# Solution manual vector mechanics engineers dynamics 8th beer chapter 12

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Chapter 12, Solution 1.

m = 20 kg, g = 3.75 m/s 2
W = mg = ( 20 )( 3.75 )

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

W = 75 N W

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Chapter 12, Solution 2.

m = 2.000 kg W

At all latitudes,

(a) φ = 0°,

(

)

g = 9.7807 1 + 0.0053 sin 2 φ = 9.7807 m/s 2
W = mg = ( 2.000 )( 9.7807 )

(

W = 19.56 N W

)

(b) φ = 45°, g = 9.7807 1 + 0.0053 sin 2 45° = 9.8066 m/s 2
W = mg = ( 2.000 )( 9.8066 )

(

W = 19.61 N W

)

(c) φ = 60°, g = 9.7807 1 + 0.0053 sin 2 60° = 9.8196 m/s2
W = mg = ( 2.000 )( 9.8196 )

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

W = 19.64 N W

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Chapter 12, Solution 3.

Assume g = 32.2 ft/s 2
m=

W
g

ΣF = ma : W − Fs =

a
W 1 −  = Fs
g

or

Fs

W =

1−

a
g

W
a
g

7

=
1−

2
32.2
W = 7.46 lb W

m=

W
7.4635
=
= 0.232 lb ⋅ s 2 /ft
g
32.2
ΣF = ma : Fs − W =

W
a
g

a
Fs = W 1 + 
g

2 

= 7.46 1 +

32.2

Fs = 7.92 lb W

For the balance system B,
ΣM 0 = 0: bFw − bFp = 0
Fw = Fp

a
a
But, Fw = Ww 1 +  and Fp = W p 1 + 
g
g

so that Ww = W p and mw =

Wp
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

mw = 0.232 lb ⋅ s 2 /ft W

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Chapter 12, Solution 4.

Periodic time:

τ = 12 h = 43200 s

R = 3960 mi = 20.9088 × 106 ft

r = 3960 + 12580 = 16540 mi = 87.33 × 106 ft

Velocity of satellite:

v=

2π r

τ

=

( 2π ) (87.33 × 106 )
43200

= 12.7019 × 103 ft/s
It is given that
(a)

mv = 750 × 103 lb ⋅ s
m=

mv
750 × 103
=
= 59.046 lb ⋅ s 2 /ft
3
v
12.7019 × 10
m = 59.0 lb ⋅ s 2 /ft W

(b)

W = mg = ( 59.046 )( 32.2 ) = 1901 lb

W = 1901 lb W

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 12, Solution 5.

+ ∑ Fy = ma y :

10 + 10 + 10 + 20 − 40 =
ay =

ay =

40
ay
32.2

( 32.2 )(10 ) = 8.05 ft/s2
40

dv dy dv
dv
=
=v
dt
dt dy
dy

v dv = a y d y
v

v

∫ 0 v dv = ∫ 0 a y d y
v = 2a y y =

1 2
v = ay y
2

( 2 )(8.05)(1.5)

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 = 4.91 ft/s W

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Chapter 12, Solution 6.

Data: v0 = 108 km/h = 30 m/s, x f = 75 m
(a)

Assume constant acceleration. a = v

dv dv
=
= constant
dx dt

0
xf
∫ v0 v dv = ∫ 0 a dx

1
− v02 = a x f
2
a=−

v02
2x f

=−

(30)
( 2)( 75)

= − 6 m/s 2

0
tf
∫ v0 dv = ∫ 0 a dt

− v0 = a t f
tf = −
(b)

v0 − 30
=
a
−6

t f = 5.00 s W

+ ∑ Fy = 0: N − W = 0
N =W

∑ Fx = ma :
µ=−

µ=−

− µ N = ma

ma
ma
a
=−
=−
N
W
g

( − 6)
9.81

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

µ = 0.612 W

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Chapter 12, Solution 7.

