CHAPTER 6
PROBLEM 6.1
Using the method of joints, determine the force in each member of the truss
shown. State whether each member is in tension or compression.
SOLUTION
Free body: Entire truss:
ΣFy = 0: By = 0
By = 0
ΣM C = 0: − Bx (3.2 m) − (48 kN)(7.2 m) = 0
Bx = −108 kN B x = 108 kN
ΣFx = 0: C − 108 kN + 48 kN = 0
C = 60 kN
C = 60 kN
Free body: Joint B:
FAB FBC 108 kN
=
=
5
4
3
FAB = 180.0 kN T 
FBC = 144.0 kN T 
Free body: Joint C:
FAC FBC 60 kN
=
=
13
12
5
FAC = 156.0 kN C 
FBC = 144 kN (checks)
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PROBLEM 6.2
Using the method of joints, determine the force in each member of the truss
shown. State whether each member is in tension or compression.
SOLUTION
Free body: Entire truss:
Fy  0:  Ay  2.8 kN  0
Ay  2.8 kN
A y  2.8 kN
M A  0: C (1.4 m)  (2.8 kN)(0.75 m)  0
C  1.500 kN C  1.500 kN
Fx  0: Ax  1.500 kN  0
Ax  1.500 kN
A x  1.500 kN
Free body: Joint C:
FBC FAC 1.500 kN


1.25
1
0.75
FAC  2.00 kN T 
FBC  2.50 kN T 
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PROBLEM 6.2 (Continued)
Free body: Joint A:
 7.5 
Fx  0: 
 FAB  1.500 kN  0
 8.5 
FAB  1.700 kN T 
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PROBLEM 6.3
Using the method of joints, determine the force in each member of
the truss shown. State whether each member is in tension or
compression.
SOLUTION
Reactions:
M A  0: C  12.6 kN
Fx  0: A x  0
Fy  0: A y  9.6 kN
Joint B:
FAB FBC
3 kN


12
13
5
FAB  7.2 kN T 
FBC  7.8 kN C 
Joint A:
Fy  0:  9.6 kN 
4
FAC  0
5
FAC  12 kN F AC  12 kN C 
Fx  0: 7.2 kN  (12 kN)
3
 0 (checks)
5
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PROBLEM 6.4
Using the method of joints, determine the force in each
member of the truss shown. State whether each member is in
tension or compression.
SOLUTION
Reactions:
M D  0: Fy (3 m)  (24 kN  8 kN)(1.5 m)  (7 kN)(3 m)  0
Fy  23.0 kN
Fx  0: Fx  0
Fy  0: D  (7  24  8  7)  23  0
D  23.0 kN
Joint A:
Fx  0: FAB  0
FAB  0 
Fy  0 : 7  FAD  0
FAD  7 kN
Joint D:
Fy  0:  7  23.0 
FAD  7.00 kN C 
8
FBD  0
17
FBD  34.0 kN
FBD  34.0 kN C 
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PROBLEM 6.4 (Continued)
Fx  0:
15
(34.0)  FDE  0
17
FDE  30.0 kN
FDE  30.0 kN T 
Joint E:
Fy  0 : FBE  8  0
FBE  8.00 kN
FBE  8.00 kN T 
Truss and loading symmetrical about cL. 
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PROBLEM 6.5
Using the method of joints, determine the force in each
member of the truss shown. State whether each member is in
tension or compression.
SOLUTION
Reactions:
M B  0: Ax (1 m)  (10 kN)(2 m)  (10 kN)(4 m)  0
A x  60.0 kN
Fx  0: Ax  B  0
B  60 kN
Fy  0: Ay  10 kN  10 kN  0
A y  20.0 kN
Joint D:
F
F
10 kN
 DC  DA
1
4
17
FDA  41.2 kN T 
 F DC  40.0 kN C 
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PROBLEM 6.5 (Continued)
Fy  0:
Joint C:
1
5
FCA  10 kN  0
FCA  22.4 kN
Fx  0:
2
5
FCA  22.4 kN T 
(22.4 kN)  40 kN  FCB  0
FCB  60.0 kN
FCB  60.0 kN C 
Joint B:
Fy  0 : FBA  0
FBA  0 
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PROBLEM 6.6
Using the method of joints, determine the force in each member
of the truss shown. State whether each member is in tension or
compression.
SOLUTION
DE 
Reactions:
 6 m 2   3.2 m 2
 6.80 m
4
M B  0:   FAC (4.5 m)  (24 kN)(12 m)  0
5
FAC  80.0 kN T 
4
Fx  0: Bx     80 kN   0
5
B x  64 kN
3
Fy  0: B y     80 kN   24 kN  0
5
B y  24 kN
Joint E:
FCE FDE 24 kN


6
6.8
3.2
FCE  45.0 kN T 
FDE  51.0 kN C 
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PROBLEM 6.6 (Continued)
Joint D:
 6 
 6 
Fx  0: 
FBD  

  51.0 kN   0
 6.8 
 6.8 
FBD  51.0 kN C 
FCD  48.0 kN T 
Joint C:
4
Fx  0:  FBC     80 kN    45.0 kN   0
5
FBC  19.00 kN C 
3
Fy  0 :    80 kN   48.0 kN  0 (checks) 
5
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2.5 m
2.5 m
5.5 m
PROBLEM 6.7
3 kN
A
Using the method of joints, determine the force in each member
of the truss shown. State whether each member is in tension or
compression.
B
C
6m
D
E
SOLUTION
AD = 2.52 + 62 = 6.5 m
BCD = 62 + 82 = 10 m
Reactions:
ΣFx = 0: Dx = 0
ΣM E = 0: D y (10.5) − (3 kN)(2.5) = 0
ΣFy = 0: 0.714 kN − 3 kN + E = 0
Joint D:
D y = 714 N
E = 2.286 kN
ΣFx = 0:
5
4
FAD + FDC = 0
13
5
(1)
ΣFy = 0:
12
3
FAD + FDC + 714 N = 0
13
5
(2)
Solving (1) and (2), simultaneously:
FAD = −1125.09 N
FAD = 1.125 kN C
FDC = +540.9 N
FDC = 541 N
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T
PROBLEM 6.7 (Continued)
Joint E:
ΣFx = 0:
5
4
FBE + FCE = 0
13
5
(3)
ΣFy = 0:
12
3
FBE + FCE + 2.286 kN = 0
13
5
(4)
Solving (3) and (4), simultaneously:
FBE = −3.602 kN
FBE = 3.602 kN C
FCE = +1.732 kN
FCE = 1.732 kN T
Joint C:
Force polygon is a parallelogram ((see Fig.
Fi 66. 11 p. 290)
FAC = 1.732 kN T
FBC = 541 N
Joint A:
ΣFx = 0:
4
5
(1.125 kN) + (1.732 kN) + FAB = 0
5
13
FAB = −1.818 kN
ΣFy = 0:
T
FAB = 1.818 kN C
12
3
(1.125 kN) − (1.732 kN) = 0
13
5
0 = 0 (Checks)
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748
PROBLEM 6.8
Using the method of joints, determine the force in each member of the truss
shown. State whether each member is in tension or compression.
SOLUTION
Free body: Entire truss:
Fx  0: C x  2(5 kN)  0
Cx  10 kN

C x  10 kN
M C  0: D (2 m)  (5 kN)(8 m)  (5 kN)(4 m)  0

D  30 kN
D  30 kN

Fy  0: C y  30 kN  0 C y  30 kN C y  30 kN
Free body: Joint A:
FAB FAD 5 kN


4
1
17
FAB  20.0 kN T 


FAD  20.6 kN C 
Free body: Joint B:
Fx  0: 5 kN 
1
5
FBD  0
FBD  5 5 kN
Fy  0: 20 kN  FBC 
2
5
FBD  11.18 kN C 
(5 5 kN)  0
FBC  30 kN
FBC  30.0 kN T 
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PROBLEM 6.8 (Continued)
Free body: Joint C:
Fx  0: FCD  10 kN  0
FCD  10 kN
FCD  10.00 kN T 
Fy  0: 30 kN  30 kN  0 (checks)
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.9
Determine the force in each member of the truss shown. State whether
each member is in tension or compression.
SOLUTION
Reactions for Entire truss:
By inspection: A  4 kN  E  4 kN 
Free body: Joint A:
FAB 
4 kN
sin 30
FAF  (4 kN)(cot 30 )
FAB  8.00 kN C 


FAF  6.93 kN T 
Free body: Joint F:
Fx  0: FFG  FAF
FFG  6.93 kN T 
Fy  0: FFB  4.00 kN T
FFB  4.00 kN T 

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PROBLEM 6.9 (Continued)
Free body: Joint B:
Fx  0: 8cos30  FBC cos30  FBG cos30  0 (1)
Fy  0 : 8sin 30  4  FBC sin 30  FBG sin 30  0 (2)
Solving (1) and (2) simultaneously yields:
FBC  4.00 kN C 


FBG  4.00 kN C 

Free body: Joint C:
FCD  FCB  4.00 kN C 
By symmetry,
Fy  0 : 4 sin 30  4 sin 30  FCG  0
FCG  4.00 kN T 
By symmetry about the truss centerline, we note:
FGD  4.00 kN C , FGH  6.93 kN T , FDH  4.00 kN T , FDE  8.00 kN C , FEH  6.93 kN T 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.10
Determine the force in each member of the truss shown. State whether each
member is in tension or compression.
SOLUTION
Free body: Entire truss:
Fx  0: Fx  0
M F  0: H  5 kN
Fy  0: Fy  0
Free body: Joint C:
FCB  5.00 kN, FCE  5 2  7.07 kN
FCB  5.00 kN T 



FCE  7.07 kN C 
By symmetry about the horizontal centerline: FHG  5.00 kN T , FHE  7.07 kN C 
Joint E: Members lie in two intersecting straight lines, thus we have:
FEG  FEC  7.07 kN C ; FEB  FEH  7.07 kN C 
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PROBLEM 6.10 (Continued)
Free body: Joint B:
Fy  0:
1
2
 7.07  
1
2
FBD  0
FBD  7.07 kN T 
Fx  0: 5  FBA 
1
2
(7.07) 
1
2
 7.07   0
FBA  5.00 kN C 
By symmetry about the horizontal centerline:
Joint D:
FDG  7.07 kN T , FGF  5.00 kN C 
FDF  FDB  FDA  FDG  7.07 kN T 
Free body: Joint A:
FAF  5.00 kN C 
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2 kN
2 kN
PROBLEM 6. 11
2 kN
D
1 kN
1 kN
A
C
4m
G
E
4m
4m
Determine the force in each member of the Gambrel roof
truss shown. State whether each member is in tension or
compression.
1.167 m
F
B
3m
H
4m
SOLUTION
Free body: Truss
ΣFx = 0: H x = 0
Because of the symmetry of the truss and loading
A = Hy =
1
2
Total load
A = Hy = 4 kN
Free body: Joint A:
FAB FAC 3 kN
=
=
5
4
3
FAB = 5 kN
C
FAC = 4 kN
T
FCE = 4 kN
T
Free body: Joint C:
BC is a zero-force member
FBC = 0
Free body: Joint B:
ΣFx = 0:
24
4
4
FBD + FBE + (5 kN) = 0
25
5
5
24 FBD + 20 FBE = −100 kN
or
ΣFy = 0:
(1)
7
3
3
FBD − FBE + (5) − 2 = 0
25
5
5
7FBD − 15FBE = −25 kN
or
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755
(2)
PROBLEM 6.11 (Continued)
Multiply (1) by (3), (2) by 4, and add:
100 FBD = −400 kN
FBE = −0.2 kN
FBD = 4 kN
C
FBE = 200 N
C
FDF = 4 kN
C
Free body: Joint D:
ΣFx = 0:
24
24
(4 kN) +
FDF = 0
25
25
FDF = −4 kN
ΣFx = 0:
7
7
(4 kN) −
(−4 kN) − 2 kN − FDE = 0
25
25
FDE = 0.24 kN
FDE = 240 N
T
FEF = FBE
FEF = 200 N
C
FEG = FCE
FEG = 4 kN
T
Because of the symmetry of the truss and loading, we deduce that
FFG = FBC
FFG = 0
FFH = FAB
FFH = 5 kN
C
FGH = FAC
FGH = 4 kN
T
Note: Compare results with those of Problem 6.9.
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756
PROBLEM 6.12
2 kN
2 kN
2 kN
D
1 kN
1 kN 3 m
F
B
H
A
E
C
4m
4m
Determine the force in each member of the Howe roof
truss shown. State whether each member is in tension or
compression.
3m
G
4m
4m
SOLUTION
Free body: Truss
ΣFx = 0: H x = 0
Because of the symmetry of the truss and loading:
A = Hy =
1
2
Total load
A = H y = 4 kN
Free body: Joint A:
FAB FAC
3 kN
=
=
5
4
3
FAB = 5 kN C
FAC = 4 kN T
Free body: Joint C:
BC is a zero-force member
FBC = 0
FCE = 4 kN T
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PROBLEM 6.12 (Continued)
Free body: Joint B:
ΣFx = 0:
4
4
4
FBD + FBE + (5 kN) = 0
5
5
5
FBD + FBE = −5 kN
or
ΣFy = 0:
(1)
3
3
3
FBD − FBE + (5 kN) − 2 kN = 0
5
5
5
FBD − FBE = −1.667 kN
or
Add Eqs. (1) and (2):
Subtract (2) from (1):
2 FBD = −6.667 kN
2 FBE = −3.333 kN
(2)
FBD = 3.333 kN C
FBE = 1.667 kN C
Free Body: Joint D:
ΣFx = 0:
4
4
(3.333 kN) + FDF = 0
5
5
FDF = −3.333 kN
ΣFy = 0:
FDF = 3.333 kN C
3
3
(3.333 kN) − (−3.333 kN) − 2 kN − FDE = 0
5
5
FDE = +2 kN
T
FDE = 2 kN
Because of the symmetry of the truss and loading, we deduce that
FEF = FBE
FEF = 1.667 kN C
FEG = FCE
FEG = 4 kN
FFG = FBC
T
FFG = 0
FFH = FAB
FFH = 5 kN
C
FGH = FAC
FGH = 4 kN
T
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754
PROBLEM 6.13
Using the method of joints, determine the force in each
member of the roof truss shown. State whether each
member is in tension or compression.
SOLUTION
Free body: Truss:
Fx  0: A x  0
From symmetry of loading:
Ay  E 
1
2
total load
A y  E  3.6 kN
We note that DF is a zero-force member and that EF is aligned with the load. Thus,
FDF  0 
FEF  1.2 kN C 
Free body: Joint A:
FAB FAC 2.4 kN


13
12
5
FAB  6.24 kN C 
FAC  2.76 kN T 
Free body: Joint B:
Fx  0:
3
12
12
FBC  FBD  (6.24 kN)  0
3.905
13
13
 Fy  0:

