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Chapter 8, Solution 1.
FBD Block B:
Tension in cord is equal to W A = 25 lb from FBD’s of block A and
pulley.
ΣFy = 0:
N − WB cos 30° = 0,
N = WB cos 30°
(a) For smallest WB , slip impends up the incline, and
F = µ s N = 0.35WB cos30°
ΣFx = 0:
F − 25 lb + WB sin 30° = 0
( 0.35cos30° + sin 30° )WB = 25 lb
WB min = 31.1 lb
(b) For largest WB , slip impends down the incline, and
F = − µ s N = − 0.35 WB cos30°
ΣFx = 0:
Fs + WB sin 30° − 25 lb = 0
( sin 30° − 0.35cos30° )WB = 25 lb
W B max = 127.0 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 8, Solution 2.
FBD Block B:
Tension in cord is equal to WA = 40 lb from FBD’s of block A and
pulley.
(a)
ΣFy = 0:
N − ( 52 lb ) cos 25° = 0,
N = 47.128 lb
Fmax = µ s N = 0.35 ( 47.128 lb ) = 16.495 lb
ΣFx = 0:
Feq − 40 lb + ( 52 lb ) sin 25° = 0
So, for equilibrium, Feq = 18.024 lb
Since Feq > Fmax , the block must slip (up since F > 0)
∴ There is no equilibrium
(b) With slip,
F = µk N = 0.25 ( 47.128 lb )
F = 11.78 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.
35°
COSMOS: Complete Online Solutions Manual Organization System
Chapter 8, Solution 3.
FBD Block:
Tension in cord is equal to P = 40 N, from FBD of pulley.
(
)
W = (10 kg ) 9.81 m/s 2 = 98.1 N
ΣF y = 0 :
N − (98.1 N ) cos 20° + (40 N ) sin 20° = 0
N = 78.503 N
Fmax = µ s N = ( 0.30 )( 78.503 N ) = 23.551 N
For equilibrium:
ΣFx = 0:
( 40 N ) cos 20° − ( 98.1 N ) sin 20° − F = 0
Feq = 4.0355 N < Fmax ,
F = Feq
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.
∴ Equilibrium exists
F = 4.04 N
20°
COSMOS: Complete Online Solutions Manual Organization System
Chapter 8, Solution 4.
Tension in cord is equal to P = 62.5 N, from FBD of pulley.
(
)
W = (10 kg ) 9.81 m/s 2 = 98.1 N
ΣFy = 0:
N − ( 98.1 N ) cos 20° + ( 62.5 N ) sin15° = 0
N = 76.008 N
Fmax = µ s N = ( 0.30 )( 76.008 N ) = 22.802 N
For equilibrium:
ΣFx = 0:
( 62.5 N ) cos15° − ( 98.1 N ) sin 20° − F = 0
Feq = 26.818 N > Fmax
so no equilibrium,
and block slides up the incline
Fslip = µ x N = ( 0.25 )( 76.008 N ) = 19.00 N
F = 19.00 N
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.
20°
COSMOS: Complete Online Solutions Manual Organization System
Chapter 8, Solution 5.
Tension in cord is equal to P from FBD of pulley.
(
)
W = (10 kg ) 9.81 m/s 2 = 98.1 N
ΣFy = 0:
N − ( 98.1 N ) cos 20° + P sin 25° = 0
(1)
ΣFx = 0:
P cos 25° − ( 98.1 N ) sin 20° + F = 0
(2)
For impending slip down the incline, F = µ s N = 0.3 N and solving
(1) and (2),
PD = 7.56 N
For impending slip up the incline, F = − µ s N = − 0.3 N and solving
(1) and (2),
PU = 59.2 N
so, for equilibrium
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.
7.56 N ≤ P ≤ 59.2 N
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Chapter 8, Solution 6.
FBD Block:
(
)
W = ( 20 kg ) 9.81 m/s 2 = 196.2 N
For θ min motion will impend up the incline, so F is downward and
F = µs N
ΣFy = 0:
ΣFx = 0:
(1) + ( 2 ):
N − ( 220 N ) sin θ − (196.2 N ) cos 35° = 0
F = µ s N = 0.3 ( 220 sin θ + 196.2 cos 35° ) N
(1)
( 220 N ) cosθ
(2)
− F − (196.2 N ) sin 35° = 0
0.3 ( 220 sin θ + 196.2cosθ ) N
= ( 220 cosθ ) N − (196.2sin 35° ) N
or
220cosθ − 66sin θ = 160.751
Solving numerically:
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.