(a)

+ ∑ F = ma :
a=−
=−

Ff
m

− F f + W sin α = ma
+

Ff
W sin α
=−
+ g sin α
m
m

(

)

7500 N
+ 9.81 m/s 2 sin 4° = − 4.6728 m/s 2
1400 kg

a = 4.6728 m/s 2

v0 = 88 km/h = 24.444 m/s
From kinematics,

a=v

dv
dx

xf
0
∫ 0 a dx = ∫ v0 v dv

1
a x f = − v02
2

( 24.444 )
v02
=−
2a
( 2 )( − 4.6728)
2

xf = −

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

x f = 63.9 m W

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Chapter 12, Solution 8.

(a) Coefficient of static friction.
ΣFy = 0:

N −W = 0

N =W

v0 = 70 mi/h = 102.667 ft/s
v 2 v02

= at ( s − s0 )
2
2
0 − (102.667 )
v 2 − v02
=
= − 31.001 ft/s 2
2 ( s − s0 )
( 2 )(170 )
2

at =

For braking without skidding µ = µ s , so that µ s N = m | at |
ΣFt = mat : − µ s N = mat

µs = −

mat
a
31.001
= − t =
W
g
32.2

µ s = 0.963 W

(b) Stopping distance with skidding.
Use µ = µk = ( 0.80 )( 0.963) = 0.770
ΣF = mat : µk N = −mat
at = −

µk N
m

= − µk g = − 24.801 ft/s 2

Since acceleration is constant,

( s − s0 ) =

0 − (102.667 )
v 2 − v02
=
2at
( 2 )( − 24.801)

2

s − s0 = 212 ft W

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 12, Solution 9.

For the thrust phase,

ΣF = ma : Ft − W = ma =

W
a
g

F

 2

a = g  t − 1 = ( 32.2 ) 
− 1 = 289.8 ft/s 2
W
0.2

At t = 1 s,
v = at = ( 289.8 )(1) = 289.8 ft/s
y =

1 2 1
2
at = ( 289.8 )(1) = 144.9 ft
2
2

For the free flight phase, t > 1 s. a = − g = − 32.2 ft/s
v = v1 + a ( t − 1) = 289.8 + ( − 32.2 )( t − 1)
At v = 0,

t −1 =

289.8
= 9.00 s, t = 10.00 s
32.2

v 2 − v12 = 2a ( y − y1 ) = −2 g ( y − y1 )
0 − ( 289.8 )
v 2 − v12
=−
= 1304.1 ft
y − y1 = −
2g
( 2 )( 32.2 )
2

(a)

ymax = h = 1304.1 + 144.9

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

h = 1449 ft W
t = 10.00 s W

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Chapter 12, Solution 10.

Kinematics: Uniformly accelerated motion. ( x0 = 0, v0 = 0 )
x = x0 + v0t +

1 2
at ,
2

a=

or

2 x ( 2 )(10 )
=
= 1.25 m/s 2
2
2
t
( 4)

ΣFy = 0: N − P sin 50° − mg cos 20° = 0
N = P sin 50° + mg cos 20°
ΣFx = ma : P cos 50° − mg sin 20° − µ N = ma
or P cos50° − mg sin 20° − µ ( P sin 50° + mg cos 20° ) = ma
P=

ma + mg ( sin 20° + µ cos 20° )
cos50° − µ sin 50°

For motion impending, set a = 0 and µ = µ s = 0.30.
P=

( 40 )( 0 ) + ( 40 )( 9.81)( sin 20° + 0.30cos 20° )
cos50° − 0.30sin 50°

= 593 N

For motion with a = 1.25 m/s 2 , use µ = µk = 0.25.
P=

( 40 )(1.25) + ( 40 )( 9.81)( sin 20° + 0.25cos 20° )
cos50° − 0.25sin 50°

P = 612 N W

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 12, Solution 11.