2.5
5
5
FBC  FBD  (6.24 kN)  2.4 kN  0
3.905
13
13
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
(1)
(2)
PROBLEM 6.13 (Continued)
Multiply Eq. (1) by 2.5, Eq. (2) by 3, and add:
45
45
FBD  (6.24 kN)  7.2 kN  0, FBD  4.16 kN, FBD  4.16 kN C 
13
13
Multiply Eq. (1) by 5, Eq. (2) by –12, and add:
45
FBC  28.8 kN  0, FBC  2.50 kN,
3.905
FBC  2.50 kN C 
Free body: Joint C:
Fy  0:
5
2.5
FCD 
(2.50 kN)  0
5.831
3.905
FCD  1.867 kN T 
Fx  0: FCE  5.76 kN 
3
3
(2.50 kN) 
(1.867 kN)  0
3.905
5.831
FCE  2.88 kN T
Free body: Joint E:
Fy  0:
5
FDE  3.6 kN  1.2 kN  0
7.81
FDE  3.75 kN FDE  3.75 kN C 
Fx  0:  FCE 
6
(3.75 kN)  0
7.81
FCE  2.88 kN FCE  2.88 kN T
(Checks)
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.14
Determine the force in each member of the Fink roof truss
shown. State whether each member is in tension or
compression.
SOLUTION
Free body: Truss:
Fx  0: A x  0
Because of the symmetry of the truss and loading,
Ay  G 
1
total load
2
A y  G  6.00 kN
Free body: Joint A:
F
FAB
4.50 kN
 AC 
2.462 2.25
1
FAB  11.08 kN C 
FAC  10.125 kN
FAC  10.13 kN T 
Free body: Joint B:
Fx  0:
3
2.25
2.25
(11.08 kN)  0
FBC 
FBD 
5
2.462
2.462
(1)
F
4
11.08 kN
FBC  BD 
 3 kN  0
5
2.462
2.462
(2)
Fy  0: 
Multiply Eq. (2) by –2.25 and add to Eq. (1):
12
FBC  6.75 kN  0
5
FBC  2.8125
FBC  2.81 kN C 
Multiply Eq. (1) by 4, Eq. (2) by 3, and add:
12
12
FBD 
(11.08 kN)  9 kN  0
2.462
2.462
FBD  9.2335 kN
FBD  9.23 kN C 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.14 (Continued)
Free body: Joint C:
Fy  0:
4
4
FCD  (2.8125 kN)  0
5
5
FCD  2.8125 kN,
FCD  2.81 kN T 
3
3
Fx  0: FCE  10.125 kN  (2.8125 kN)  (2.8125 kN)  0
5
5
FCE  6.7500 kN FCE  6.75 kN T 
Because of the symmetry of the truss and loading, we deduce that
FDE  FCD
FCD  2.81 kN T 
FDF  FBD
FDF  9.23 kN C 
FEF  FBC
FEF  2.81 kN C 
FEG  FAC
FEG  10.13 kN T 
FFG  FAB
FFG  11.08 kN C 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
3m
6m
D
B
PROBLEM 6.15
3m
6m
F
4m
G
A
C
6m
Determine the force in each member of the Warren bridge
truss shown. State whether each member is in tension or
compression.
E
6m
6 kN
6m
6 kN
SOLUTION
ΣFx = 0: Ax = 0
Free body: Truss
Due to symmetry of truss and loading
Ay = G =
Free body: Joint A:
1
Total load = 6 kN
2
FAB FAC
6 kN
=
=
5
3
4
Free body: Joint B:
FBC FBD 7.5 kN
=
=
5
6
5
FAB = 7.50 kN
C
FAC = 4.50 kN
T
FBC = 7.50 kN
T
FBD = 9.00 kN
C
Free body: Joint C:
ΣFy = 0:
4
4
(7.5) + FCD − 6 = 0
5
5
FCD = 0
3
ΣFx = 0: FCE − 4.5 − (7.5) = 0
5
FCE = 9.00 kN
FCE = +9 kN
Truss and loading symmetrical about cL
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
761
T
3m
6m
D
B
PROBLEM 6. 16
3m
6m
F
Solve Problem 6.15 assuming that the load applied at E has
been removed.
4m
G
A
C
6m
PROBLEM 6.15 Determine the force in each member of the
Warren bridge truss shown. State whether each member is in
tension or compression.
E
6m
6 kN
6m
6 kN
SOLUTION
Free body: Truss
ΣFx = 0: Ax = 0
ΣM G = 0: 6(12) − Ay (18) = 0 A y = 4 kN
ΣFy = 0: 4 − 6 + G = 0 G = 2 kN
Free body: Joint A:
Free body Joint B:
FAB FAC 4 kN
=
=
5
3
4
FBC FBD 5 kN
=
=
5
6
5
FAB = 5.00 kN
C
FAC = 3.00 kN
T
FBC = 5.00 kN
T
FBD = 6.00 kN
C
FCD = 2.50 kN
T
FCE = 4.50 kN
T
Free body Joint C:
ΣFy = 0:
4
4
(5) + FCD − 6 = 0
5
5
3
3
ΣFx = 0: FCE + (2.5) − (5) − 3 = 0
5
5
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
762
PROBLEM 6. 16 (Continued)
Free body: Joint D:
4
4
ΣFy = 0: − (2.5) − FDE = 0
5
5
FDE = −2.5 kN
FDE = 2.50 kN
C
FDF = −3 kN
FDF = 3.00 kN
C
FEF FFG 3 kN
=
=
5
5
6
FEF = 2.50 kN
T
FFG = 2.50 kN
C
FEG = 1.500 kN
T
3
3
ΣFx = 0: FDF + 6 − (2.5) − (2.5) = 0
5
5
Free body: Joint F:
Free body: Joint G:
FEG 2 kN
=
3
4
Also:
FFG 2 kN
=
5
4
FFG = 2.50 kN
C
(Checks)
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.17
Determine the force in each member of the Pratt
roof truss shown. State whether each member is
in tension or compression.
SOLUTION
Free body: Truss:
Fx  0: Ax  0
Due to symmetry of truss and load,
Ay  H 
1
total load  21 kN
2
Free body: Joint A:
FAB FAC 15.3 kN


37
35
12
FAB  47.175 kN
FAC  44.625 kN
FAB  47.2 kN C 
FAC  44.6 kN T 
Free body: Joint B:
From force polygon:
FBD  47.175 kN, FBC  10.5 kN
FBC  10.50 kN C 
FBD  47.2 kN C 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.17 (Continued)
Free body: Joint C:
Fy  0:
3
FCD  10.5  0
5
Fx  0: FCE 
4
(17.50)  44.625  0
5
FCE  30.625 kN
Free body: Joint E:
FCD  17.50 kN T 
DE is a zero-force member.
FCE  30.6 kN T 
FDE  0 
Truss and loading symmetrical about cL .
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.18
The truss shown is one of several supporting an advertising panel.
Determine the force in each member of the truss for a wind load
equivalent to the two forces shown. State whether each member is in
tension or compression.
SOLUTION
Free body: Entire truss:
M F  0: (800 N)(7.5 m)  (800 N)(3.75 m)  A(2 m)  0
A  2250 N
A  2250 N
Fy  0: 2250 N  Fy  0
Fy  2250 N Fy  2250 N
Fx  0: 800 N  800 N  Fx  0
Fx  1600 N Fx  1600 N
Joint D:
800 N FDE FBD


8
15
17
FBD  1700 N C 
FDE  1500 N T 
Joint A:
2250 N FAB FAC


15
17
8
FAB  2250 N C 
FAC  1200 N T 
Joint F:
Fx  0: 1600 N  FCF  0
FCF  1600 N
FCF  1600 N T 
Fy  0: FEF  2250 N  0
FEF  2250 N
FEF  2250 N T 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.18 (Continued)
Joint C:
Fx  0:
8
FCE  1200 N  1600 N  0
17
FCE  850 N
Fy  0: FBC 
15
FCE  0
17
FBC  
15
15
FCE   (850 N)
17
17
FBC  750 N
Joint E:
FCE  850 N C 
Fx  0:  FBE  800 N 
FBC  750 N T 
8
(850 N)  0
17
FBE  400 N
Fy  0: 1500 N  2250 N 
FBE  400 N C 
15
(850 N)  0
17
0  0 (checks)
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.19
Determine the force in each member of the Pratt bridge
truss shown. State whether each member is in tension or
compression.
SOLUTION
Free body: Truss:
Fz  0: A x  0
M A  0 : H (12 m)  (4 kN)(3 m)
 (4 kN)(6 m)  (4 kN)(9 m)  0
H  6 kN
Fy 0: Ay  6 kN  12 kN  0
A y  6 kN
Free body: Joint A:
FAB FAC
6 kN


5
3
4
FAB  7.50 kN
C 
FAC  4.50 kN
T 
Fx  0:
FCE  4.50 kN
T 
Fy  0:
FBC  4.00 kN
T 
FBE  2.50 kN
T 
FBD  6.00 kN
C 
Free body: Joint C:
Free body: Joint B:
Fy  0: 
Fx  0:
4
4
FBE  (7.50 kN)  4.00 kN  0 
5
5
8
3
(7.50 kN)  (2.50 kN)  FBD  0
5
5
FBD  6.00 kN
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.19 (Continued)
Free body: Joint D:
We note that DE is a zero-force member:
FDE  0 
FDF  6.00 kN C 
Also,
From symmetry:
FFE  FBE
FEF  2.50 kN T 
FEG  FCE
FEG  4.50 kN T 
FFG  FBC
FFG  4.00 kN T 
FFH  FAB
FFH  7.50 kN C 
FGH  FAC
FGH  4.50 kN T 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.20
Solve Problem 6.19 assuming that the load applied at G has
been removed.
PROBLEM 6.19 Determine the force in each member of
the Pratt bridge truss shown. State whether each member is
in tension or compression.
SOLUTION
Free body: Truss:
Fx  0: A x  0
M A  0: H (12 m)  (4 kN)(3 m)  (4 kN)(6 m)  0
H  3.00 kN
Fy  0: A y  5.00 kN
We note that DE and FG are zero-force members.
FDE  0,
Therefore,
FFG  0 
Also,
FBD  FDF
(1)
and
FEG  FGH
(2)
Free body: Joint A:
FAB FAC
5 kN


5
3
4
FAB  6.25 kN C 
FAC  3.75 kN T 

Free body: Joint C:
Fx  0:
FCE  3.75 kN T 
Fy  0:
FBC  4.00 kN T 
Free body: Joint B:
Fx  0:
4
4
(6.25 kN)  4.00 kN  FBE  0
5
5
FBE  1.250 kN T 
3
3
Fx  0: FBD  (6.25 kN)  (1.250 kN)  0 
5
5
FBD  4.50 kN
FBD  4.50 kN C 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.20 (Continued)
Free body: Joint F:
We recall that FFG  0, and from Eq. (1) that
FDF  FBD
FDF  4.50 kN C 
FEF FFH
4.50 kN


5
5
6
FEF  3.75 kN T 
FFH  3.75 kN C 
Free body: Joint H:
FGH
3.00 kN

3
4
FGH  2.25 kN T 
Also,
FFH
3.00 kN

5
4
FFH  3.75 kN C (checks)
From Eq. (2):
FEG  FGH
FEG  2.25 kN T 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.21
Determine the force in each of the members located to the
left of FG for the scissors roof truss shown. State whether
each member is in tension or compression.
SOLUTION
Free Body: Truss:
Fx  0: A x  0
M L  0: (1 kN)(12 m)  (2 kN)(10 m)  (2 kN)(8 m)  (1 kN)(6 m)  Ay (12 m)  0
A y  4.50 kN
FBC  0 
We note that BC is a zero-force member:
Also,
FCE  FAC
Free body: Joint A:
Fx  0:
1
Fy  0:
1
(1)
2
2
FAB 
2
FAB 
1
5
5
FAC  0
(2)
FAC  3.50 kN  0
(3)
Multiply Eq. (3) by –2 and add Eq. (2):

1
2
FAB  7 kN  0
FAB  9.90 kN C 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.21 (Continued)
Subtract Eq. (3) from Eq. (2):
1
5
FAC  3.50 kN  0 FAC  7.826 kN
FCE  FAC  7.826 kN
From Eq. (1):
FAC  7.83 kN T 
FCE  7.83 kN T 
Free body: Joint B:
Fy  0:
1
2
FBD 
1
2
(9.90 kN)  2 kN  0
FBD  7.071 kN
Fx  0: FBE 
1
2
(9.90  7.071) kN  0
FBE  2.000 kN
Free body: Joint E:
Fx  0:
2
5
FBD  7.07 kN C 
FBE  2.00 kN C 
( FEG  7.826 kN)  2.00 kN  0
FEG  5.590 kN FEG  5.59 kN T 
Fy  0: FDE 
1
5
(7.826  5.590) kN  0
FDE  1.000 kN
FDE  1.000 kN T 
Free body: Joint D:
Fx  0:
2
5
( FDF  FDG ) 
1
2
(7.071 kN)
FDF  FDG  5.590 kN
or
Fy  0:
1
5
( FDF  FDG ) 
1
2
(4)
(7.071 kN)  2 kN  1 kN  0
FDE  FDG  4.472
or
(5)
Add Eqs. (4) and (5):
2 FDF  10.062 kN
FDF  5.03 kN C 
Subtract Eq. (5) from Eq. (4):
2 FDG  1.1180 kN
FDG  0.559 kN C 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
5.76 m
400 N
5.76 m
800 N
A
5.76 m
800 N
B
6.72 m
5.76 m
800 N
D
PROBLEM 6.22
Determine the force in member DE and in each of the
members located to the left of DE for the inverted Howe
roof truss shown. State whether each member is in tension or
compression.
400 N
F
C
H
E
10.54 m
G
12.5 m
SOLUTION
Free body: Truss:
ΣFx = 0: A x = 0
ΣMH = 0:
(400 N)(4d) + (800 N)(3d) + (800 N)(2d) + (800 N)d − Ay(4d) = 0
Ay = 1600 N
Angles:
6.72
α = 32.52°
10.54
6.72
β = 16.26°
tan β =
23.04
tan α =
Free body: Joint A:
FAC
FAB
1200 N
=
=
sin 57.48° sin106.26° sin16.26°
FAB = 3613.8 N C
FAC = 4114.3 N T
FAB = 3.61 kN C,
FAC = 4.11 kN
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
764
T
PROBLEM 6.22 (Continued)
Free body: Joint B:
ΣFy = 0: − FBC − (800 N) cos16.26° = 0
FBC = −768.0 N
FBC = 768 N C
ΣFx = 0: FBD + 3613.8 N + (800 N) sin16.26° = 0
FBD = −3837.8 N
FBD = 3.84 kN C
Free body: Joint C:
ΣFy = 0: − FCE sin 32.52° + (4114.3 N) sin 32.52° − (768 N) cos16.26° = 0
FCE = 2742.9 N
FCE = 2.74 kN T
ΣFx = 0: FCD − (4114.3 N) cos 32.52° + (2742.9 N) cos 32.52° − (768 N) sin16.26° = 0
FCD = 1371.4 N
FCD = 1.37 kN T
Free body: Joint E:
FDE
2742.9 N
=
sin 32.52° sin73.74°
FDE = 1.54 kN C
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
5.76 m
400 N
5.76 m
800 N
A
5.76 m
800 N
B
800 N
D
6.72 m
5.76 m
PROBLEM 6.23
Determine the force in each of the members located to the
right of DE for the inverted Howe roof truss shown. State
whether each member is in tension or compression.
400 N
F
C
H
E
10.54 m
G
12.5 m
SOLUTION
Free body: Truss
ΣM A = 0: H (4d ) − (800 N)d − (800 N)(2d ) − (800 N)(3d ) − (400 N)(4d ) = 0
H = 1600 N
Angles:
6.72
α = 32.52°
10.54
6.72
tan β =
β = 16.26°
23.04
tan α =
Free body: Joint H:
FGH = (1200 N)cot16.26°
FGH = 4114.3 N T
FFH =
1200 N
= 4285.8 N
sin16.26°
FGH = 4.11 kN T
FFH = 4.29 kN C
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
766
PROBLEM 6.23 (Continued)
Free body Joint F:
ΣFy = 0: − FFG − (800 N) cos16.26° = 0
FFG = −768.0 N
FFG = 768 N C
ΣFx = 0: − FDF − 4285.8 N + (800 N) sin 16.26° = 0
FDF = −4061.8 N
FDF = 4.06 kN C
Free body: Joint G:
ΣFy = 0: FDG sin 32.52° − (768.0 N) cos16.26° = 0
FDG = +1371.4 N
FDG = 1.37 kN T
ΣFx = 0: − FEG + 4114.3 N − (768.0 N)sin16.26° − (1371.4 N) cos 32.52° = 0
FEG = +2742.9 N
FEG = 2.74 kN T
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
767
PROBLEM 6.24
The portion of truss shown represents the upper
part of a power transmission line tower. For the
given loading, determine the force in each of the
members located above HJ. State whether each
member is in tension or compression.
SOLUTION
Free body: Joint A:
F
FAB
1.2 kN
 AC 
2.29 2.29
1.2
FAB  2.29 kN T 
FAC  2.29 kN C 
Free body: Joint F:
FDF
F
1.2 kN
 EF 
2.29 2.29
2.1
FDF  2.29 kN T 
FEF  2.29 kN C 
Free body: Joint D:
FBD FDE 2.29 kN


