A = 3i − 2 j + 4k
mB v B = 8i + 6 j
mC vC = 6i + 15j − 9k

Position vectors, (meters):

rA = 3j + k ,

rB = 3i + 2.5k ,

rC = 4i + 2 j + k

Angular momentum about O, ( kg ⋅ m 2 /s ) .
HO = rA × ( mA v A ) + rB × ( mB v B ) + rC × ( mC vC )
i j k
i j k
i j k
= 0 3 1 + 3 0 2.5 + 4 2 1
3 −2 4
8 6 0
6 15 − 9
= (14i + 3j − 9k ) + ( −15i + 20 j + 18k ) + ( −33i + 42 j + 48k )
= − 34i + 65j + 57k

(

) (

) (

)

HO = − 34 kg ⋅ m 2 /s i + 65 kg ⋅ m 2 /s j + 57 kg ⋅ m 2 /s k

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

COSMOS: Complete Online Solutions Manual Organization System

Chapter 14, Solution 14.

Position vectors, (meters):

rA = 3j + k ,

(a) Mass center:

6r

rB = 3i + 2.5k ,

rC = 4i + 2 j + k

( mA + mB + mC ) r = mArA + mBrB + mCrC
= (1)( 3j + k ) + ( 2 )( 3i + 2.5k ) + ( 3)( 4i + 2 j + k )

r = 3i + 1.5j + 1.5k
r = ( 3.00 m ) i + (1.500 m ) j + (1.500 m ) k

Linear momentum of each particle, ( kg ⋅ m/s ) .
mA v A = 3i − 2 j + 4k
mB v B = 8i + 6 j

mC vC = 6i + 15j − 9k

(b) Linear momentum of the system, ( kg ⋅ m/s.)

mv = mA v A + mB v B + mC vC = 17i + 19 j − 5k

mv = (17.00 kg ⋅ m/s ) i + (19.00 kg ⋅ m/s ) j − ( 5.00 kg ⋅ m/s ) k

Position vectors relative to the mass center, (meters).
rA′ = rA − r = − 3i + 1.5j − 0.5k

rB′ = rB − r = −1.5j + k
rC′ = rC − r = i + 0.5j − 0.5k
(c) Angular momentum about G, ( kg ⋅ m 2 /s ) .

HG = rA′ × mA v A + rB′ × mB v B + rC′ × mC vC
i
j
k
i
j k
i j
k
= −3 1.5 −0.5 + 0 −1.5 1 + 1 0.5 −0.5
3 −2 4
8 6 0
6 15 −9
= ( 5i + 10.5j + 1.5k ) + ( −6i + 8 j + 12k ) + ( 3i + 6 j + 12k )
= 2i + 24.5j + 25.5k

(

) (

) (

)

HG = 2.00 kg ⋅ m 2 /s i + 24.5 kg ⋅ m 2/s j + 25.5 kg ⋅ m 2 /s k

i j k
r × mv = 3 1.5 1.5
17 19 −5
= − ( 36 kg ⋅ m 2 /s ) i + ( 40.5 kg ⋅ m 2 /s ) j + ( 31.5 kg ⋅ m 2 /s ) k

HG + r × mv = − ( 34 kg ⋅ m 2 /s ) i + ( 65 kg ⋅ m 2/s ) j + ( 57 kg ⋅ m 2/s ) k

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

COSMOS: Complete Online Solutions Manual Organization System

HO = rA × ( mA v A ) + rB × ( mB v B ) + rC × ( mC vC )
i j k
i j k
i j k
= 0 3 1 + 3 0 2.5 + 4 2 1
3 −2 4
8 6 0
6 15 −9

= (14i + 3j − 9k ) + ( −15i + 20 j + 18k ) + ( − 33i + 42 j + 48k )

(

)

= − ( 34 kg ⋅ m 2 /s ) i + 65 kg ⋅ m 2 /s j + ( 57 kg ⋅ m 2/s ) k
Note that
H O = H G + r × mv

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

COSMOS: Complete Online Solutions Manual Organization System

Chapter 14, Solution 15.

The mass center moves as if the projectile had not exploded.
2
1

1
r = ( v 0t ) −  gt 2  j = ( 60i )( 2 ) −  ( 9.81)( 2 )  j
2
2

= (120 m ) i − (19.62 m ) j

( mA + mB ) r
rB =
=

= mArA + mBrB

1
( mA + mB ) r − mArA 
mB 
1
( 20 )(120i − 19.62 j) − 8 (120i − 10 j − 20k ) 
12 

= 120i − 26.033j + 13.333k

rB = (120.0 m ) i − ( 26.0 m ) j + (13.33 m ) k

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

COSMOS: Complete Online Solutions Manual Organization System

Chapter 14, Solution 16.

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

r = v 0t = ( 450i )( 4 ) = (1800 m ) i

( mA + mB + mC ) r
rC =
=

= mArA + mBrB + mC rC

1
( mA + mB + mC ) r − mArA − mBrB 
mC 
1
( 500 )(1800i ) − ( 300 )(1200i − 350 j − 600k )
50 

− (150 )( 2500i + 450 j + 900k ) 
= 3300i + 750 j + 900k

rC = ( 3300 m ) i + ( 750 m ) j + ( 900 m ) k

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

COSMOS: Complete Online Solutions Manual Organization System

Chapter 14, Solution 17.

Mass center at time of first collision.