Calculation of braking force/mass ( Fb / m ) from data for level pavement.
v0 = 100 km/hr = 27.778 m/s
v 2 v02

= a ( x − x0 )
2
2
a=

0 − ( 27.778 )
v 2 − v02
=
2 ( x − x0 )
( 2 )( 60 )

2

= −6.43 m/s 2
ΣFx = ma : − Fbr = ma
Fbr
= −a = 6.43 m/s 2
m
(a) Going up a 6° incline. (θ = 6° )
ΣF = ma : − Fbr − mg sin θ = ma
a=−

Fbr
− g sin θ
m

= −6.43 − 9.81sin 6° = −7.455 m/s 2
0 − ( 27.778 )
v 2 − v02
=
2a
( 2 )( −7.455)

2

x − x0 =

x − x0 = 51.7 m W
(b) Going down a 2% incline. ( tan θ = −0.02, θ = −1.145° )
ΣF = ma : − Fbr − mg sin θ = ma
F
a = − br − g sin θ
m
= − 6.43 − 9.81sin ( −1.145° ) = − 6.234 m/s 2
0 − ( 27.778 )
v 2 − v02
=
2a
( 2 )( −6.234 )

2

x − x0 =

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

x − x0 = 61.9 m W

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Chapter 12, Solution 12.

Let the positive directions of x A and xB be down the incline.
Constraint of the cable:

x A + 3xB = constant

a A + 3aB = 0

1
aB = − a A
3

or

For block A:

ΣF = ma : mA g sin 30° − T = mAa A

For block B:

ΣF = ma : mB g sin 30° − 3T = mB aB = − mB a A (2)

Eliminating T and solving for

(1)

aA
,
g

( 3mA g − mB g ) sin 30° =  3mA +

mB 
 aA
3 

( 3mA − mB ) sin 30° = ( 30 − 8) sin 30° = 0.33673
aA
=
g
3m A + mB / 3
30 + 2.667
(a) a A = ( 0.33673)( 9.81) = 3.30 m/s 2
aB = −

1
( 3.30 ) = −1.101 m/s2
3

a A = 3.30 m/s 2

30° W

a B = 1.101 m/s 2

30° W

(b) Using equation (1),

a 
T = mA g  sin 30° − A  = (10 )( 9.81)( sin 30° − 0.33673)
g 

T = 16.02 N W

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 12, Solution 13.

Let the positive directions of x A and xB be down the incline.
Constraint of the cable:

x A + 3xB = constant
1
aB = − a A
3

a A + 3aB = 0

ΣFy = 0: N A − mA g cos 30° = 0

Block A:

ΣFx = ma : mA g sin 30° − µ N A − T = m Aa A
Eliminate N A.
mA g ( sin 30° − µ cos 30° ) − T = mAa A
ΣFy = 0: N B − mB g cos 30° = 0

Block B:

ΣF = ma : mB g sin 30° + µ N B − 3T = mB aB = −

mB a A
3

Eliminate N B .
mB g ( sin 30° + µ cos30° ) − 3T = −

mB a A
3

Eliminate T.

( 3mA g − mB g ) sin 30° − µ ( 3mA g + mB g ) cos 30° =  3mA +

mB 
 aA
3 

Check the value of µ s required for static equilibrium. Set a A = 0 and
solve for µ.

µ =

( 3mA − mB ) sin 30°
( 3mA + mB ) cos 30°

=

( 75 − 20 ) tan 30° = 0.334.
( 75 + 20 )

Since µ s = 0.25 < 0.334, sliding occurs.
Calculate

aA
for sliding. Use µ = µk = 0.20.
g
continued

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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( 3mA − mB ) sin 30° − µ ( 3mA + mB ) cos 30°
aA
=
g
3mA + mB / 3
=

( 30 − 8) sin 30° − ( 0.20 )( 30 + 8) cos 30°
30 + 2.667

(a) a A = ( 0.13525 )( 9.81) = 1.327 m/s 2

1
aB = −   (1.327 ) = − 0.442 m/s 2
3

= 0.13525

a A = 1.327 m/s 2

30° W

a B = 0.442 m/s 2

30° W

(b) T = mA g ( sin 30° − µ cos 30° ) − mAa A
= (10 )( 9.81)( sin 30° − 0.20cos 30° ) − (10 )(1.327 )
T = 18.79 N W

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 12, Solution 14.