θ = 28.9°
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Chapter 8, Solution 7.
FBD Block:
For Pmin motion will impend down the incline, and the reaction force R
will make the angle
φ s = tan −1 µ s = tan −1 ( 0.35 ) = 19.2900°
with the normal, as shown.
Note, for minimum P, P must be ⊥ to R, i.e. β = φs (angle between
P and x equals angle between R and normal).
β = 19.29°
(b)
then P = (160 N ) cos ( β + 40° )
= (160 N ) cos 59.29° = 81.71 N
(a)
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.
Pmin = 81.7 N
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Chapter 8, Solution 8.
FBD block (impending motion
downward)
φ s = tan −1 µ s = tan −1 ( 0.25 ) = 14.036°
(a) Note: For minimum P,
P⊥R
So
β = α = 90° − ( 30° + 14.036° ) = 45.964°
and
P = ( 30 lb ) sin α = ( 30 lb ) sin ( 45.964° ) = 21.567 lb
P = 21.6 lb
(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
© 2007 The McGraw-Hill Companies.
β = 46.0°
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Chapter 8, Solution 9.
FBD Block:
For impending motion. φ s = tan −1 µ s = tan −1 ( 0.40 )
φ s = 21.801°
Note β1,2 = θ1,2 − φ s
10 lb
15 lb
=
sinφs sinβ1,2
From force triangle:
15 lb
33.854°
sin ( 21.801° ) =
10 lb
146.146°
β1,2 = sin −1
55.655°
So θ1,2 = β1,2 + φ s =
167.947°
So
(a)
equilibrium for
0 ≤ θ ≤ 55.7°
(b)
equilibrium for
167.9° ≤ θ ≤ 180°
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 8, Solution 10.
FBD A with pulley:
Tension in cord is T throughout from pulley FBD’s
ΣFy = 0:
2T − 20 lb = 0,
T = 10 lb
FBD E with pulley:
For θ max , motion impends to right, and
φ s = tan −1 µ s = tan −1 ( 0.35 ) = 19.2900°
From force triangle,
20 lb
10 lb
=
,
sin (θ − φs ) sinφ s
2sin φ s = sin (θ − φ s )
θ = sin −1 ( 2sin19.2900° ) + 19.2900° − 60.64°
θ max = 60.6°
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 8, Solution 11.
FBD top block:
ΣFy = 0:
N1 − 196.2 N = 0
N1 = 196.2 N
(a) With cable in place, impending motion of bottom block requires
impending slip between blocks, so F1 = µ s N1 = 0.4 (196.2 N )
F1 = 78.48 N
FBD bottom block:
ΣFy = 0:
N 2 − 196.2 N − 294.3 N = 0
N 2 = 490.5 N
F2 = µ s N 2 = 0.4 ( 490.5 N ) = 196.2 N
ΣFx = 0:
− P + 78.48 N + 196.2 N = 0
P = 275 N
FBD block:
(b) Without cable AB, top and bottom blocks will move together
ΣFy = 0:
N − 490.5 N = 0,
Impending slip:
ΣFx = 0:
N = 490.5 N
F = µ s N = 0.40 ( 490.5 N ) = 196.2 N
− P + 196.2 N = 0
P = 196.2 N
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 8, Solution 12.
FBD top block:
Note that, since φ s = tan −1 µ s = tan −1 ( 0.40 ) = 21.8° > 15°, no motion
will impend if P = 0, with or without cable AB.
(a) With cable, impending motion of bottom block requires impending
slip between blocks, so F1 = µ s N
ΣFy′ = 0:
N1 − W1 cos15° = 0,
N1 = W1 cos15° = 189.515 N
F1 = µ s N1 = ( 0.40 )W1 cos15° = 0.38637 W1
F1 = 75.806 N
FBD bottom block:
ΣFx′ = 0:
T − F1 − W1 sin15° = 0
T = 75.806 N + 50.780 N = 126.586 N
(
)
W2 = ( 30 kg ) 9.81 m/s 2 = 294.3 N
ΣFy = 0 :
N 2 − (189.515 N ) cos (15° ) − 294.3 N
+ ( 75.806 N ) sin15° = 0
N 2 = 457.74 N
F2 = µ s N 2 = ( 0.40 )( 457.74 N ) = 183.096 N
FBD block:
ΣFx = 0:
− P + (189.515 N ) + ( 75.806 N ) cos15°
+ 126.586 N + 183.096 N = 0
P = 361 N
(b) Without cable, blocks remain together
ΣFy = 0:
N − W1 − W2 = 0
N = 196.2 N + 294.3 N
= 490.5 N
F = µ s N = ( 0.40 )( 490.5 N ) = 196.2 N
ΣFx = 0:
− P + 196.2 N = 0
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.