2.21 0.6
2.29
FBD  2.21 kN T 
FDE  0.600 kN C 
Free body: Joint B:
Fx  0:
4
2.21
FBE  2.21 kN 
(2.29 kN)  0
5
2.29
FBE  0 
3
0.6
Fy  0:  FBC  (0) 
(2.29 kN)  0
5
2.29
FBC   0.600 kN
FBC  0.600 kN C 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.24 (Continued)
Free body: Joint C:
Fx  0: FCE 
2.21
(2.29 kN)  0
2.29
FCE   2.21 kN
Fy  0:  FCH  0.600 kN 
FCH  1.200 kN
FCE  2.21 kN C 
0.6
(2.29 kN)  0
2.29
FCH  1.200 kN C 
Free body: Joint E:
Fx  0: 2.21 kN 
2.21
4
(2.29 kN)  FEH  0
2.29
5
FEH  0 
Fy  0:  FEJ  0.600 kN 
FEJ   1.200 kN
0.6
(2.29 kN)  0  0
2.29
FEJ  1.200 kN C 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.25
For the tower and loading of Problem 6.24 and
knowing that FCH  FEJ  1.2 kN C and FEH  0,
determine the force in member HJ and in each of
the members located between HJ and NO. State
whether each member is in tension or
compression.
PROBLEM 6.24 The portion of truss shown
represents the upper part of a power transmission
line tower. For the given loading, determine the
force in each of the members located above HJ.
State whether each member is in tension or
compression.
SOLUTION
Free body: Joint G:
FGH
F
1.2 kN
 GI 
3.03 3.03
1.2
FGH  3.03 kN T 
FGI  3.03 kN C 
Free body: Joint L:
FJL
F
1.2 kN
 KL 
3.03 3.03
1.2
FJL  3.03 kN T 
FKL  3.03 kN C 
Free body: Joint J:
Fx  0:  FHJ 
2.97
(3.03 kN)  0
3.03
FHJ  2.97 kN T 
Fy  0:  FJK  1.2 kN 
FJK  1.800 kN
0.6
(3.03 kN)  0
3.03
FJK  1.800 kN C 
Free body: Joint H:
Fx  0:
4
2.97
FHK  2.97 kN 
(3.03 kN)  0
5
3.03
FHK  0 
0.6
3
(3.03) kN  (0)  0
3.03
5
 1.800 kN
FHI  1.800 kN C 
Fy  0:  FHI  1.2 kN 
FHI
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.25 (Continued)
Free body: Joint I:
Fx  0: FIK 
2.97
(3.03 kN)  0
3.03
FIK  2.97 kN
Fy  0:  FIN  1.800 kN 
FIN  2.40 kN
FIK  2.97 kN C 
0.6
(3.03 kN)  0
3.03
FIN  2.40 kN C 
Free body: Joint K:
Fx  0: 
4
2.97
(3.03 kN)  0
FKN  2.97 kN 
5
3.03
FKN  0 
Fy  0:  FKO 
0.6
3
(3.03 kN)  1.800 kN  (0)  0
3.03
5
FKO  2.40 kN
FKO  2.40 kN C 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.26
Solve Problem 6.24 assuming that the cables
hanging from the right side of the tower have
fallen to the ground.
PROBLEM 6.24 The portion of truss shown
represents the upper part of a power transmission
line tower. For the given loading, determine the
force in each of the members located above HJ.
State whether each member is in tension or
compression.
SOLUTION
Zero-Force Members:
Considering joint F, we note that DF and EF are zero-force
members:
FDF  FEF  0 
Considering next joint D, we note that BD and DE are zero-force
members:
FBD  FDE  0 
Free body: Joint A:
F
FAB
1.2 kN
 AC 
2.29 2.29
1.2
FAB  2.29 kN T 
FAC  2.29 kN C 
Free body: Joint B:
Fx  0:
4
2.21
FBE 
(2.29 kN)  0
5
2.29
FBE  2.7625 kN
Fy  0:  FBC 
FBE  2.76 kN T 
0.6
3
(2.29 kN)  (2.7625 kN)  0
2.29
5
FBC  2.2575 kN
FBC  2.26 kN C 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.26 (Continued)
Free body: Joint C:
Fx  0: FCE 
2.21
(2.29 kN)  0
2.29
FCE  2.21 kN C 
Fy  0:  FCH  2.2575 kN 
FCH  2.8575 kN
0.6
(2.29 kN)  0
2.29
FCH  2.86 kN C 
Free body: Joint E:
Fx  0: 
4
4
FEH  (2.7625 kN)  2.21 kN  0
5
5
FEH  0 
3
3
Fy  0:  FEJ  (2.7625 kN)  (0)  0
5
5
FEJ  1.6575 kN
FEJ  1.658 kN T 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.27
Determine the force in each member of the truss shown. State whether
each member is in tension or compression.
SOLUTION
FCH = 0 t
First note that, by inspection of joint H:
First,฀note฀that฀by฀inspection฀of฀joint฀H:
and
FDH = FGH
and
FBC = FCD
and
FFG = FGH
and
FAB = FBC
FCG = 0 t
then, by inspection of joint C:
FBG = 0 t
then, by inspection of joint G:
Joint FBDs:
Joint D:
FCD FDH 10 kN
=
=
12
13
5
Joint A:
FCD = 24.0 kN T t
so
FDH = 26.0 kN C t
FAB = FBC = 24.0 kN T t
and, from above:
Joint F:
FBF = 0 t
then, by inspection of joint B:
FGH = FFG = 26.0 kN C t
FAF
F
24 kN
= AE =
5
4
41
FAF = 30.0 kN C t
FAE = 38.4
25.6 kN T t
ΣFx = 0: FEF −
12
(26 kN ) = 0
13
FEF = 24.0 kN C t
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.28
Determine the force in each member of the truss shown.
State whether each member is in tension or compression.
SOLUTION
Free body: Truss
Fx  0: H x  0
M H  0: 48(16)  G (4)  0 G  192 kN
Fy  0: 192  48  H y  0
H y  144 kN H y  144 kN
Zero-Force Members:
Examining successively joints C, B, E, D, and F, we note that the following are zero-force members: BC,
BE, DE, DG, FG
Thus,
FBC  FBE  FDE  FDG  FFG  0
Also note:
FAB  FBD  FDF  FFH
(1)
FAC  FCE  FEG
(2)
Free body: Joint A:
From Eq. (1):
From Eq. (2):
FAB FAC 48 kN


8
3
73
FAB  128 kN
FAB  128.0 kN T 
FAC  136.704 kN
FAC  136.7 kN C 
FBD  FDF  FFH  128.0 kN T 
FCE  FEG  136.7 kN C 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.28 (Continued)
Free body: Joint H
Also
FGH
145

144 kN
9
FGH  192.7 kN C 
FFH 144 kN

8
9
FFH  128.0 kN T
(Checks) 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.29
Determine whether the trusses of Problems 6.31a,
6.32a, and 6.33a are simple trusses.
SOLUTION
Truss of Problem 6.31a:
Starting with triangle HDI and adding two members at a time, we
obtain successively joints A, E, J, and B, but cannot go further. Thus,
this truss
is not a simple truss. 
Truss of Problem 6.32a:
Starting with triangle ABC and adding two members at a time, we obtain
joints D, E, G, F, and H, but cannot go further. Thus, this truss
is not a simple truss. 
Truss of Problem 6.33a:
Starting with triangle ABD and adding two members at a time, we
obtain successively joints H, G, F, E, I, C, and J, thus completing the
truss.
Therefore, this is a simple truss. 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.30
Determine whether the trusses of Problems 6.31b, 6.32b, and 6.33b are simple trusses.
SOLUTION
Truss of Problem 6.31b:
Starting with triangle CGM and adding two members at a time, we obtain
successively joints B, L, F, A, K, J, then H, D, N, I, E, O, and P, thus
completing the truss.
Therefore, this truss is a simple truss. 
Truss of Problem 6.32b:
Starting with triangle ABC and adding two members at a time, we obtain
successively joints E, D, F, G, and H, but cannot go further. Thus, this truss
is not a simple truss. 
Truss of Problem 6.33b:
Starting with triangle GFH and adding two members at a time, we obtain
successively joints D, E, C, A, and B, thus completing the truss.
Therefore, this is a simple truss. 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.31
For the given loading, determine the zero-force members in
each of the two trusses shown.
SOLUTION
Truss (a):
FB: Joint B: FBC  0
FB: Joint C: FCD  0
FB: Joint J : FIJ  0
FB: Joint I : FIL  0
FB: Joint N : FMN  0
(a)
FB: Joint M : FLM  0
The zero-force members, therefore, are
Truss (b):
BC, CD, IJ , IL, LM , MN 
FB: Joint C: FBC  0
FB: Joint B: FBE  0
FB: Joint G: FFG  0
FB: Joint F : FEF  0
FB: Joint E: FDE  0
FB: Joint I : FIJ  0
(b)
FB: Joint M : FMN  0
FB: Joint N : FKN  0
The zero-force members, therefore, are
BC, BE, DE, EF , FG, IJ , KN , MN 
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PROBLEM 6.32
For the given loading, determine the zero-force members
in each of the two trusses shown.
SOLUTION
Truss (a):
FB: Joint B: FBJ  0
FB: Joint D: FDI  0
FB: Joint E: FEI  0
FB: Joint I : FAI  0
FB: Joint F : FFK  0
(a)
FB: Joint G: FGK  0
FB: Joint K : FCK  0
AI , BJ , CK , DI , EI , FK , GK 
The zero-force members, therefore, are
Truss (b):
FB: Joint K : FFK  0
FB: Joint O : FIO  0
(b)
The zero-force members, therefore, are
FK and IO 
All other members are either in tension or compression.
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.33
For the given loading, determine the
zero-force members in each of the
two trusses shown.
SOLUTION
Truss (a):
Note: Reaction at F is vertical ( Fx  0).
Joint G:
Joint D:
F  0,
F  0,
FDG  0 
Joint F :
Joint G:
F  0,
F  0,
FFG  0 
Joint J :
Joint I :
F  0,
F  0,
FHI  0 
Joint A:
F  0,
FAC  0 
Joint C:
Joint E:
Joint F :
F  0,
F  0,
F  0,
FCE  0 
Joint G:
F  0,
FGH  0 
FDB  0 
FGH  0 
FIJ  0 
Truss (b):
FEF  0 
FFG  0 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.34
Determine the zero-force members in the truss of (a) Problem 6.21, (b) Problem 6.27.
SOLUTION
(a)
Truss of Problem 6.21:
FB : Joint C: FBC  0
FB : Joint J : FIJ  0
FB : Joint K : FJK  0
FB : Joint I : FHI  0
The zero-force members, therefore, are
(b)
BC , HI , IJ , JK 
Truss of Problem 6.27:
FB : Joint B: FBF  0
FB : Joint B: FBG  0
FB : Joint C: FCG  0
FB : Joint C: FCH  0
The zero-force members, therefore, are
BF , BG, CG, CH 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
y
3.5 m
3.5 m
PROBLEM 6.35*
A
5m
B O
The truss shown consists of six members and is supported by a
short link at A, two short links at B, and a ball and socket at D.
Determine the force in each of the members for the given loading.
D
z
12 m
C
x
4 kN
SOLUTION
Free body: Truss
From symmetry:
Dx = Bx
ΣMz = 0:
and
Dy = By
−A(5 m) − (4 kN)(12 m) = 0
A = −9.6 kN
ΣFx = 0: Bx + Dx + A = 0
2Bx − 9.6 kN = 0, Bx = 4.8 kN
ΣFy = 0:
By + Dy − 4 kN = 0
2 By = 4 kN
By = +2 kN
B = (4.8 kN)i + (2 kN)j
Thus
Free body: C
FCA = FAC
FCB = FBC
FCD = FCD
CA FAC
( −24i + 10 j)
=
CA 26
CB FBC
(−24i + 7k )
=
CB
25
CD FCD
(−24i − 7k )
=
25
CD
ΣF = 0: FCA + FCB + FCD − (4 kN)j = 0
*Advanced or specialty topic
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
796
PROBLEM 6.35* (Continued)
Substituting for FCA , FCB , FCD , and equating to zero the coefficients of i, j, k :
−
i:
24
24
FAC − ( FBC + FCD ) = 0
26
25
j:
10
FAC − 4 kN = 0
26
k:
7
( FBC − FCD ) = 0 FCD = FBC
25
(1)
FAC = 10.4 kN
T
Substitute for FAC and FCD in Eq. (1):
−
24
24
(10.40 kN) − (2 FBC) = 0
26
25
FBC = −5 kN
FBC = FCD = 5 kN
C
Free body: B
CB
= −(4.8 kN)i + (1.4 kN)k
CB
BA FAB
= FAB
=
(5j − 3.5k)
BA 6.103
= − FBD k
FBC = (5 kN)
FBA
FBD
ΣF = 0: FBA + FBD + FBC + (4.8 kN)i + (2 kN)j = 0
Substituting for FBA , FBD , FBC and equating to zero the coefficients of j and k:
j:
5
FAB + 2 kN = 0
6.103
k: −
3.5
FAB − FBD + 1.4 kN = 0
6.103
FAB = −2.44 kN
FBD = −
From symmetry:
3.5
(−2.44 kN) + 1.4 kN = +2.8 kN
6.103
FAD = FAB
FAB = 2.44 kN C
FBD = 2.8 kN T
FAD = 2.44 kN C
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
797
PROBLEM 6.36*
The truss shown consists of six members and is supported by
a ball and socket at B, a short link at C, and two short links at
D. Determine the force in each of the members for
P  (2184 N)j and Q  0.
SOLUTION
Free body: Truss:
From symmetry:
Dx  Bx and Dy  By
Fx  0: 2 Bx  0
Bx  D x  0
Fz  0: Bz  0
M c z  0:  2 By (2.8 m)  (2184 N)(2 m)  0
By  780 N
B  (780 N) j 
Thus,
Free body: A:

AB FAB

FAB  FAB
(0.8i  4.8 j  2.1k )
AB 5.30

AC FAC
FAC  FAC

(2i  4.8 j)
AC 5.20

AD FAD
FAD  FAD

(0.8i  4.8 j  2.1k )
AD 5.30
F  0: FAB  FAC  FAD  (2184 N) j  0
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.36* (Continued)
Substituting for FAB , FAC , FAD , and equating to zero the coefficients of i, j, k :
i:

0.8
2
FAC  0
( FAB  FAD ) 
5.30
5.20
(1)
j:

4.8
4.8
FAC  2184 N  0
( FAB  FAD ) 
5.30
5.20
(2)
2.1
( FAB  FAD )  0
5.30
k:
FAD  FAB
Multiply Eq. (1) by –6 and add Eq. (2):
 16.8 

 FAC  2184 N  0, FAC  676 N
 5.20 
FAC  676 N C 
Substitute for FAC and FAD in Eq. (1):
 0.8 
 2 

 2 FAB  
 (676 N)  0, FAB  861.25 N
 5.30 
 5.20 
Free body: B:
FAB  FAD  861 N C 

AB
 (130 N)i  (780 N) j  (341.25 N)k
FAB  (861.25 N)
AB
 2.8i  2.1k 
FBC  FBC 
  FBC (0.8i  0.6k )
3.5


FBD   FBD k
F  0: FAB  FBC  FBD  (780 N) j  0
Substituting for FAB , FBC , FBD and equating to zero the coefficients of i and k,
i:
k:
130 N  0.8 FBC  0
FBC  162.5 N
FBC  162.5 N T 
341.25 N  0.6 FBC  FBD  0
FBD  341.25  0.6(162.5)  243.75 N FBD  244 N T 
From symmetry:
FCD  FBC
FCD  162.5 N T 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.37*
The truss shown consists of six members and is supported by
a ball and socket at B, a short link at C, and two short links at
D. Determine the force in each of the members for P  0 and
Q  (2968 N)i.
SOLUTION
Free body: Truss:
From symmetry:
Dx  Bx and Dy  By
Fx  0: 2 Bx  2968 N  0
Bx  Dx  1484 N
M cz  0:  2 By (2.8 m)  (2968 N)(4.8 m)  0
By  2544 N
B  (1484 N)i  (2544 N) j 
Thus,
Free body: A:
FAB  FAB

AB
AB
FAB
(0.8i  4.8 j  2.1k )
5.30

AC FAC
 FAC

(2i  4.8 j)
AC 5.20

AD
 FAD
AD
FAD

(0.8i  4.8 j  2.1k )
5.30

FAC
FAD
F  0: FAB  FAC  FAD  (2968 N)i  0
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.37* (Continued)
Substituting for FAB , FAC , FAD , and equating to zero the coefficients of i, j, k ,
0.8
2
FAC  2968 N  0
( FAB  FAD ) 
5.30
5.20
(1)
4.8
4.8
FAC  0
( FAB  FAD ) 
5.30
5.20
(2)
i:

j:

k:
2.1
( FAB  FAD )  0
5.30
FAD  FAB
Multiply Eq. (1) by –6 and add Eq. (2):
 16.8 

 FAC  6(2968 N)  0, FAC  5512 N
 5.20 
FAC  5510 N C 
Substitute for FAC and FAD in Eq. (2):
 4.8 
 4.8 

 2 FAB  
 (5512 N)  0, FAB  2809 N
 5.30 
 5.20 
FAB  FAD  2810 N T 
Free body: B:
FAB
FBC
FBD

BA
 (2809 N)
 (424 N)i  (2544 N) j  (1113 N)k
BA
 2.8 i  2.1k 
 FBC 
  FBC (0.8i  0.6k )
3.5


  FBD k
F  0: FAB  FBC  FBD  (1484 N)i  (2544 N) j  0
Substituting for FAB , FBC , FBD and equating to zero the coefficients of i and k,
i:
24 N  0.8 FBC  1484 N  0, FBC  1325 N
k:
1113 N  0.6 FBC  FBD  0
FBC  1325 N T 
FBD  1113 N  0.6(1325 N)  1908 N, FBD  1908 N C 
From symmetry:
FCD  FBC
FCD  1325 N T 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
y
PROBLEM 6.38*
6m
The truss shown consists of nine members and is
supported by a ball and socket at A, two short links at B,
and a short link at C. Determine the force in each of the
members for the given loading.
D
8m
E
A
C
8 kN
B
z
6m
6m
6m
7.5 m
x
SOLUTION
Free body: Truss
From symmetry:
Az = Bz = 0
ΣFx = 0: Ax = 0
ΣM BC = 0: Ay (6 m) + (8 kN)(7.5 m) = 0
Ay = −10 kN
A = −(10 kN)j
By = C
From symmetry:
ΣFy = 0: 2By − 10 kN − 8 kN = 0
By = 9 kN
Free body: A
B = (9 kN)j
ΣF = 0: FAB + FAC + FAD − (10 kN)j = 0
FAB
i+k
+ FAC
2
i−k
2
+ FAD (0.6 i + 0.8 j) − (10 kN)j = 0
Factoring i, j, k and equating their coefficient to zero:
i:
j:
k:
1
2
FAB +
1
2
FAC + 0.6 FAD = 0
(1)
FAD = 12.5 kN T
0.8FAD − 10 kN = 0
1
2
FAB −
1
2
FAC = 0
FAC = FAB
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802
PROBLEM 6.38* (Continued)
Substitute for FAD and FAC into (1):
2
2
Free body: B
FAB + 0.6(12.5 kN) = 0, FAB = −5.3 kN,
FAB = FAC = 5.3 kN
C
BA
i+k
= +(5.3 kN)
= (3.75 kN)(i + k)
BA
2
= − FBC k
FBA = FAB
FBC
FBD = FBD (0.8 j − 0.6k )
FBE = FBE
BE FBE
(7.5i + 8 j − 6k )
=
BE 12.5
ΣF = 0: FBA + FBC + FBD + FBE + (9 kN)j = 0
Substituting for FBA , FBC , FBD + FBE and equate to zero the coefficients of i, j, k :
i:
3.75 kN +
7.5
FBE = 0,
12.5
j:
0.8 FBD +
8
(−6.25 kN) + 9 kN = 0, FBD = −6.25 kN
12.5
k:
3.75 kN − FBC − 0.6(−6.25 kN) −
FBE = 6.25 kN
C
FBD = 6.25 kN
C
FBC = 10.5 kN
T
FBD = FCD = 6.25 kN
C
FBE = −6.25 kN
6
(−6.25 kN) = 0, FBC = 10.5 kN
12.5
From symmetry:
Free body: D
ΣF = 0: FDA + FDB + FDC + FDE i = 0
We now substitute for FDA , FDB , FDC and equate to zero the coefficient of i. Only FDA contains i and its
coefficient is
−0.6 FAD = −0.6(12.5 kN) = −7.5 kN
i:
−7.5 kN + FDE = 0
FDE = 7.5 kN
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T
PROBLEM 6.39*
The truss shown consists of nine members and is supported by a
ball and socket at B, a short link at C, and two short links at D.
(a) Check that this truss is a simple truss, that it is completely
constrained, and that the reactions at its supports are statically
determinate. (b) Determine the force in each member for
P  (1200 N)j and Q  0.
SOLUTION
Free body: Truss:
M B  0: 1.8i  Cj  (1.8i  3k )  ( Dy j  Dk )
 (0.6i  0.75k )  (1200 j)  0
1.8 Ck  1.8 Dy k  1.8 Dz j
 3Dy i  720k  900i  0
Equate to zero the coefficients of i, j, k :
i:
3Dy  900  0, Dy  300 N
j:
Dz  0,
D  (300 N) j 
k:
1.8C  1.8(300)  720  0
C  (100 N) j 
F  0: B  300 j  100 j  1200 j  0
B  (800 N) j 
Free body: B:
F  0: FBA  FBC  FBE  (800 N) j  0, with

BA FAB
FBA  FAB

(0.6i  3j  0.75k )
BA 3.15
FBC  FBC i
FBE   FBE k
Substitute and equate to zero the coefficient of j, i, k :
j:
 3 

 FAB  800 N  0, FAB  840 N,
 3.315 
i:
 0.6 

 (840 N)  FBC  0
 3.15 
FBC  160.0 N T 
k:
 0.75 

 (840 N)  FBE  0
 3.15 
FBE  200 N T 
FAB  840 N C 
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PROBLEM 6.39* (Continued)
Free body: C:
F  0: FCA  FCB  FCD  FCE  (100 N) j  0, with

F
CA
FCA  FAC
 AC (1.2i  3j  0.75 k )
CA 3.317
FCB  (160 N) i
FCD   FCD k FCE  FCE

F
CE
 CE (1.8i  3k )
CE 3, 499
Substitute and equate to zero the coefficient of j, i, k :
 3 

 FAC  100 N  0, FAC  110.57 N
 3.317 
j:

i:
k:

FAC  110.6 N C 
1.2
1.8
FCE  0, FCE  233.3
(110.57)  160 
3.317
3.499
FCE  233 N C 
0.75
3
(110.57)  FCD 
(233.3)  0
3.317
3.499
FCD  225 N T 
Free body: D:
F  0: FDA  FDC  FDE  (300 N) j  0, with

F
DA
FDA  FAD
 AD (1.2i  3j  2.25k )
DA 3.937
FDC  FCD k  (225 N)k
FDE   FDE i
Substitute and equate to zero the coefficient of j, i, k :
j:
 3 

 FAD  300 N  0,
 3.937 
i:
 1.2 

 (393.7 N)  FDE  0
 3.937 
k:
 2.25 

 (393.7 N)  225 N  0 (Checks)
 3.937 
FAD  393.7 N,
FAD  394 N C 
FDE  120.0 N T 
Free body: E:
Member AE is the only member at E which does not lie in the xz plane.
Therefore, it is a zero-force member.
FAE  0 
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PROBLEM 6.40*
Solve Prob. 6.39 for P  0 and Q  (900 N)k.
PROBLEM 6.39* The truss shown consists of nine
members and is supported by a ball and socket at B, a short
link at C, and two short links at D. (a) Check that this truss
is a simple truss, that it is completely constrained, and that
the reactions at its supports are statically determinate. (b)
Determine the force in each member for P  (1200 N)j
and Q  0.
SOLUTION
Free body: Truss:
M B  0: 1.8i  Cj  (1.8i  3k )  ( Dy j  Dz k )
 (0.6i  3j  0.75k )  (900N)k  0
1.8 Ck  1.8Dy k  1.8Dz j
 3Dy i  540 j  2700i  0
Equate to zero the coefficient of i, j, k :
3Dy  2700  0
Dy  900 N
1.8 Dz  540  0
Dz  300 N
1.8C  1.8Dy  0
Thus,
C   Dy  900 N
C  (900 N) j D  (900 N) j  (300 N)k
F  0: B  900 j  900 j  300k  900k  0

B  (600 N)k 
Free body: B:
Since B is aligned with member BE,
FAB  FBC  0,
FBE  600 N T 
Free body: C:

F  0: FCA  FCD  FCE  (900 N) j  0, with
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PROBLEM 6.40* (Continued)

FCA
FCD

F
CA
 FAC
 AC (1.2i  3j  0.75k ) 
CA 3.317

F
CE
  FCD k FCE  FCE
 CE (1.8i  3k )
CE 3.499
Substitute and equate to zero the coefficient of j, i, k:
j:
 3 

 FAC  900 N  0,
 3.317 
FAC  995.1 N
FAC  995 N T 
FCE  699.8 N
FCE  700 N C 
i:

1.2
1.8
FCE  0,
(995.1) 
3.317
3.499
k:

0.75
3
(995.1)  FCD 
(699.8)  0
3.317
3.499
FCD  375 N T 
Free body: D:
F  0: FDA  FDE  (375 N)k +(900 N)j  (300 N) k  0

F
DA
FDA  FAD
 AD (1.2i  3j  2.25k )
DA 3.937
with
FDE   FDE i
and
Substitute and equate to zero the coefficient j, i, k:
j:
 3 