Data:

mA =

55000 lb
= 1708.1 lb ⋅ s 2 / ft
32.2 ft/s 2

mB =

44000 lb
= 1366.5 lb ⋅ s 2 / ft
32.2 ft/s 2

v0 = − 55 mi/h = − 80.667 ft/s
(a) Use both cars together as a free body. Consider horizontal force components only. Both cars have same
acceleration.

∑ Fx = ∑ max : − Fb − Fb = mAax + mB ax
ax =

Fb + Fb
7000 + 7000
=
= 4.5534 m/s 2
mA + mB 1708.1 + 1366.5

ax = v

dv
dx

xf
0
∫ 0 ax dx = ∫ v0 v dv

ax x f =

v02
2

( −80.667 ) = − 751 ft
v2
xf = − 0 = −
2ax
( 2 )( 4.5534 )
2

715 ft to the left W
(b) Use car A as free body. Fc = coupling force.

∑ Fx = ∑ max : Fc − Fb = mAax
Fc = mAax − Fb = (1708.1)( 4.5534 ) + 7000 = 778 lb
Fc = 778 lb tension W

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 12, Solution 15.

mA =

55000 lb
= 1708.1 lb ⋅ s 2 / ft
32.2 ft/s 2

mB =

44000 lb
= 1366.5 lb ⋅ s 2 / ft
32.2 ft/s 2

Data:

v0 = − 55 mi/h = − 80.667 ft/s
(a) Use both cars together as a free body. Consider horizontal force components only. Both cars have same
acceleration.

∑ Fx = ∑ max : − Fb − Fb = mAax + mB ax
ax =

Fb
7000
=
= 2.2767 m/s 2
mA + mB 1708.1 + 1366.5

ax = v

xf
0

dv
dx
0

ax dx = ∫ v v dv
0

ax x f =

v02
2

( −80.667 ) = 1429 ft
v02
=−
2ax
( 2 )( 2.2767 )
2

xf = −

1429 ft to the left W
(b) Use car B as a free body. Fc = coupling force.

∑ Fx = ∑ max : − Fc = mB ax
− Fc = (1366.5)( 2.2767 ) = 3110 lb
Fc = 3110 lb. compression W

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 12, Solution 16.

Constraint of cable:

2 x A + ( xB − x A ) = x A + xB = constant.

a A + aB = 0,

or

aB = −a A

Assume that block A moves down and block B moves up.
Block B:

ΣFy = 0: N AB − WB cosθ = 0
ΣFx = ma : − T + µ N AB + WB sin θ =

WB
aB
g

Eliminate N AB and aB .
−T + WB ( sin θ + µ cosθ ) = WB

Block A:

aB
a
= −WB A
g
g

ΣFy = 0: N A − N AB − WA cosθ = 0
N A = N AB + WA cosθ = (WB + WA ) cosθ
ΣFx = m Aa A : − T + WA sin θ − FAB − FA =
−WB ( sin θ + µ cosθ ) − WB

aA
+ WA sin θ − µWB cosθ
g

− µ (WB + WA ) cosθ = WA

(WA − WB ) sinθ

WA
aA
g

aA
g

− µ (WA + 3WB ) cosθ = (WA + WB )

aA
g

Check the condition of impending motion.

µ = µs = 0.20,

a A = aB = 0,

θ = θs

(WA − WB ) sin θ s − 0.20 (WA + 3WB ) cosθ s

=0

continued

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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tan θ s =

0.20 (WA + 3WB ) ( 0.20 )(128 )
=
= 0.40
64
WA − WB

θ s = 21.8° < θ = 25°. The blocks move.
Calculate

aA
using µ = µ k = 0.15 and θ = 25°.
g

(WA − WB ) sin θ − µk (WA + 3WB ) cosθ
aA
=
g
WA + WB
=

64sin 25° − ( 0.15 )(128 ) cos 25°
96

= 0.10048

a A = ( 0.10048 )( 32.2 ) = 3.24 ft/s 2

(a) aB = −3.24 ft/s 2
(b)

T = WB ( sin θ + µ cosθ ) + WB

a B = 3.24 ft/s 2

25° !

aA
g

= 16 ( sin 25° + 0.15cos 25° ) + (16 )( 0.10048 )
T = 10.54 lb !