P = 196.2 N
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Chapter 8, Solution 13.
FBD A:
Note that slip must impend at both surfaces simultaneously.
N1 + T sin θ − 16 lb = 0
ΣFy = 0:
N1 = 16 lb − T sin θ
Impending slip:
F1 = µ s N1 = ( 0.20 )(16 lb − T sin θ )
F1 = 3.2 lb − ( 0.2 ) T sin θ
(1)
F1 − T cosθ = 0
ΣFx = 0:
(2)
FBD B:
ΣFy = 0:
N 2 − N1 − 24 lb = 0,
N 2 = N1 + 24 lb
= 30 lb − T sin θ
Impending slip:
F2 = µ s N 2 = ( 0.20 )( 30 lb − T sin θ )
= 6 lb − 0.2 T sin θ
ΣFx = 0:
10 lb − F1 − F2 = 0
10 lb = µ s ( N1 + N 2 ) = ( 0.2 ) N1 + ( N1 + 24 lb )
10 lb = 0.4 N1 + 4.8 lb,
Then
Then
N1 = 13 lb
F1 = µ s N1 = ( 0.2 )(13 lb ) = 2.6 lb
(1):
T sin θ = 3.0 lb
( 2 ):
T cosθ = 2.6 lb
Dividing tan θ =
3
,
2.6
θ = tan −1
3
= 49.1°
2.6
θ = 49.1°
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 8, Solution 14.
FBD’s:
Note: Slip must impend at both surfaces simultaneously.
A:
ΣFy = 0:
N1 − 20 lb = 0,
Impending slip:
N1 = 20 lb
F1 = µ s N1 = ( 0.25 )( 20 lb ) = 5 lb
ΣFx = 0:
− T + 5 lb = 0,
T = 5 lb
ΣFy′ = 0:
N 2 − ( 20 lb + 40 lb ) cosθ − ( 5 lb ) sin θ = 0
N 2 = ( 60 lb ) cosθ − ( 5 lb ) sin θ
B:
Impending slip:
ΣFx′ = 0:
F2 = µ s N 2 = ( 0.25 )( 60cosθ − 5sin θ ) lb
− F2 − 5 lb − ( 5 lb ) cosθ + ( 20 lb + 40 lb ) sin θ = 0
− 20cosθ + 58.75sin θ − 5 = 0
Solving numerically,
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.
θ = 23.4°
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Chapter 8, Solution 15.
FBD:
For impending tip the floor reaction is at C.
(
)
W = ( 40 kg ) 9.81 m/s 2 = 392.4 N
For impending slip φ = φs = tan −1 µ s = tan −1 ( 0.35 )
φ = 19.2900°
tan φ =
0.8 m
,
EG
EG =
0.4 m
= 1.14286 m
0.35
EF = EG − 0.5 m = 0.64286 m
(a)
α s = tan −1
EF
0.64286 m
= tan −1
= 58.109°
0.4 m
0.4 m
α s = 58.1°
(b)
P
W
=
sin19.29° sin128.820
P = ( 392.4 N )( 0.424 ) = 166.379 N
P = 166.4 N
Once slipping begins, φ will reduce to φk = tan −1 µk .
Then α max will increase.
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.
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Chapter 8, Solution 16.
First assume slip impends without tipping, so F = µ s N
FBD
ΣFy = 0:
N + P sin 40° − W = 0,
N = W − P sin 40°
F = µ s N = 0.35 (W − P sin 40° )
ΣFx = 0:
F − P cos 40° = 0
0.35W = P ( cos 40° + 0.35sin 40° )
Ps = 0.35317 W
(1)
Next assume tip impends without slipping, R acts at C.