 FAD  900 N  0, FAD  1181.1 N
 3.937 
i:
 1.2 

 (1181.1 N)  FDE  0
 3.937 
k:
 2.25 

 (1181.1 N  375 N  300 N  0)
 3.937 
FAD  1181 N C 
FDE  360 N T 
(Checks)
Free body: E:
Member AE is the only member at E which
does not lie in the x z plane. Therefore, it is
a zero-force member.
FAE  0 
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y
(240 N) k
PROBLEM 6.41*
(275 N) i
The truss shown consists of 18 members and is supported by a
ball and socket at A, two short links at B, and one short link at G.
(a) Check that this truss is a simple truss, that it is completely
constrained, and that the reactions at its supports are statically
determinate. (b) For the given loading, determine the force in
each of the six members joined at E.
H
G
E
D
3m
F
C
x
A
2.88 m
B
z
3.3 m
SOLUTION
(a)
Check simple truss.
(1) start with tetrahedron BEFG
(2) Add members BD, ED, GD joining at D.
(3) Add members BA, DA, EA joining at A.
(4) Add members DH, EH, GH joining at H.
(5) Add members BC, DC, GC joining at C.
Truss has been completed: It is a simple truss
Free body: Truss
Check constraints and reactions:
Six unknown reactions-ok-however supports at A and B
constrain truss to rotate about AB and support at G
prevents such a rotation. Thus,
Truss is completely constrained and reactions are statically determinate
Determination of Reactions:
ΣMA = 0: 3.3i × (Byj + Bzk) + (3.3i − 2.88k) × Gj
+ (3j − 2.88k) × (275i + 240k) = 0
3.3Byk − 3.3Bz j + 3.3Gk + 2.88Gi − (3)(275)k
+ (3)(240)i − (2.88)(275)j = 0
Equate to zero the coefficient of i, j, k:
i: 2.88G + (3)(240) = 0
G = −250 N
G = (−250 N)j
j: −3.3Bz − (2.88)(275) = 0 Bz = −240 N
k: 3.3By + 3.3(−250) − (3)(275) = 0,
By = 500 N
B = (500 N)j − (240 N)k
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808
PROBLEM 6.41* (Continued)
ΣF = 0: A + (500 N) j − (240 N)k − (250 N) j
+ (275 N)i + (240 N)k = 0
A = −(275 N)i − (250 N) j
Zero-force members
The determination of these members will facilitate our solution.
FB: C. Writing
ΣFx = 0, ΣFy = 0, ΣFz = 0
Yields FBC = FCD = FCG = 0
FB: F. Writing
ΣFx = 0, ΣFy = 0, ΣFz = 0
Yields FBF = FEF = FFG = 0
FB: A: Since
FB: H: Writing
Az = 0, writing ΣFz = 0
ΣFy = 0
Yields FAD = 0
Yields FDH = 0
FB: D: Since FAD = FCD = FDH = 0, we need
consider only members DB, DE, and DG.
Since FDE is the only force not contained in plane BDG, it must be
zero. Simple reasonings show that the other two forces are also zero.
FBD = FDE = FDG = 0
The results obtained for the reactions at the supports and for the zero-force members are shown on the
figure below. Zero-force members are indicated by a zero (“0”).
(b)
Force in each of the members joined at E
FDE = FEF = 0
We already found that
Free body: A
ΣFy = 0
Yields FAE = 250 N T
Free body: H
ΣFz = 0
Yields FEH = 240 N C
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PROBLEM 6.41* (Continued)
Free body: E
ΣF = 0: FEB + FEG + (240 N)k − (250 N) j = 0
F
FBE
(3.3i − 3j) + EG (3.3i − 2.88k) + 240k − 250j = 0
4.46
4.38
Equate to zero the coefficient of j and k:
j: −
k:
−
3
4.46
FBE − 250 = 0
FBE = 372 N C
2.88
FEG + 240 = 0
4.38
FEG = 365 N T
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810
y
PROBLEM 6.42*
(240 N) k
(275 N) i
The truss shown consists of 18 members and is supported by a
ball and socket at A, two short links at B, and one short link at G.
(a) Check that this truss is a simple truss, that it is completely
constrained, and that the reactions at its supports are statically
determinate. (b) For the given loading, determine the force in each
of the six members joined at G.
H
G
E
D
3m
F
C
x
A
2.88 m
B
z
3.3 m
SOLUTION
See solution to Prob. 6.41 for Part (a) and for reactions and zero-force members.
(b)
Force in each of the members joined at G.
We already know that
FCG = FDG = FFG = 0
Free body: H
ΣFx = 0
Free body: G
ΣF = 0: FGB + FGE + (275 N)i − (250 N) j = 0
Yields FGH = 275 N C
FBG
F
(−3j + 2.88k) + EG (−3.3i + 2.88k) + 275i − 250j = 0
4.16
4.38
Equate to zero the coefficient of i, j, k:
i: −
3.3
FEG + 275 = 0, FEG = 365 .0 N
4.38
FEG = 365N T
j: −
3
FBG − 250 = 0, FBG = −346.67 N
4.16
FBG = 347 N C
k:
2.88
2.88
(−346.67) +
(365) = 0
4.16
4.38
(Checks)
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811
PROBLEM 6.43
A Mansard roof truss is loaded as shown. Determine the force in
members DF, DG, and EG.
SOLUTION
Reactions:
1
Because of the symmetry of the truss and loadings, Ax  0, Ay  L  (1.2 kN)  5   3 kN
2
We pass a section through DF, DG, and EG and use the free body shown:
Member DF:
M G  0: (1.2 kN)  8 m   1.2 kN  4 m 
  3 kN 10.25 m   FDF (3 m)  0
FDF  5.45 kN
FDF  5.45 kN C 
Member DG:
3
Fy  0: 3 kN  1.2 kN  1.2 kN    FDG  0
5
FDG  1.000 kN
FDG  1.000 kN T 
Member EG:
M D  0:
1.2 kN  4 m   FEG (3 m)  (3 kN)(6.25 m)  0
FEG  4.65 kN T 
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PROBLEM 6.44
A Mansard roof truss is loaded as shown. Determine the force in
members GI, HI, and HJ.
SOLUTION
Reactions:
1
Because of the symmetry of the truss and loadings, Ax  0, Ay  L  (1.2 kN)  5   3 kN
2
We pass a section through GI, HI, and HJ and use the free body shown:
Member DF:
M H  0:
 3 kN  6.25 m   1.2 kN  4 m   FGI (3 m)  0
FGI  4.65 kN
FGI  4.65 kN T 
Member HI:
Fy  0: 3 kN  1.2 kN  FHI  0
FHI  1.800 kN
FHI  1.800 kN C 
Member HJ:
M I  0:
 3 kN  6.25 m   FHJ (3 m)  (1.2 kN)(4 m)  0
FHJ  4.65 kN
FHJ  4.65 kN C 
Check Fx  0 : 4.65  4.65  0 o.k.
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PROBLEM 6.45
Determine the force in members BD and CD of the truss
shown.
SOLUTION
Reactions from Free body of entire truss:
A  A y  27 kN 
H  45 kN 
We pass a section through members BD, CD, and CE and use the
free body shown.
M C  0:  FBD (1.5 m)  (27 kN)(2 m)  0
FBD  36.0 kN
FBD  36.0 kN C 
3
Fy  0: 27 kN +   FCD  0
5
FCD  45.0 kN
FCD  45.0 kN C 
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PROBLEM 6.46
Determine the force in members DF and DG of the truss
shown.
SOLUTION
Reactions from Free body of entire truss:
A  A y  27 kN 
H  45 kN 
We pass a section through members DF, DG, and EG and use the
free body shown.
M G  0: FFD (1.5 m)  (45 kN)(10 m)  0
FFD  60.0 kN
FFD  60.0 kN C 
3
Fy  0: 45 kN +   FGD  36 kN  0
5
FGD  15.00 kN
FGD  15.00 kN C 
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PROBLEM 6.47
Determine the force in members CD and DF of the truss shown.
SOLUTION
Reactions:
M J  0: (12 kN)(4.8 m)  (12 kN)(2.4 m)  B (9.6 m)  0
B  9.00 kN
Fy  0: 9.00 kN  12.00 kN  12.00 kN  J  0
J  15.00 kN
Member CD:
Fy  0: 9.00 kN  FCD  0
FCD  9.00 kN C 
Member DF:
M C  0: FDF (1.8 m)  (9.00 kN)(2.4 m)  0
FDF  12.00 kN T 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.48
Determine the force in members FG and FH of the truss shown.
SOLUTION
Reactions:
M J  0: (12 kN)(4.8 m)  (12 kN)(2.4 m)  B (9.6 m)  0
B  9.00 kN
Fy  0: 9.00 kN  12.00 kN  12.00 kN  J  0
J  15.00 kN
Member FG:
3
Fy  0:  FFG  12.00 kN  15.00 kN  0
5
FFG  5.00 kN T 
Member FH:
M G  0: (15.00 kN)(2.4 m)  FFH (1.8 m)  0
FFH  20.0 kN T 
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PROBLEM 6.49
Determine the force in members CD and DF of the truss shown.
SOLUTION
tan  
5
  22.62
12
sin  
5
12
cos  
13
13
Member CD:
M I  0: FCD (9 m)  (10 kN)(9 m)  (10 kN)(6 m)  (10 kN)(3 m)  0
FCD  20.0 kN C 
FCD  20 kN
Member DF:
M C  0: ( FDF cos  )(3.75 m)  (10 kN)(3 m)  (10 kN)(6 m)  (10 kN)(9 m)  0
FDF cos   48 kN
 12 
FDF    48 kN
 13 
FDF  52.0 kN
FDF  52.0 kN C 
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PROBLEM 6.50
Determine the force in members CE and EF of the truss shown.
SOLUTION
Member CE:
M F  0: FCE (2.5 m)  (10 kN)(3 m)  (10 kN)(6 m)  0
FCE  36 kN
FCE  36.0 kN T 
Member EF:
M I  0: FEF (6 m)  (10 kN)(6 m)  (10 kN)(3 m)  0
FEF  15 kN
FEF  15.00 kN C 
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PROBLEM 6.51
Determine the force in members DE and DF of the truss
shown when P  20 kN.
SOLUTION
Reactions:
C  K  2.5P
1.5
3.6
  22.62
tan  
Member DE:
M A  0: (2.5 P )(1.2 m)  FDE (2.4 m)  0
FDE  1.25 P
For P  20 kN,
FDE  1.25(20)  25 kN
FDE  25.0 kN T 
Member DF:
M E  0: P (2.4 m)  (2.5 P )(1.2 m)  FDF cos  (1 m)  0
12 P  15 P  FDF cos 22.62(1 m)  0
FDF  0.65 P
For P  20 kN,
FDF  0.65(20)  13 kN
FDF  13.00 kN C 
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PROBLEM 6.52
Determine the force in members EG and EF of the truss
shown when P  20 kN.
SOLUTION
Reactions:
C  K  2.5P
1.5
tan  
1.2
  51.34
Member EG:
M F  0: P (3.6 m)  2.5P (3.4 m)  P (1.2 m)  FEG (1.5 m)  0
FEG  0.8 P;
For P  20 kN,
FEG  0.8(20)  16 kN
FEG  16.00 kN T 
Member EF:
M A  0: 2.5 P (1.2 m)  P (2.4 m)  FEF sin 51.34 (2.4 m)  0
FEF  0.320 P;
For P  20 kN,
FEF  0.320(20)  6.4 kN
FEF  6.40 kN C 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.53
Determine the force in members DF and DE of the truss shown.
SOLUTION
1
 
  tan 1    7.1
8
Member CE:
M F  0:  FDF (1.75 m)  (30 kN)(4 m)  (20 kN)(2 m)  0
FDF  91.4 kN
FDF  91.4 kN T 
Member EF:
M O  0:  FDE (14 m)  (30 kN)(10 m)  20 kN)(12 m)  0
FDE  38.6 kN
FDE  38.6 kN C 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.54
Determine the force in members CD and CE of the truss shown.
SOLUTION
1
 
 1.5 
  tan 1    36.9
 2 
  tan 1    7.1
8
Member CE:


M O  0: FCD sin 36.9 (14 m)  (30 kN)(10 m)  (20 kN)(12 m)  0
FCD  64.2 kN
FCD  64.2 kN T 
Member EF:
 
M O  0: FCE cos 7.1 (1.75 m)  (30 kN)(4 m)  (20 kN)(2 m)  0
FCE  92.1 kN
FCE  92.1 kN C 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.55
A monosloped roof truss is loaded as shown. Determine the
force in members CE, DE, and DF.
SOLUTION
Reactions at supports: Because of the symmetry of the loading,
Ax  0
Ay  I 
1
1
(Total load)  (8 kN)
2
2
A  I  4 kN 
We pass a section through members CD, DE, and DF, and use the free body shown.
(We moved FDE to E and FDF to F )
Slope
BJ 
Slope
DE 
2.16 m 9

9.6 m 40
1 m 5

2.4 m 12
0.46 m
0.46 m
a

 2.0444 m
9
Slope BJ
40
M D  0: FCE (1 m)  (1 kN)(2.4 m)  (4 kN)(2.4 m)  0
FCE  7.20 kN
FCE  7.20 kN T 
M K  0: (4 kN)(2.0444 m)  (1 kN)(2.0444 m)
 5

 (2 kN)(4.4444 m)   FDE  (6.8444 m)  0
13


FDE  1.047 kN
FDE  1.047 kN C 
M E  0: (1 kN)(4.8 m)  (2kN)(2.4 m)  (4 kN)(4.8 m)
 40

  FDF  (1.54 m)  0
 41

FDF  6.39 kN
FDF  6.39 kN C 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.56
A monosloped roof truss is loaded as shown. Determine the
force in members EG, GH, and HJ.
SOLUTION
Reactions at supports: Because of the symmetry of the loading,
Ax  0
Ay  I 
1
1
(Total load)  (8 kN)
2
2
A  I  4 kN 
We pass a section through members EG, GH, and HJ, and use the free body shown.
M H  0: (4 kN)(2.4 m)  (1 kN)(2.4 m)  FEG (2.08 m)  0
FEG  3.4615 kN
FEG  3.46 kN T 
M J  0:  FGH (2.4 m)  FEG (2.62 m)  0
FGH  
2.62
(3.4615 kN)
2.4
FGH  3.7788 kN
Fx  0:  FEG 
FGH  3.78 kN C 
2.4
FHJ  0
2.46
FHJ  
2.46
2.46
(3.4615 kN)
FEG  
2.4
2.4
FHJ  3.548 kN
FHJ  3.55 kN C 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
1.6 kN
1.6 kN
1.6 kN
0.8 kN
F
A
H
D
B
C
4m
G
E
4m
PROBLEM 6.57
1.6 kN
4m
I
4m
1.6 kN
0.8 kN 3 m
J
L 2.25 m
K
4m
A Howe scissors roof truss is loaded as shown. Determine
the force in members DF, DG, and EG.
4m
SOLUTION
Reactions at supports.
Because of symmetry of loading.
Ax = 0,
Ay = L =
1
1
(Total load) = (9.60 kN) = 4.80 kN
2
2
A = L = 4.80 kN
We pass a section through members DF, DG, and EG, and use the free body shown.
We slide FDF to apply it at F:
ΣMG = 0:
(0.8 kN)(12 m) + (1.6 kN)(8 m) + (1.6 kN)(4 m)
−(4.8 kN)(12 m) −
4FDF
42 + 1.752
(3 m) = 0
FDF = −10.5 kN
FDF = 10.5 kN
C
ΣM A = 0: −(1.6 kN)(4 m) − (1.6 kN)(8 m)
−
1.25FDG
4 + 1.25
2
2
(8 m) −
4FDG
42 + 1.252
(3.5 m) = 0
FDG = −3.35 kN
ΣMD = 0:
(0.8 kN)(8 m) + (1.6 kN)(4 m) − (4.8 kN)(8 m) +
4FEG
42 + 0.752
FDG = 3.35 kN
(2 m) = 0
FEG = +13.02 kN
FEG = 13.02 kN
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
820
C
T
PROBLEM 6.58
1.6 kN
1.6 kN
1.6 kN
0.8 kN
F
H
D
B
A
1.6 kN
E
C
4m
4m
G
4m
I
4m
1.6 kN
0.8 kN 3 m
J
L 2.25 m
K
4m
A Howe scissors roof truss is loaded as shown. Determine
the force in members GI, HI, and HJ.
4m
SOLUTION
Reactions at supports. Because of symmetry of loading:
1
(Total load)
2
1
= (9.60 kN)
2
Ay = L =
Ax = 0,
= 4.80 kN
A = L = 4.80 kN
We pass a section through members GI, HI, and HJ, and use the free body shown.
8FGI
ΣMH = 0: −
82 +1.52
(2 m) + (4.8 kN)(8 m) − (0.8 kN)(8 m) − (1.6 kN)(4 m) = 0
FGI = +13.02 kN
ΣML = 0:
FGI = 13.02 kN
T
FHI = 0.800 kN
T
(1.6 kN)(4 m) − FHI (8 m) = 0
FHI = +0.800 kN
We slide FHJ to apply it at H.
ΣMI = 0:
4FHJ
42 +1.752
(2 m) + (4.8 kN)(8 m) − (1.6 kN)(4 m) − (0.8 kN)(8 m) = 0
FHJ = −13.97 kN
FHJ = 13.97 kN
C
,
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
821
PROBLEM 6.59
Determine the force in members AD, CD, and CE of the
truss shown.
SOLUTION
Reactions:
ΣM k = 0: 36(2.4) − B(13.5) + 20(9) + 20(4.5) = 0
ΣFx = 0: −36 + K x = 0
B = 26.4 kN
K x = 36 kN
ΣFy = 0: 26.4 − 20 − 20 + K y = 0 K y = 13.6 kN
ΣM C = 0: 36(1.2) − 26.4(2.25) − FAD (1.2) = 0
FAD = −13.5 kN
 8

ΣM A = 0:  FCD  (4.5) = 0
17


FAD = 13.5 kN C 
FCD = 0 
 15

ΣM D = 0:  FCE  (2.4) − 26.4(4.5) = 0
 17

FCE = +56.1 kN
FCE = 56.1 kN T 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.60
Determine the force in members DG, FG, and FH of the
truss shown.
SOLUTION
See the solution to Problem 6.49 for free-body diagram and analysis to determine the reactions at the supports
B and K.
B = 26.4 kN ;
K x = 36.0 kN
;
K y = 13.60 kN
ΣM F = 0: 36(1.2) − 26.4(6.75) + 20(2.25) − FDG (1.2) = 0
FDG = −75 kN
FDG = 75.0 kN C 
 8