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 12, Solution 17.

Constraint of cable:

2 x A + ( xB − x A ) = x A + xB = constant.
a A + aB = 0,

or

aB = −a A

Assume that block A moves down and block B moves up.
Block B:

ΣFy = 0: N AB − WB cosθ = 0
ΣFx = max : − T + µ N AB + WB sin θ =

WB
aB
g

Eliminate N AB and aB .
−T + WB ( sin θ + µ cosθ ) = WB

Block A:

aB
a
= −WB A
g
g

ΣFy = 0: N A − N AB − WA cosθ + P sin θ = 0
N A = N AB + WA cosθ − P sin θ
= (WB + WA ) cosθ − P sin θ

ΣFx = mAa A : − T + WA sin θ − FAB − FA + P cosθ =
−WB ( sin θ + µ cosθ ) − WB

aA
+ WA sin θ − µWB cosθ
g

− µ (WB + WA ) cosθ + µ P sin θ + P cosθ = WA

(WA − WB ) sin θ

WA
aA
g

aA
g

− µ (WA + 3WB ) cosθ + P ( µ sin θ + cosθ ) = (WA + WB )

Check the condition of impending motion.

µ = µ s = 0.20, a A = aB = 0, θ = 25°

(WA

− WB ) sin θ − µ s (WA + 3WB ) cosθ + Ps ( µ s sin θ + cosθ ) = 0
continued

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

aA
g

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Ps =

=

µ s (WA + 3WB ) cosθ − (WA − WB ) sin θ
µ s sin θ + cosθ

( 0.20 )(128) cos 25° − 64 sin 25°
0.20 sin 25° + cos 25°

= −3.88 lb < 10 lb

Blocks will move with P = 10 lb.
Calculate

aA
using µ = µ k = 0.15, θ = 25°, and P = 10 lb.
g

(WA − WB ) sin θ − µk (WA + 3WB ) cosθ + P ( µk sinθ + cosθ )
aA
=
g
WA + WB
=

64 sin 25° − ( 0.15 )(128 ) cos 25° + (10 )( 0.15sin 25° + cos 25° )
96

= 0.20149

a A = ( 0.20149 )( 32.2 ) = 6.49 ft/s 2

(a) aB = −6.49 ft/s 2 ,
(b)

T = WB ( sin θ + µ cosθ ) + WB

a B = 6.49 ft/s 2

25° !

aA
g

= 16 ( sin 25° + 0.15cos 25° ) + (16 )( 0.20149 )
T = 12.16 lb. !

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 12, Solution 18.

Assume a B > a A so that the boxes separate. Boxes are slipping.

µ = µk
ΣFy = 0: N − mg cos15° = 0
N = mg cos15°
ΣFx = ma : µ k N − mg sin15° = ma

µ k mg cos15° − mg sin15° = ma
a = g ( µ k cos15° − sin15° ) ,

independent of m.

For box A, µ k = 0.30
a A = 9.81( 0.30cos15° − sin15° )

or a A = 0.304 m/s 2

15° W

For box B, µ k = 0.32
aB = 9.81( 0.32cos15° − sin15° )

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

or a B = 0.493 m/s 2

15° W

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Chapter 12, Solution 19.

Let y be positive downward position for all blocks.
Constraint of cable attached to mass A: y A + 3 yB = constant
a A + 3aB = 0
Constraint of cable attached to mass C:

a A = −3aB

yC + yB = constant

aC + aB = 0

For each block

or
or

aC = − aB

ΣF = ma :

Block A:

WA − TA = mAa A ,

or TA = WA − mAa A = WA − 3m AaB

Block C:

WC − TC = mC aC ,

or TC = WC − mC aC = WC − mC aB

Block B: WB − 3TA − TC = mB aB
WB − 3 (WA − 3mAaB ) − (WC − mC aB ) = mB aB
or
(a) Accelerations.