ΣM A = 0:
( 0.8 m ) P sin 40° + ( 0.5 m ) P cos 40° − ( 0.4 m )W
=0
Pt = 0.4458W > Ps from (1)
(
∴ Pmax = Ps = 0.35317 ( 40 kg ) 9.81 m/s 2
)
= 138.584 N
(a)
Pmax = 138.6 N
(b) Slip is impending
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 8, Solution 17.
FBD Cylinder:
For maximum M, motion impends at both A and B
FA = µ A N A;
ΣFx = 0:
FB = µ B N B
N A = FB = µ B N B
N A − FB = 0
FA = µ A N A = µ Aµ B N B
ΣFy = 0:
N B + FA − W = 0
1
or
NB =
and
FB = µ B N B =
1 + µ Aµ B
W
µB
W
1 + µ Aµ B
FA = µ Aµ B N B =
µ Aµ B
W
1 + µ Aµ B
ΣM C = 0: M − r ( FA + FB ) = 0
(a) For
µA = 0
N B (1 + µ Aµ B ) = W
and
M = Wr µ B
1 + µA
1 + µ Aµ B
µ B = 0.36
M = 0.360Wr
(b) For
µ A = 0.30
and
µ B = 0.36
M = 0.422Wr
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 8, Solution 18.
FBD’s:
FBD Drum:
(a)
ΣM D = 0:
10
ft F − 50 lb ⋅ ft = 0
12
F = 60 lb
Impending slip: N =
F
µs
=
60 lb
= 150 lb
0.40
FBD arm:
ΣM A = 0:
( 6 in.) C + ( 6 in.) F − (18 in.) N
=0
C = − 60 lb + 3 (150 lb ) = 390 lb
Ccw = 390 lb
(b) Reversing the 50 lb ⋅ ft couple reverses the direction of F, but the magnitudes of F and N are not changed.
Then, using the FBD arm:
ΣM A = 0:
( 6 in.) C − ( 6 in.) F − (18 in.) N
=0
C = 60 lb + 3 (150 lb ) = 510 lb
Cccw = 510 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 8, Solution 19.
For slipping, F = µ k N = 0.30 N
FBD’s:
(a) For cw rotation of drum, the friction force F is as shown.
From FBD arm:
ΣM A = 0:
( 6 in.)( 600 lb ) + ( 6 in.) F − (18 in.) N
600 lb + F − 3
F =
=0
F
=0
0.30
600
lb
9
Moment about D = (10 in.) F = 666.67 lb ⋅ in.
M cw = 55.6 lb ⋅ ft
(b) For ccw rotation of drum, the friction force F is reversed
ΣM A = 0:
( 6 in.)( 600 lb ) − ( 6 in.) F − (18 in.) N
600 lb − F − 3
=0
F
=0
0.30
F =
600
lb
11
10 600
Moment about D = ft
lb = 45.45 lb ⋅ ft
12 11
M ccw = 45.5 lb ⋅ ft
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 8, Solution 20.
FBD:
(a)
ΣM C = 0:
r ( F − T ) = 0,
T = F
Impending slip: F = µ s N or N =
ΣFx = 0:
F
µs
=
T
µs
F + T cos ( 25° + θ ) − W sin 25° = 0
T 1 + cos ( 25° + θ ) = W sin 25°
ΣFy = 0:
(1)
N − W cos 25° + T sin ( 25° + θ ) = 0
1
+ sin ( 25° + θ ) = W cos 25°
T
0.35
Dividing (1) by (2):
(2)
1 + cos ( 25° + θ )
= tan 25°
1
+ sin ( 25° + θ )
0.35
Solving numerically, 25° + θ = 42.53°
θ = 17.53°
(b) From (1)
T (1 + cos 42.53° ) = W sin 25°
T = 0.252W
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 8, Solution 21.
FBD ladder:
Note: slope of ladder =
L = 6.5 m, so AC =
4.5 m
12
13
=
, so AC = ( 4.5 m )
= 4.875
1.875 m
5
12
4.875 m
3
= L,
6.5 m
4
and DC = BD =
AD =
1
L
2
1
L
4
For impending slip: FA = µ s N A ,
FC = µ s NC
12
Also θ = tan −1 − 15° = 52.380°
5
FA − W sin15° + FC cosθ − NC sin θ = 0
ΣFx = 0:
FA = W sin15° − µ s
10
10
W cosθ +
W sin θ
39
39
= ( 0.46192 − 0.15652µ s )W
ΣFy = 0:
N A − W cos15° + FC sin θ + NC cosθ = 0
N A = W cos15° − µ s
10
10
W sin θ − W cosθ
39
39
= ( 0.80941 − 0.20310µ s )W
But FA = µ N A :
0.46192 − 0.15652µ s = 0.80941µ s − 0.20310µ s2
µ s2 − 4.7559µ s + 2.2743
µ s = 0.539, 4.2166
µ s min = 0.539
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 8, Solution 22.