( mA + mB + mC ) r1 = mA ( rA )1 + mB ( rB )1 + mC ( rC )1
(WA + WB + WC ) r1 = WA ( rA )1 + WB ( rB )1 + WC ( rC )1
9600 r1 = ( 2800 )( − 27.8j) + ( 3600 )( − 38.4 j) + ( 3200 )(120i )
r1 = ( 40 ft ) i − ( 22.508 ft ) j
Mass center at time of photo.

( mA + mB + mC ) r2
(WA + WB + WC ) r2

= mA ( rA )2 + mB ( rB )2 + mC ( rC )2

= WA ( rA )2 + WB ( rB )2 + WC ( rC )2

9600 r2 = ( 2800 )( − 30.3i + 50.7 j) + ( 3600 )( − 30.3i + 61.2 j)
+ ( 3200 )( − 59.4i − 45.6 j)
r2 = − ( 40 ft ) i + ( 22.5375 ft ) j

Since no external horizontal forces act, momentum is conserved, and the mass center moves at constant
velocity.

( mA + mB + mC ) v = mA ( v A )1 + mB ( v B )1 + mC ( vC )1

(1)

r2 − r1 = vt

(2)

Combining (1) and (2),

( mA + mB + mC ) ( r2 − r1 ) =  mA ( v A )1 + mB ( v B )1 + mC ( vC )1  t
(WA + WB + WC ) ( r2 − r1 ) = WA ( v A )1 + WB ( v B )1 + WC ( vC )1  t

( 9600 )( − 80i + 45.0455j) = 0 + ( 3600 )( vB )1 j + ( 3200 )( − 66i ) t
Components.

i: − 768000 = −211200t

t = 3.64 s

j: 432440 = 3600 vB t

vB =

( 432440 )
( 3600 )( 3.6363)

= 30.034

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

vB = 30.0 ft/s

COSMOS: Complete Online Solutions Manual Organization System

Chapter 14, Solution 18.

Mass center at time of first collision.

( mA + mB + mC ) r1 = mA ( rA )1 + mB ( rB )1 + mC ( rC )1
(WA + WB + WC ) r1 = WA ( rA )1 + WB ( rB )1 + WC ( rC )1
9600 r1 = ( 2800 )( − 27.8j) + ( 3600 )( − 38.4 j) + ( 3200 )(120i )
r1 = ( 40 ft ) i − ( 22.508 ft ) j
Mass center at time of photo.

( mA + mB + mC ) r2
(WA + WB + WC ) r2

= mA ( rA )2 + mB ( rB )2 + mC ( rC )2

= WA ( rA )2 + WB ( rB )2 + WC ( rC )2

9600 r2 = ( 2800 )( − 30.3i + 50.7 j) + ( 3600 )( − 30.3i + 61.2 j)
+ ( 3200 )( − 59.4i − 45.6 j)
r2 = − ( 40 ft ) i + ( 22.5375 ft ) j
Since no external horizontal forces act, momentum is conserved, and the mass center moves at constant
velocity.

( mA + mB + mC ) v = mA ( v A )1 + mB ( v B )1 + mC ( vC )1

(1)
(2)

r2 − r1 = vt

( mA + mB + mC ) ( r2 − r1 ) =  mA ( v A )1 + mB ( v B )1 + mC ( vC )1  t

Combining (1) and (2),

(WA + WB + WC ) ( r2 − r1 ) = WA ( v A )1 + WB ( v B )1 + WC ( vC )1  t
( 9600 )( − 80i + 45.0455j) = 0 + ( 3600 )( vB )1 j + ( 3200 ) ( vC )1 i  ( 3.4 )
Components.

j: 432440 = 12240 ( vB )1 ,

( vB )1 = 35.33 ft/s,
vB = 24.1 mi/h

i:

− 768000 = −10880 ( vC )1 ,

( vC )1 = 70.588 ft/s,
vC = 48.1 mi/h

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

COSMOS: Complete Online Solutions Manual Organization System

Chapter 14, Solution 19.

ax = 0,

Projectile motion

( v x )0

= 165 m/s,

a y = − g = − 9.81 m/s 2 ,

( v y )0 = 0,

( v z )0

az = 0
=0

After the chain breaks the mass center continues the original projectile motion.
At t = 1.5 s,

x = x0 + ( vx )0 t = 0 + (165 )(1.5 ) = 247.5 m

( )0 t − 12 gt 2 = 15 + 0 − 12 ( 9.81)(1.5)2 = 3.9638 m

y = y0 + v y

z = z0 + ( v z ) 0 t = 0
Position of first cannon ball at this time is

x1 = 240 m, y1 = 0, z1 = 7 m
Definition of mass center: ( m1 + m2 ) r = m1r1 + m2r2

r2 =

( m1 + m2 ) r −
m2

m1
r1
m2

30
15
( 247.5i + 3.9638j) − ( 240i + 7k )
15
15
= ( 255 m ) i + ( 7.9276 m ) j − 7k
=

Position of second cannon ball:

x2 = 255 m, y2 = 7.93 m, z2 = − 7 m

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

COSMOS: Complete Online Solutions Manual Organization System

Chapter 14, Solution 20.

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

a y = − g = − 9.81 m/s 2

v y = v0 + a yt = 28 − 9.81t
1 2
a yt = 60 + 28t − 4.905t 2
2
x = 0, y = 55.939 m

y = y0 + v0t +
At t = 5.85 s,

Definition of mass center :

m r = m ArA + mBrB

3x = 1x A + 2 xB

x component:

0 = − 74.4 + 2 xB
y component:

xB = 37.2 m

3y = 1y A + 2 yB

( 3)( 55.939 ) = 0 + 2 yB

yB = 83.9 m

Position of part B.

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

37.2 m (east), 83.9 m (up)

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