ΣM D = 0: − 26.4(4.5) +  FFG  (4.5) = 0
17


FFG = +56.1 kN


 15
ΣM G = 0: 20(4.5) − 26.4(9) +  FFH
 17
FFG = 56.1 kN T 

 (2.4) = 0 

FFH = +69.7 kN 
FFH = 69.7 kN T 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.61
Determine the force in members DG and FI of the truss shown. (Hint:
Use section aa.)
SOLUTION
M F  0: FDG (4 m)  (5 kN)(3 m)  0
FDG  3.75 kN
FDG  3.75 kN T 
Fy  0:  3.75 kN  FFI  0
FFI  3.75 kN
FFI  3.75 kN C 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.62
Determine the force in members GJ and IK of the truss shown. (Hint:
Use section bb.)
SOLUTION
M I  0: FGJ (4 m)  (5 kN)(6 m)  (5 kN)(3 m)  0
FGJ  11.25 kN
FGJ  11.25 kN T 
Fy  0:  11.25 kN  FIK  0
FIK  11.25 kN FIK  11.25 kN C 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
a H
E
B
b
F
C
J
1.6 m
K
N
A
D
3m
3m
G
a
3m
b
I
3m
12 kN
PROBLEM 6.63
M
L
Determine the force in members EH and GI of the
truss shown. (Hint: Use section aa.)
O
3m
12 kN
P
1.6 m
3m
12 kN
SOLUTION
Reactions:
9m
3m
3m
3m
12 kN 12 kN 12 kN
ΣFx = 0: Ax = 0
ΣM P = 0: 12(9) + 12(6) + 12(3) − Ay (18) = 0
A y = 12 kN
3.2 m
6m
ΣFy = 0: 12 − 12 − 12 − 12 + P = 0
P = 24 kN
ΣM G = 0: −(12 kN)(6 m) − FEH (3.2 m) = 0
FEH = −22.5 kN
12 kN
ΣFx = 0: FGI −22.5 kN = 0
FEH = 22.5 kN
C
FGI = 22.5 kN
T
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
834
a H
E
B
b
F
C
J
1.6 m
K
N
A
D
3m
G
3m
a
b
I
3m
3m
12 kN
PROBLEM 6.64
M
L
Determine the force in members HJ and IL of the truss
shown. (Hint: Use section bb.)
O
3m
12 kN
P
1.6 m
3m
12 kN
SOLUTION
See the solution to Prob. 6.61 for free body diagram and analysis to determine the reactions at supports A
and P.
A x = 0; Ay = 12.00 kN ;
P = 24.0 kN
ΣM L = 0: FHJ (3.2 m) − (12 kN)(3 m) + (24 kN)(6 m) = 0
3.2 m
3m
12 kN
12 kN
3m
24 kN
FHJ = −33.75 kN
FHJ = 33.8 kN
C
FIL = 33.8 kN
T
ΣFx = 0 : 33.75 kN − FIL = 0
FIL = +33.75 kN
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
835
PROBLEM 6.65
The diagonal members in the center
panels of the power transmission line
tower shown are very slender and can act
only in tension; such members are known
as counters. For the given loading,
determine (a) which of the two counters
listed below is acting, (b) the force in
that counter.
Counters CJ and HE
SOLUTION
Free body: Portion ABDFEC of tower.
We assume that counter CJ is acting and show the forces exerted by that counter and by members CH and
EJ.
Fx  0:
4
FCJ  2(1.2 kN)sin 20  0 FCJ  1.026 kN
5
Since CJ is found to be in tension, our assumption was correct. Thus, the answers are
(a) CJ

(b) 1.026 kN T 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.66
The diagonal members in the center
panels of the power transmission line
tower shown are very slender and can act
only in tension; such members are known
as counters. For the given loading,
determine (a) which of the two counters
listed below is acting, (b) the force in that
counter.
Counters IO and KN
SOLUTION
Free body: Portion of tower shown.
We assume that counter IO is acting and show the forces exerted by that counter and by members IN and
KO.
Fx  0 :
4
FIO  4(1.2 kN)sin 20  0 FIO  2.05 kN
5
Since IO is found to be in tension, our assumption was correct. Thus, the answers are
(a) IO 
(b) 2.05 kN T 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.67
The diagonal members in the center panels of the truss shown are
very slender and can act only in tension; such members are known
as counters. Determine the force in member DE and in the counters
that are acting under the given loading.
SOLUTION
Reactions from free body of entire truss:
M H  0: A  12 kN 
F
y  0 : H  15 kN 
Portion to the left of vertical cut through panel BCDE:
Required vertical component of bar forces FBE and FCD is downward. We must have
FBE  tension and FCD  0.
3
Fy  0: 12 kN  6 kN    FBE  0
5
FBE  10.00 kN T 

Required vertical component of bar forces FEF and FDG is downward. We must have
FEF  tension and FDG  0.

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PROBLEM 6.67 (Continued)
3
Fy  0:  12 kN + 15 kN    FEF  0
5
FEF  5.00 kN
Member DE: Free body of joint D
Joint E: (check)
Fy  0 :
FEF  5.00 kN T 
Fy  0 : FDE  0
 3
3
  10 kN      5 kN   9 kN + FDE  0
5
5
FDE  0 (checks)
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.68
Solve Prob. 6.67 assuming that the 9-kN load has been removed.
Problem 6.67 The diagonal members in the center panels of the
truss shown are very slender and can act only in tension; such
members are known as counters. Determine the force in member
DE and in the counters that are acting under the given loading.
SOLUTION
Reactions from free body of entire truss:
M H  0: A  7.5 kN 
Fy  0 : H  10.5 kN 
Portion to the left of vertical cut through panel BCDE:
Required vertical component of bar forces FBE and FCD is downward. We must have
FBE  tension and FCD  0.
 3
Fy  0: 7.5 kN  6 kN    FBE  0
5
FBE  2.50 kN T 
Portion to the right of vertical cut through panel DFEG:

Required vertical component of bar forces FEF and FDG is downward. We must have
FDG  tension and FEF  0.

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PROBLEM 6.68 (Continued)

 3
Fy  0:  12 kN + 10.5 kN +   FDG  0
5
FDG  2.5 kN
Member DE: Free body of joint D
Fy  0 : FDE 
FDG  2.50 kN T 
3
 2.5 kN   0
5
FDE  1.500 kN
FDE  1.500 kN C 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.69
Classify each of the structures shown as completely, partially, or improperly constrained; if completely
constrained, further classify as determinate or indeterminate. (All members can act both in tension and in
compression.)
SOLUTION
Structure (a)
Number of members:
m  16
Number of joints:
n  10
Reaction components:
r4
m  r  20, 2n  20
(a)
m  r  2n 
Thus,
To determine whether the structure is actually completely constrained and determinate, we must try to
find the reactions at the supports. We divide the structure into two simple trusses and draw the free-body
diagram of each truss.
This is a properly supported simple truss – O.K.
This is an improperly supported simple
truss. (Reaction at C passes through B. Thus,
Eq. M B  0 cannot be satisfied.)
Structure is improperly constrained. 
Structure (b)
m  16
n  10
r4
m  r  20, 2n  20
Thus,
(b)
m  r  2n 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.69 (Continued)
We must again try to find the reactions at the supports dividing the structure as shown.
Both portions are simply supported simple trusses.
Structure is completely constrained and determinate. 
Structure (c)
m  17
n  10
r4
m  r  21, 2n  20
Thus,
(c)
m  r  2n 
This is a simple truss with an extra support which causes reactions (and forces in members) to be
indeterminate.
Structure is completely constrained and indeterminate. 
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PROBLEM 6.70
Classify each of the structures shown as completely, partially, or improperly constrained; if completely
constrained, further classify as determinate or indeterminate. (All members can act both in tension and in
compression.)
SOLUTION
Structure (a):
Nonsimple truss with r  4, m  16, n  10,
so m  r  20  2 n, but we must examine further.
FBD Sections:
FBD
I:
M A  0

T1
II:
Fx  0

T2
I:
Fx  0

Ax
I:
Fy  0

Ay
II:
M E  0

Cy
II:
Fy  0

Ey
Since each section is a simple truss with reactions determined,
structure is completely constrained and determinate. 
Structure (b):
Nonsimple truss with r  3, m  16, n  10,
so
m  r  19  2n  20
Structure is partially constrained. 
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PROBLEM 6.70 (Continued)
Structure (c):
Simple truss with r  3, m  17, n  10, m  r  20  2 n, but the horizontal reaction forces Ax and E x
are collinear and no equilibrium equation will resolve them, so the structure is improperly constrained
and indeterminate.

Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.71
Classify each of the structures shown as completely, partially, or improperly constrained; if completely
constrained, further classify as determinate or indeterminate. (All members can act both in tension and in
compression.)
SOLUTION
Structure (a):
Nonsimple truss with r  4, m  12, n  8 so r  m  16  2n.
Check for determinacy:
One can solve joint F for forces in EF, FG and then solve joint
E for E y and force in DE.
This leaves a simple truss ABCDGH with
r  3, m  9, n  6 so r  m  12  2n
Structure is completely constrained and determinate. 
Structure (b):
Simple truss (start with ABC and add joints alphabetically to complete truss) with r  4, m  13, n  8
so
r  m  17  2n  16
Constrained but indeterminate 
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PROBLEM 6.71 (Continued)
Structure (c):
Nonsimple truss with r  3, m  13, n  8 so r  m  16  2n. To further examine, follow procedure in
part (a) above to get truss at left.
Since F1  0 (from solution of joint F),
M A  aF1
 0 and there is no equilibrium.
Structure is improperly constrained. 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.72
Classify each of the structures shown as completely, partially, or improperly constrained; if completely
constrained, further classify as determinate or indeterminate. (All members can act both in tension and in
compression.)
SOLUTION
Structure (a)
Number of members:
Number of joints:
Reaction components:
Thus,
m  12
n8
r 3
m  r  15,
2n  16
m  r  2n

Structure is partially constrained. 
Structure (b)
m  13, n  8
r 3
m  r  16, 2n  16
Thus,
m  r  2n

To verify that the structure is actually completely constrained and determinate, we observe that it is a
simple truss (follow lettering to check this) and that it is simply supported by a pin-and-bracket and a
roller. Thus,
structure is completely constrained and determinate. 
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PROBLEM 6.72 (Continued)
Structure (c)
m  13,
r4
n8
m  r  17, 2n  16
Thus,
m  r  2n

Structure is completely constrained and indeterminate. 
This result can be verified by observing that the structure is a simple truss (follow lettering to check this),
therefore it is rigid, and that its supports involve four unknowns.
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PROBLEM 6.73
Classify each of the structures shown as completely, partially, or improperly constrained; if completely
constrained, further classify as determinate or indeterminate. (All members can act both in tension and in
compression.)
SOLUTION
Structure (a):
Rigid truss with r  3, m  14, n  8,
so
r  m  17  2n  16
so completely constrained but indeterminate 
Structure (b):
Simple truss (start with ABC and add joints alphabetically), with
r  3, m  13, n  8, so r  m  16  2n
so completely constrained and determinate 
Structure (c):
Simple truss with r  3, m  13, n  8, so r  m  16  2 n, but horizontal reactions ( Ax and Dx ) are
collinear, so cannot be resolved by any equilibrium equation.
Structure is improperly constrained. 
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PROBLEM 6.74
Classify each of the structures shown as completely, partially, or improperly constrained; if completely
constrained, further classify as determinate or indeterminate. (All members can act both in tension and in
compression.)
SOLUTION
Structure (a):
m  12
No. of members
No. of joints
n  8 m  r  16  2n
No. of reaction components
r  4 unknows  equations
FBD of EH:
M H  0
FDE ; Fx  0
FGH ; Fy  0
Hy
Then ABCDGF is a simple truss and all forces can be determined.
This example is completely constrained and determinate. 
Structure (b):
No. of members
m  12
No. of joints
n  8 m  r  15  2n  16
No. of reaction components
r  3 unknows  equations
partially constrained 
Note: Quadrilateral DEHG can collapse with joint D moving downward; in (a), the roller at F prevents
this action.
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PROBLEM 6.74 (Continued)
Structure (c):
No. of members
m  13
No. of joints
n  8 m  r  17  2n  16
No. of reaction components
r  4 unknows  equations
completely constrained but indeterminate 
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PROBLEM 6.75
Determine the force in member BD and the components of the reaction
at C.
SOLUTION
We note that BD is a two-force member. The force it exerts on ABC, therefore, is directed along line BD.
Free body: ABC:
We have BD 
1.92
2

 0.562  2 m
M C  0:
 0.56

 1.92

(310 N)(2.1 m)  
FBD  (1.4 m)  
FBD  1.4 m   0
2
2




FBD  375 N
Fx  0: Cx  310 N 
0.56
(375 N)  0
2
C x  205 N
Fy  0: C y 
FBD  375 N C 
C x  205 N

1.92
(375 N)  0
2
C y  360 N
C y  360 N 
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PROBLEM 6.76
Determine the force in member BD and the components
of the reaction at C.
SOLUTION
We note that BD is a two-force member. The force it exerts on ABC, therefore, is directed along line BD.
Free body: ABC:
BD  (24) 2  (10) 2  26 cm
 10

M C  0: (160 N)(30 cm)  
FBD  (16 cm)  0
26


FBD  780 N
M x  0: C x 
FBD  780 N T 
24
(780 N)  0
26
C x  720 N
Fy  0: C y  160 N 
C x  720 N

10
(780 N)  0
26
C y  140.0 N
C y  140.0 N 
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PROBLEM 6.77
For the frame and loading shown, determine the force
acting on member ABC (a) at B, (b) at C.
SOLUTION
FBD ABC:
Note: BD is two-force member
(a)
3

M C  0: (0.09 m)(200 N)  (2.4 m)  FBD   0
5

FBD  125.0 N
(b)
4
Fx  0: 200 N  (125 N)  C x  0
5
Fy  0:
3
FBD  C y  0
5
36.9 
C x  100 N
3
C y  (125 N)  75 N
5
C  125.0 N
36.9 
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PROBLEM 6.78
Determine the components of all forces acting on member ABCD of the
assembly shown.
SOLUTION
Free body: Entire assembly:
ΣM B = 0: D (120 mm) − (480 N)(80 mm) = 0
D = 320 N 
ΣFx = 0: Bx + 480 N = 0
B x = 480 N

ΣFy = 0: By + 320 N = 0
B y = 320 N 
Free body: Member ABCD:
ΣM A = 0: (320 N)(200 mm) − C (160 mm) − (320 N)(80 mm)
− (480 N)(40 mm) = 0
C = 120.0 N 
ΣFx = 0: Ax − 480 N = 0
A x = 480 N

ΣFy = 0: Ay − 320 N − 120 N + 320 N = 0
A y = 120.0 N 
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PROBLEM 6.79
For the frame and loading shown, determine the components of all forces
acting on member ABC.
SOLUTION
Free body: Entire frame:
Fx  0: Ax  18 kN  0
Ax  18 kN
A x  18.00 kN

M E  0: (18 kN)(4 m)  Ay (3.6 m)  0
Ay  20 kN
A y  20.0 kN 
Fy  0:  20 kN  F  0
F  20 kN
F  20 kN
Free body: Member ABC
Note: BE is a two-force member, thus B is directed along line BE.
M C  0: B(4 m)  (18 kN)(6 m)  (20 kN)(3.6 m)  0
B  9 kN
B  9.00 kN

Cx  9.00 kN

Fx  0: Cx  18 kN  9 kN  0
Cx  9 kN
Fy  0: C y  20 kN  0
C y  20 kN
C y  20.0 kN 
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PROBLEM 6.80
Solve Problem 6.79 assuming that the 18-kN load is replaced by a clockwise
couple of magnitude 72 kN · m applied to member CDEF at Point D.
PROBLEM 6.79 For the frame and loading shown, determine the
components of all forces acting on member ABC.
SOLUTION
Free body: Entire frame
Fx  0: Ax  0
M F  0: 72 kN  m  Ay (3.6 m)  0
Ay  20 kN A y  20 kN
A  20.0 kN 
Free body: Member ABC
Note: BE is a two-force member. Thus B is directed along line BE.
M C  0: B (4 m)  (20 kN)(3.6 m)  0
B  18 kN
B  18.00 kN