FBD ladder:
Slip impends at both A and B, FA = µ s N A , FB = µ s N B
ΣFx = 0:
ΣFy = 0:
FA − N B = 0,
N B = FA = µ s N A
N A − W + FB = 0,
N A + FB = W
N A + µs N B = W
(
)
N A 1 + µ s2 = W
ΣM O = 0:
( 6 m ) N B +
5
5
m W − m N A = 0
4
2
6µ s N A +
µ s2 +
(
)
5
5
N A 1 + µ s2 − N A = 0
4
2
24
µs − 1 = 0
5
µ s = − 2.4 ± 2.6
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 s min = 0.200
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Chapter 8, Solution 23.
FBD rod:
(a) Geometry:
BE =
L
cosθ
2
EF = L sin θ
So
or
Also,
or
L
DE = cosθ tan β
2
DF =
L cosθ
2 tan φ s
1
L cosθ
L cosθ tan β + sin θ =
2
2 tan φs
tan β + 2 tan θ =
1
1
1
=
=
= 2.5
tan φ s
µ s 0.4
(1)
L sin θ + L sin β = L
sin θ + sin β = 1
Solving Eqs. (1) and (2) numerically
θ1 = 4.62°
(2)
β1 = 66.85°
θ 2 = 48.20° β 2 = 14.75°
θ = 4.62° and θ = 48.2°
Therefore,
(b) Now
and
or
φ s = tan −1 µ s = tan −1 0.4 = 21.801°
T
W
=
sin φs
sin ( 90 + β − φ s )
T =W
sin φs
sin ( 90 + β − φ s )
For
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.
θ = 4.62°
T = 0.526W
θ = 48.2°
T = 0.374W
COSMOS: Complete Online Solutions Manual Organization System
Chapter 8, Solution 24.
FBD:
Assume the weight of the slender rod is negligible compared to P.
First consider impending slip upward at B. The friction forces will be
directed as shown and FB,C = µ s N B,C
ΣM B = 0:
( L sinθ ) P −
a
sin θ
NC = P
ΣFx = 0:
NC = 0
L 2
sin θ
a
NC sin θ + FC cosθ − N B = 0
NC ( sin θ + µ s cosθ ) = N B
so
ΣFy = 0:
NB = P
L 2
sin θ ( sin θ + µ s cosθ )
a
− P + NC cosθ − FC sin θ − FB = 0
P = NC cosθ − µ s NC sin θ − µ s N B
so P = P
L 2
L
sin θ ( cosθ − µ s sin θ ) − µ s P sin 2 θ ( sin θ + µ s cosθ )
a
a
Using θ = 35° and µ s = 0.20, solve for
(1)
L
= 13.63.
a
To consider impending slip downward at B, the friction forces will be
reversed. This can be accomplished by substituting µ s = − 0.20 in
L
= 3.46.
equation (1). Then solve for
a
Thus, equilibrium is maintained for
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.
3.46 ≤
L
≤ 13.63
a
COSMOS: Complete Online Solutions Manual Organization System
Chapter 8, Solution 25.
FBD ABC:
ΣM C = 0:
0.045 m + ( 0.30 m ) sin 30° ( 400 N ) sin 30°
+ 0.030 m + ( 0.30 m ) cos 30° ( 400 N ) cos 30°
12
5
− ( 0.03 m ) FBD − ( 0.045 m ) FBD = 0
13
13
FBD = 3097.64 N
FBD Blade:
ΣFx = 0:
N −
25
( 3097.6 N ) = 0
65
N = 1191.4
F = µ s N = 0.20 (1191.4 N ) = 238.3 N
ΣFy = 0:
P+F−
60
( 3097.6 N ) = 0
65
P = 2859.3 − 238.3 = 2621.0 N
Force by blade
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.
P = 2620 N