COSMOS: Complete Online Solutions Manual Organization System
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
© 2007 The McGraw-Hill Companies.
W = 75 N W
COSMOS: Complete Online Solutions Manual Organization System
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
© 2007 The McGraw-Hill Companies.
W = 19.64 N W
COSMOS: Complete Online Solutions Manual Organization System
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
© 2007 The McGraw-Hill Companies.
mw = 0.232 lb ⋅ s 2 /ft W
COSMOS: Complete Online Solutions Manual Organization System
Chapter 12, Solution 4.
Periodic time:
τ = 12 h = 43200 s
Radius of Earth:
R = 3960 mi = 20.9088 × 106 ft
Radius of orbit:
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
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
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
© 2007 The McGraw-Hill Companies.
v = 4.91 ft/s W
COSMOS: Complete Online Solutions Manual Organization System
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
© 2007 The McGraw-Hill Companies.
µ = 0.612 W
COSMOS: Complete Online Solutions Manual Organization System
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
4°
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
© 2007 The McGraw-Hill Companies.
x f = 63.9 m W
COSMOS: Complete Online Solutions Manual Organization System
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
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
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
(b) As already determined,
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
© 2007 The McGraw-Hill Companies.
h = 1449 ft W
t = 10.00 s W
COSMOS: Complete Online Solutions Manual Organization System
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
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
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
© 2007 The McGraw-Hill Companies.
x − x0 = 61.9 m W
COSMOS: Complete Online Solutions Manual Organization System
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
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
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
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
( 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
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
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
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
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
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
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
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
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
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
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
© 2007 The McGraw-Hill Companies.
aA
g
COSMOS: Complete Online Solutions Manual Organization System
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
© 2007 The McGraw-Hill Companies.
COSMOS: Complete Online Solutions Manual Organization System
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
© 2007 The McGraw-Hill Companies.
or a B = 0.493 m/s 2
15° W
COSMOS: Complete Online Solutions Manual Organization System
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
© 2007 The McGraw-Hill Companies.
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
© 2007 The McGraw-Hill Companies.
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
© 2007 The McGraw-Hill Companies.
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
© 2007 The McGraw-Hill Companies.