C x  18.00 kN

Fx  0:  18 kN  C x  0
C x  18 kN
Fy  0: C y  20 kN  0
C y  20 kN
C y  20.0 kN 
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PROBLEM 6.81
Determine the components of all forces acting on member
ABCD when θ = 0.
SOLUTION
Free body: Entire assembly:
ΣM B = 0: A(200) − (150 N)(500) = 0
A = +375 N
A = 375 N

B x = 375 N

ΣFx = 0: Bx + 375 N = 0
Bx = −375 N
ΣFy = 0: By − 150 N = 0
By = +150 N
B y = 150 N 
Free body: Member ABCD:
We note that D is directed along DE, since DE is a two-force member.
ΣM C = 0: D (300) − (150 N)(100) + (375 N)(200) = 0
D = −200 N
D = 200 N 
ΣFx = 0: C x + 375 − 375 = 0
Cx = 0
ΣFy = 0: C y + 150 − 200 = 0
C y = +50.0 N
C = 50.0 N 
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PROBLEM 6.82
Determine the components of all forces acting on member
ABCD when θ = 90°.
SOLUTION
Free body: Entire assembly:
ΣM B = 0: A(200) − (150 N)(200) = 0
A = +150.0 N
A = 150.0 N

ΣFx = 0: Bx + 150 − 150 = 0
Bx = 0
B = 0 
ΣFy = 0: By = 0
Free body: Member ABCD:
We note that D is directed along DE, since DE is a two-force member.
ΣM C = 0: D (300) + (150 N)(200) = 0
D = −100.0 N
D = 100.0 N 
ΣFx = 0: C x + 150 N = 0
C x = −150 N
Cx = 150.0 N

ΣFy = 0: C y − 100 N = 0
C y = +100.0 N
C y = 100.0 N 
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PROBLEM 6.83
Determine the components of the reactions at A and
E, (a) if the 800-N load is applied as shown, (b) if
the 800-N load is moved along its line of action and
is applied at point D.
SOLUTION
Free body: Entire assembly. The analysis is valid for either (a) or (b), since position of the 800-N force
on its line of action is immaterial.
M A  0: E y (1200)  (800 N)(900)  0
E y  600 N

A y  200 N

Ex  2700 N

A x  2700 N

Fx  0: Ax  E x  0 (1)
Fy  0: Ay  600  800  0
(a) Free body: Member BE:
We note that E is directed along EB, since BE is a two-force member.
Ey
Ex
 900 

or E x  
 600 N
900 200
 200 
From equation (1): Ax  2700 N  0
Ax  2700 N
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PROBLEM 6.83 (Continued)
(b) Free body: Member ABC:
We note that A is directed along AB, since ABC is a two-force member.
Ay
Ax
 300 

or Ax  
 200 N
300 200
 200 
A x  300 N

From equation (1): 300 N  E x  0
E x  300 N
E x  300 N
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
PROBLEM 6.84
Determine the components of the reactions at D and E if the
frame is loaded by a clockwise couple of magnitude 150 N·m
applied (a) at A, (b) at B.
SOLUTION
Free body: Entire assembly. The analysis is valid for either (a) or (b), since position of the couple is
immaterial.
M D  0: E y (0.6 m)  (150 N  m)  0
E y  250 N

D y  250 N

Fx  0: Dx  E x  0 (1)
Fy  0: Dy  250 N  0
(a) Free body: Member BCE:
We note that E is directed along EC, since BCE is a two-force member.
Ey
Ex
 1.2 
or E x  

 250 N
1.2 0.40
 0.40 
E x  750 N

From equation (1): D x  750 N  0
Dx  750 N
Dx  750 N
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
PROBLEM 6.84 (Continued)
(b) Free body: Member ACD:
We note that D is directed along DC, since ACD is a two-force member.
Dx D y
 0.6 

or D x  
 250 N
0.6 0.4
 0.4 
Dx  375 N

E x  375 N

From equation (1): 375 N  E x  0
E x  375 N
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PROBLEM 6.85
Determine the components of the reactions at A and E if a 750-N force
directed vertically downward is applied (a) at B, (b) at D.
SOLUTION
Free body: Entire frame:
The following analysis is valid for both parts (a) and (b) since position of load on its line of action is
immaterial.
M E  0:  (750 N)(80 mm)  Ax (200 mm)  0
Ax   300 N
Fx  0: E x  300 N  0
A x  300 N
E x  300 N E x  300 N
Fy  0: Ay  E y  750 N  0
(a)
(1)
Load applied at B.
Free body: Member CE:
CE is a two-force member. Thus, the reaction at E must be directed along CE.
Ey
300 N
From Eq. (1):

75 mm
250 mm
Ay  90 N  750 N  0
E y  90 N
Ay  660 N
Thus, reactions are
(b)
A x  300 N
, A y  660 N 
E x  300 N
, E y  90.0 N 
Load applied at D.
Free body: Member AC:
AC is a two-force member. Thus, the reaction at A must be directed along AC.
Ay
300 N

125 mm
250 mm
Ay  150 N
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PROBLEM 6.85 (Continued)
Ay  E y  750 N  0
From Eq. (1):
150 N  E y  750 N  0
E y  600 N E y  600 N
Thus, reactions are
A x  300 N
, A y  150.0 N 
E x  300 N
,
E y  600 N 
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PROBLEM 6.86
Determine the components of the reactions at A and E if the frame is
loaded by a clockwise couple of magnitude 36 N · m applied (a) at B,
(b) at D.
SOLUTION
Free body: Entire frame:
The following analysis is valid for both parts (a) and (b) since the point of application of the couple is
immaterial.
M E  0: 36 N  m  Ax (0.2 m)  0
Ax   180 N
A x  180 N
Fx  0:  180 N + E x  0
E x  180 N
E x  180 N
Fy  0: Ay  E y  0
(a)
(1)
Couple applied at B.
Free body: Member CE:
AC is a two-force member. Thus, the reaction at E must be directed along EC.
Ey
180 N
From Eq. (1):

0.075 m
0.25 m
E y  54 N
Ay  54 N  0
Ay   54 N A y  54.0 N
Thus, reactions are
(b)
A x  180.0 N
, A y  54.0 N 
E x  180.0 N
, E y  54.0 N 
Couple applied at D.
Free body: Member AC:
AC is a two-force member. Thus, the reaction at A must be
directed along EC.
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PROBLEM 6.86 (Continued)
Ay
180 N
From Eq. (1):

0.125 m
0.25 m
Ay  90 N
Ay  E y  0
90 N  E y  0
E y   90 N E y  90 N
Thus, reactions are
A x  180.0 N
E x   180.0 N
,
A y  90.0 N 
, E y  90.0 N 
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PROBLEM 6.87
10 cm
A
B
F
C
5 cm
Determine the components of the reactions at A and B, (a) if the
100-N load is applied as shown, (b) if the 100-N load is moved
along its line of action and is applied at Point F.
5 cm
E
100 N
D
4 cm
SOLUTION
Free body: Entire frame
Analysis is valid for either (a) or (b), since position of 100-N load on its line of action is immaterial.
ΣM A = 0: By (10) − (100 N )(6) = 0 By = + 60 N
ΣFy = 0: Ay + 60 − 100 = 0
Ay = + 40 N
ΣFx = 0: Ax + Bx = 0
(a)
(1)
Load applied at E.
Free body: Member AC
Since AC is a two-force member, the reaction at A must be directed along CA. We have
Ax
40 N
=
10 cm 5 cm
From Eq. (1):
A x = 80.0 N
,
A y = 40.0 N
B x = 80.0 N
,
B y = 60.0 N
− 80 + Bx = 0 Bx = + 80 N
Thus,
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866
PROBLEM 6.87 (Continued)
(b)
Load applied at F.
Free body: Member BCD
Since BCD is a two-force member (with forces applied at B and C only),
the reaction at B must be directed along CB. We have therefore
Bx = 0
The reaction at B is
From Eq. (1):
The reaction at A is
Bx = 0
Ax + 0 = 0
B y = 60.0 N
Ax = 0
Ax = 0
A y = 40.0 N
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867
PROBLEM 6.88
The 48-N load can be moved along the line of action shown and applied at
A, D, or E. Determine the components of the reactions at B and F if the
48-N load is applied (a) at A, (b) at D, (c) at E.
SOLUTION
Free body: Entire frame
The following analysis is valid for (a), (b) and (c) since the position of the load
along its line of action is immaterial.
M F  0: (48N)(8 cm)  Bx (12 cm)  0
Bx  32 N
B x  32 N
Fx  0: 32 N+ Fx  0
Fx   32 N
Fx  32 N
Fy  0: By  Fy  48 N  0
(a)
(1)
Load applied at A.
Free body: Member CDB
CDB is a two-force member. Thus, the reaction at B must be directed along BC.
By
32 N
From Eq. (1):

5 cm
16 cm
B y  10 N
10 N  Fy  48 N  0
Fy  38 N Fy  38 N
Thus reactions are:
B x  32.0 N
Fx  32.0 N
, B y  10.00 N 
, Fy  38.0 N 
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PROBLEM 6.88 (Continued)
(b)
Load applied at D.
Free body: Member ACF.
ACF is a two-force member. Thus, the reaction at F must be
directed along CF.
Fy
32 N
From Eq. (1):

7 cm
16 cm
Fy  14 N
By  14 N  48 N  0
By  34 N
By  34 N
Thus, reactions are:
(c)
B x  32.0 N
, B y  34.0 N 
Fx  32.0 N
, Fy  14.00 N 
Load applied at E.
Free body: Member CDB
This is the same free body as in Part (a).
Reactions are same as (a) 
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PROBLEM 6.89
The 48-N load is removed and a 2.88-N · m clockwise couple is applied
successively at A, D, and E. Determine the components of the reactions at B
and F if the couple is applied (a) at A, (b) at D, (c) at E.
SOLUTION
Free body: Entire frame
The following analysis is valid for (a), (b), and (c), since the point of
application of the couple is immaterial.
M F  0:  288 N  cm  Bx (12 cm)  0
Bx   24 N
B x  24 N
Fx  0:  24 N+ Fx  0
Fx  24 N
Fx  24 N
Fy  0: By  Fy  0
(a)
(1)
Couple applied at A.
Free body: Member CDB
CDB is a two-force member. Thus, reaction at B must be directed along BC.
By
24 N
From Eq. (1):

5 cm
16 cm
B y  7.5 N
 7.5 N  Fy  0
Fy  7.5 N Fy  7.5 N
Thus, reactions are:
B x  24.0 N
Fx  24.0 N
, B y  7.50 N 
, Fy  7.50 N 
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PROBLEM 6.89 (Continued)
(b)
Couple applied at D.
Free body: Member ACF.
ACF is a two-force member. Thus, the reaction at F must be directed along CF.
Fy
24 N
From Eq. (1):

7 cm
16 cm
Fy  10.5 N
By  10.5 N:
By  10.5 N B y  10.5 N
Thus, reactions are:
B x  24.0 N
Fx  24.0 N
(c)
, B y  10.50 N 
, Fy  10.50 N 
Couple applied at E.
Free body: Member CDB
This is the same free body as in Part (a).
Reactions are same as in (a) 
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PROBLEM 6.90
(a) Show that when a frame supports a pulley at A, an equivalent loading of the frame and of each of its
component parts can be obtained by removing the pulley and applying at A two forces equal and parallel
to the forces that the cable exerted on the pulley. (b) Show that if one end of the cable is attached to the
frame at a Point B, a force of magnitude equal to the tension in the cable should also be applied at B.
SOLUTION
First note that, when a cable or cord passes over a frictionless, motionless pulley,
the tension is unchanged.
M C  0: rT1  rT2  0 T1  T2
(a)
Replace each force with an equivalent force-couple.
(b)
Cut the cable and replace the forces on pulley with equivalent pair of forces at A as above.

Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.91
Knowing that each pulley has a radius of 250 mm,
determine the components of the reactions at D and E.
SOLUTION
Free body: Entire assembly:
M E  0: (4.8 kN)(4.25 m)  Dx (1.5 m)  0
Dx  13.60 kN
Dx  13.60 kN

Ex  13.60 kN

Fx  0: Ex  13.60 kN  0
Ex  13.60 kN
 Fy  0: Dy  E y  4.8 kN  0
(1)
Free body: Member ACE:
M A  0 : (4.8 kN)(2.25 m)  E y (4 m)  0
E y  2.70 kN
From Eq. (1):
E y  2.70 kN 
Dy  2.70 kN  4.80 kN  0
Dy  7.50 kN
D y  7.50 kN 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.92
Knowing that the pulley has a radius of 75 mm, determine the
components of the reactions at A and B.
SOLUTION
Free body: Entire frame and pulley:
M B  0:  (240 N)(75 mm)  Ay (600 mm)  0
Fx  0: Bx  Ax  0 Bx  Ax
Ay  30.0 N A y  30.0 N

B y  270 N

(1)
Fy  0: By  30 N  240 N  0
Free body: Member ACD:
M D  0 : Ax (200)  (30 N)(300)  (240 N)(75)  0
Ax  45.0 N
From Eq. (1):
AX  Bx  45.0 N
A x  45.0 N

B x  45.0 N

Thus, reactions are
A x  45.0 N
,
A y  30.0 N 
B x  45.0 N
,
B y  270 N 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.93
A 1.5-m-diameter pipe is supported every 8 m by a small frame like
that shown. Knowing that the combined weight of the pipe and its
contents is 500 N/m and assuming frictionless surfaces, determine the
components (a) of the reaction at E, (b) of the force exerted at C on
member CDE.
SOLUTION
Free body: 16-ft length of pipe:
W = (500 N/m)(8 m) = 4 kN
Force Triangle
4 kN
B D 4 kN
= =
3 5
4
B = 3 kN
D = 5 kN
Determination of CB = CD.
 

