Solution manual vector mechanics engineers dynamics 8th beer chapter 08

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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

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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

35°

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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

∴ Equilibrium exists
F = 4.04 N

20°

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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

20°

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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

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

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

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

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

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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

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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

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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

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

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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

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

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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

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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

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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

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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

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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

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Chapter 8, Solution 21.

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 =

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

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Chapter 8, Solution 22.
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

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

θ = 4.62°

T = 0.526W

θ = 48.2°

T = 0.374W

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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

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

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