(b) Tensions.

aB
W − 3WA − WC
60 − 60 − 20
= B
=
= − 0.076923
g
WB + 9WA + WC
60 + 180 + 20

aB = ( − 0.076923)( 32.2 ) = − 2.477 ft/s 2

a B = 2.48 ft/s 2 W

a A = − ( 3)( − 2.477 ) = 7.431 ft/s 2

a A = 7.43 ft/s 2 W

aC = − ( − 2.477 ) = 2.477 ft/s 2

aC = 2.48 ft/s 2 W

20
( 7.43)
32.2
20
TC = 20 −
( 2.477 )
32.2
TA = 20 −

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

TA = 15.38 lb W
TC = 18.46 lb W

COSMOS: Complete Online Solutions Manual Organization System

Chapter 12, Solution 20.

Let y be positive downward for both blocks.
Constraint of cable: y A + yB = constant
a A + aB = 0

For blocks A and B,

aA =

aB = −a A

ΣF = ma :

Block A: WA − T =

Solving for a A ,

or

WA
aA
g

or

T = WA −

Block B: P + WB − T =

WB
W
aB = − B a A
g
g

P + WB − WA +

WA
W
a A = − B aA
g
g

WA
aA
g

WA − WB − P
g
WA + WB

(1)

2
v A2 − ( v A )0 = 2a A  y A − ( y A )0 

with

( v A )0 = 0

v A = 2a A  y A − ( y A )0 
v A − ( v A )0 = a At
t=

with

(2)

( v A )0 = 0

vA
aA

(3)
continued

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

(a) Acceleration of block A.
System (1): WA = 100 lb,

( aA )1

By formula (1),

=

WB = 50 lb, P = 0

100 − 50
( 32.2 )
100 + 50

( a A )1 = 10.73 ft/s2

!

System (2): WA = 100 lb, WB = 0, P = 50 lb

( a A )2

By formula (1),

=

100 − 50
( 32.2 )
100

( a A )2

= 16.10 ft/s 2 !

( a A )3

= 0.749 ft/s 2 !

System (3): WA = 1100 lb, WB = 1050 lb, P = 0
By formula (1),

( a A )3

=

1100 − 1050
( 32.2 )
1100 + 1050

(b) v A at y A − ( y A )0 = 5 ft. Use formula (2).
System (1):

( v A )1

=

( 2 )(10.73)( 5)

( v A )1

= 10.36 ft/s !

System (2):

( v A )2

=

( 2 )(16.10 )( 5)

( v A )2

= 12.69 ft/s !

System (3):

( v A )3

=

( 2 )( 0.749 )( 5)

( v A )3

= 2.74 ft/s !

(c) Time at v A = 10 ft/s. Use formula (3).
System (1): t1 =

10
10.73

t1 = 0.932 s !

System (2): t2 =

10
16.10

t2 = 0.621 s !

System (3): t3 =

10
0.749

t3 = 13.35 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

COSMOS: Complete Online Solutions Manual Organization System

Chapter 12, Solution 21.

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

For the upper beam, ΣFy = 0: N1 − W = 0
N1 = W = mg

For the lower beam, ΣFy = 0: N 2 − N1 − W = 0

or

N 2 = 2W

ΣFx = ma : 0.25 N1 + 0.30 N 2 = ( 0.25 + 0.60 )W = ma

a = 0.85

W
= ( 0.85 )( 32.2 ) = 23.37 ft/s 2
m

For the upper beam,

a = 27.4 ft/s 2

!

ΣFx = ma : T − 0.25 N1 = ma

 3000 
T = 0.25W + ma = ( 0.25 )( 3000 ) + 
 ( 23.37 ) = 2927 lb
 32.2 

T = 2930 lb !

(b) Maximum deceleration of trailer.
Case 1: Assume that only the top beam slips. As in Part (a) N1 = mg.

ΣF = ma : 0.25W = ma
a = 0.25g = 8.05 ft/s 2
continued

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