We note that horizontal projection of BO + OD = horizontal projection of CD
3
4
r + r = (CD )
5
5
8
CB = CD = r = 2(0.75 m) = 1.5 m
4
Thus,
Free body: Member ABC:
ΣM A = 0: Cx h − (3 kN)(h − 1.5)
3 kN
Cx =
h − 1.5
(3 kN)
h
Cx =
4. 5 − 1. 5
(3 kN) = 2 kN
4.5
For h = 4.5 m
1.5 m
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
(1)
PROBLEM 6.93 (Continued)
Free body: Member CDE:
From above, we have
Cx = 2 kN

ΣM E = 0: (5 kN)(3.5 m) − (2 kN)(3 m) − C y (4 m) = 0
C y = +2.875 kN,
5 kN 3.5 m
3m
1.5 m
2 kN
3
(5 kN) + Ex = 0
5
Ex = −1 kN,
C y = 2.875 kN 
ΣFx = 0: − 2 kN +
E x = 1 kN

4m
ΣFy = 0: 2.875 kN −
4
(5 kN) + E y = 0,
5
E y = 1.125 kN 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.94
Solve Problem 6.93 for a frame where h = 3 m.
PROBLEM 6.93 A 1.5-m-ft-diameter pipe is supported every 8 m by a
small frame like that shown. Knowing that the combined weight of the
pipe and its contents is 500 N/m and assuming frictionless surfaces,
determine the components (a) of the reaction at E, (b) of the force
exerted at C on member CDE.
SOLUTION
See solution of Problem 6.91 for derivation of Eq. (1).
For h = 3 m, Cx =
3 − 1.5
h − 1.5
(3 kN) =
(3 kN) = 1.5 kN
3
h
Free body: Member CDE:
From above, we have
Cx = 1.5 kN

ΣM E = 0: (5 kN)(3.5 m) − (1.5 kN)(3 m) − C y (4 m) = 0
5 kN
C y = +3.25 kN,
3.5 m
C y = 3.25 kN 
3m
1.5 m
3
(5 kN) + E x = 0
5
Ex = −1.5 kN,
ΣFx = 0: −1.5 kN +
1.5 kN
4m
4
(5 kN) + E y = 0
5
E y = 0.75 kN
E x = 1.5 kN

ΣFy = 0: 3.25 kN −
E y = 0.75 kN 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.95
A trailer weighing 12 kN is attached to a 14.5-kN pickup truck by a ball-and-socket truck hitch at D.
Determine (a) the reactions at each of the six wheels when the truck and trailer are at rest, (b) the additional
load on each of the truck wheels due to the trailer.
12 kN
14.5 kN
D
B
A
0.6 m
2.7 m
0.9 m
C
1.5 m
1.2 m
SOLUTION
(a)
Free body: Trailer
(We shall denote by A, B, C the reaction at one wheel)
ΣM A = 0: −(12 kN)(0.6 m) + D(3.3 m) = 0
D = 2.18 kN
ΣFy = 0: 2A − 12 kN + 2.18 kN = 0
A = 4.91 kN
A = 4.91 kN
Free body: Truck
ΣM B = 0: (2.18 kN)(0.9 m) − (14.5 kN)(1.5 m) + 2C(2.7 m) = 0
C = 3.66 kN
C = 3.66 kN
ΣFy = 0: 2B − 2.18 kN − 14.5 kN + 2(3.66 kN) = 0
B = 4.68 kN
(b)
B = 4.68 kN
Additional load on truck wheels
Use free body diagram of truck without 14.5 kN.
ΣM B = 0: (2.18 kN)(0.9 m) + 2C(2.7 m) = 0
C = –0.363 kN
∆C = – 0.36 kN
ΣFy = 0: 2B − 2.18 kN − 2(0.363 kN) = 0
B = 1.453 kN
∆B = +1.45 kN
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
879
PROBLEM 6.96
E
Chain under
tension T
In order to obtain a better weight distribution over the
four wheels of the pickup truck of Prob . 6.95, a
compensating hitch of the type shown is used to attach
the trailer to the truck. The hitch consists of two bar
springs (only one is shown in the figure) that fit into
bearings inside a support rigidly attached to the truck.
The springs are also connected by chains to the trailer
frame, and specially designed hooks make it possible to
place both chains in tension. (a) Determine the tension T
required in each of the two chains if the additional load
due to the trailer is to be evenly distributed over the four
wheels of the truck. (b) What are the resulting reactions at
each of the six wheels of the trailer-truck combination?
D
F
Bar spring
0.5 m
PROBLEM 6.95 A trailer weighing 12 kN is attached
to a 14.5-kN pickup truck by a ball-and-socket truck hitch
at D. Determine (a) the reactions at each of the six wheels
when the truck and trailer are at rest, (b) the additional
load on each of the truck wheels due to the trailer.
SOLUTION
(a)
We small first find the additional reaction ∆ at each wheel due the trailer.
Free body diagram (Same ∆ at each truck wheel)
ΣM A = 0: −(12 kN)(0.6 m) + 2∆(4.2 m) + 2∆(6.9 m) = 0
∆ = 0.324 kN
ΣFY = 0: 2A − 12 kN + 4(0.324 kN) = 0;
A = 5.352 kN;
A = 5.32 kN
Free body: Truck
(Trailer loading only)
ΣM D = 0: 2∆(3.6 m) + 2∆(0.9 m) − 2T(0.5 m) = 0
T = 9∆
= 9(0.324 N)
T = 2.916 kN
(b)
T = 2.92 kN
Free body: Truck
(Truck weight only)
ΣM B = 0: −(14.5 kN)(1.5 m) + 2C′(2.7 m) = 0
C′ = 4.028 kN
C′ = 4.03 kN
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
880
PROBLEM 6.96 (Continued)
ΣFy = 0: 2B′ − 14.5 kN + 2(4.028 kN) = 0
B′ = 3.222 kN
B′ = 3.22 kN
Actual reactions
B = B′ + ∆ = 3.222 kN + 0.324 kN = 3.546 kN
B = 3.55 kN
C = C ′ + ∆ = 4.028 kN + 0.324 kN = 4.352 kN
C = 4.35 kN
(From Part a):
A = 5.32 kN
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
881
PROBLEM 6.97
The cab and motor units of the front-end loader shown are connected by a vertical pin located 2 m behind
the cab wheels. The distance from C to D is 1 m. The center of gravity of the 300-kN motor unit is
located at Gm, while the centers of gravity of the 100-kN cab and 75-kN load are located, respectively, at
Gc and Gl. Knowing that the machine is at rest with its brakes released, determine (a) the reactions at each
of the four wheels, (b) the forces exerted on the motor unit at C and D.
SOLUTION
(a)
Free body: Entire machine:
A  Reaction at each front wheel
B  Reaction at each rear wheel
M A  0: 75(3.2 m)  100(1.2 m)  2 B (4.8 m)  300(5.6 m)  0
2 B  325 kN
B  162.5 kN 
Fy  0: 2 A  325  75  100  300  0
2 A  150 kN
A  75.0 kN 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.97 (Continued)
(b)
Free body: Motor unit:
M D  0: C (1 m)  2 B (2.8 m)  300(3.6 m)  0
C  1080  5.6 B
Recalling
B  162.5 kN, C  1080  5.6(162.5)  170 kN
Fx  0: Dx  170  0



(1)
Fy  0: 2(162.5)  Dy  300  0
C  170.0 kN

D x  170.0 kN

D y  25.0 kN 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.98
Solve Problem 6.97 assuming that the 75-kN load has been removed.
PROBLEM 6.97 The cab and motor units of the front-end loader shown are connected by a vertical pin
located 2 m behind the cab wheels. The distance from C to D is 1 m. The center of gravity of the 300-kN
motor unit is located at Gm, while the centers of gravity of the 100-kN cab and 75-kN load are located,
respectively, at Gc and Gl. Knowing that the machine is at rest with its brakes released, determine (a) the
reactions at each of the four wheels, (b) the forces exerted on the motor unit at C and D.
SOLUTION
(a)
Free body: Entire machine:
A  Reaction at each front wheel
B  Reaction at each rear wheel
M A  0: 2 B (4.8 m)  100(1.2 m)  300(5.6 m)  0
2 B  375 kN
B  187.5 kN 
Fy  0: 2 A  375  100  300  0
2 A  25 kN
(b)
A  12.50 kN 
Free body: Motor unit:
See solution of Problem 6.97 for free body diagram and derivation of Eq. (1). With B  187.5 kN,
we have
C  1080  5.6(187.5)  30 kN
Fx  0: Dx  30  0
Fy  0: 2(187.5)  Dy  300  0
C  30.0 kN

D x  30.0 kN

D y  75.0 kN 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.99
Knowing that P = 90 N and Q = 60 N, determine the
components of all forces acting on member BCDE of the
assembly shown.
SOLUTION
Free body: Entire assembly:
M C  0: A(12 cm)  (60 N)(4 cm)  90 N 12 cm  0
A  110.0 N
A  110.0 N 
Fx  0: C x  60 N  0
Cx  60 N
C x  60.0 N

C y  200 N

Fy  0: C y  110 N  90 N  0
C y  200 N
Free body: Member BCDE:

M D  0: B (10 cm)  200 N(4 cm)   90 N8 cm  0
B  152.0 N
B  152.0 N
Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 6.99 (Continued)
Fx  0: Dx  60 N  0
Dx  60.0 N
D  60.0 N

Fy  0: D y  200 N  152 N  90 N  0
D y  42.0 N
D y  42.0 N
Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 6.100
Knowing that P = 60 N and Q = 90 N, determine the
components of all forces acting on member BCDE of the
assembly shown.
SOLUTION
Free body: Entire assembly:
M C  0: A(12 cm)  (60 N)(12 cm)  90 N 4 cm.  0
A  90.0 N
A  90.0 N 
Fx  0: C x  90 N  0
Cx  90 N
C x  90.0 N

C y  150.0 N

Fy  0: C y  60 N  90 N  0
C y  150 N
Free body: Member BCDE:

Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.100 (Continued)
M D  0: B (10 cm)   150 N  (4 cm)   60 N  8 cm  0
B  108.0 N B  108.0 N

D  90.0 N

Fx  0: Dx  90 N  0
Dx  90.0 N
Fy  0: D y  150 N  108 N  60 N  0
D y  18.00 N
D y  18.00 N
Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 6.101
For the frame and loading shown, determine the components of all forces
acting on member ABE.
SOLUTION
FBD Frame:
M E  0: (1.8 m) Fy  (2.1 m)(12 kN)  0
Fy  14.00 kN
Fy  0:  E y  14.00 kN  12 kN  0
E y  2 kN
E y  2.00 kN 
FBD member BCD:
M B  0: (1.2 m)C y  (12 kN)(1.8 m)  0 Cy  18.00 kN
But C is  ACF, so Cx  2C y ; Cx  36.0 kN
Fx  0:  Bx  C x  0 Bx  C x  36.0 kN
Bx  36.0 kN
on BCD
Fy  0:  By  18.00 kN  12 kN  0 By  6.00 kN on BCD
B x  36.0 kN
On ABE:

B y  6.00 kN 
FBD member ABE:
M A  0: (1.2 m)(36.0 kN)  (0.6 m)(6.00 kN)
 (0.9 m)(2.00 kN)  (1.8 m)( Ex )  0
Fx  0:  23.0 kN  36.0 kN  Ax  0
Fy  0:  2.00 kN  6.00 kN  Ay  0
E x  23.0 kN

A x  13.00 kN

A y  4.00 kN 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.102
For the frame and loading shown, determine the components of all
forces acting on member ABE.
SOLUTION
FBD Frame:
M F  0: (1.2 m)(2400 N)  (4.8 m) E y  0
E y  600 N 
FBD member BC:
Cy 
4.8
8
Cx  Cx
5.4
9
M C  0: (2.4 m) By  (1.2 m)(2400 N)  0 By  1200 N
B y  1200 N 
On ABE:
Fy  0: 1200 N  C y  2400 N  0 C y  3600 N
Cx 
so
Fx  0:  Bx  C x  0
On ABE:
9
Cy
8
Bx  4050 N
C x  4050 N
on BC
B x  4050 N
Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 6.102 (Continued)
FBD member AB0E:
M A  0: a (4050 N)  2 aE x  0
E x  2025 N
Fx  0 :  Ax  (4050  2025) N  0
Fy  0: 600 N  1200 N  Ay  0
E x  2025 N

A x  2025 N

A y  1800 N 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.103
For the frame and loading shown, determine the components of all forces
acting on member ABD.
SOLUTION
Free body: Entire frame:
cm
cm
N
N
cm
Σ M A = 0: E (12 cm) − (360 N)(15 cm) − (240 N)(33 cm) = 0
E = +1110 N
E = + 1.11 kN
ΣFx = 0: Ax + 1110 N = 0
Ax = −1110 N
A x = 1.11 kN

ΣFy = 0: Ay − 360 N − 240 N = 0
A y = 600 N 
Ay = +600 N
Free body: Member CDE:
Σ M C = 0: (1110 N)(24 cm)− Dx (12 cm) = 0
cm
Dx = +2220 N
cm
N
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.103 (Continued)
Free body: Member ABD:
From above:
D x = 2.22 kN

ΣM B = 0: Dy (18 cm) − (600 N)(6 cm) = 0
Dy = +200 N
D y = 200 N 
ΣFx = 0: Bx + 2220 N − 1110 N = 0
N
Bx = −1110 N
N
cm
cm
B x = 1.11 kN

N
ΣFy = 0: By + 200 N + 600 N = 0
By = −800 N
B y = 800 N 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.104
Solve Prob. 6.103 assuming that the 360-N load has been removed.
PROBLEM 6.103 For the frame and loading shown, determine the
components of all forces acting on member ABD.
SOLUTION
Free body diagram of entire frame.
cm
cm
N
cm
ΣM A = 0: E (12 cm) − (240 N)(33 cm) = 0
E = +660 N
E = 660 N
ΣFx = 0: Ax + 660 N = 0
Ax = −660 N
A x = 660 N

ΣFy = 0: Ay − 240 N = 0
A y = 240 N 
Ay = +240 N
Free body: Member CDE:
ΣM C = 0: (660 N)(24 cm) − Dx (12 cm) = 0
cm
Dx = +1320 N
cm
N
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.104 (Continued)
Free body: Member ABD:
From above:
D x = 1.32 kN

ΣM B = 0: Dy (18 cm) − (240 N)(6 cm) = 0
Dy = +80 N
N
N
N
cm
cm
D y = 80.0 N 
ΣFx = 0: Bx + 1320 N − 660 N = 0
Bx = −660 N
B x = 660 N

ΣFy = 0: By + 80 N + 240 N = 0
By = −320 N
B y = 320 N 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.105
For the frame and loading shown, determine the components of
the forces acting on member DABC at B and D.
SOLUTION
Free body: Entire frame:
M G  0: H (0.6 m)  (12 kN)(1 m)  (6 kN)(0.5 m)  0
H  25 kN H  25 kN
Free body: Member BEH:
M F  0: Bx (0.5 m)  (25 kN)(0.2 m)  0
Bx  10 kN
Free body: Member DABC:
B x  10.00 kN
From above:

M D  0:  B y (0.8 m)  (10 kN  12 kN)(0.5 m)  0
B y   13.75 kN
B y  13.75 kN 
Fx  0:  Dx  10 kN  12 kN  0
Dx   22 kN
Dx  22.0 kN

Fy  0:  D y  13.75 kN  0
D y   13.75 kN
D y  13.75 kN 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.106
Solve Problem 6.105 assuming that the 6-kN load has been
removed.
PROBLEM 6.105 For the frame and loading shown, determine
the components of the forces acting on member DABC at B and D.
SOLUTION
Free body: Entire frame
M G  0: H (0.6 m)  (12 kN)(1 m)  0
H  20 kN H  20 kN
Free body: Member BEH
M E  0: Bx (0.5 m)  (20 kN)(0.2 m)  0
Bx   8 kN
Free body: Member DABC
B x  8.00 kN
From above:

M D  0:  B y (0.8 m)  (8 kN  12 kN)(0.5 m)  0
B y   12.5 kN
B y  12.50 kN 
Fx  0:  Dx  8 kN  12 kN  0
Dx   20 kN
Fy  0:  D y  12.5 kN  0; D y   12.5 kN
D x  20.0 kN

D y  12.50 kN 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.107
The axis of the three-hinge arch ABC is a
parabola with the vertex at B. Knowing that
P  112 kN and Q  140 kN, determine (a) the
components of the reaction at A, (b) the
components of the force exerted at B on
segment AB.
SOLUTION
Free body: Segment AB:
M A  0: Bx (3.2 m)  B y (8 m)  P (5 m)  0
(1)
Bx (2.4 m)  B y (6 m)  P (3.75 m)  0
(2)
0.75 (Eq. 1):
Free body: Segment BC:
M C  0: Bx (1.8 m)  B y (6 m)  Q (3 m)  0
Add Eqs. (2) and (3):
(3)
4.2 Bx  3.75 P  3Q  0
Bx  (3.75 P  3Q )/4.2
From Eq. (1):
(3.75 P  3Q )
(4)
3.2
 8B y  5 P  0
4.2
B y  ( 9P  9.6Q )/33.6
(5)
given that P  112 kN and Q  140 kN.
(a)
Reaction at A.
Considering again AB as a free body,
Fx  0: Ax  Bx  0;
Ax  Bx  200 kN
A x  200 kN

Fy  0: Ay  P  B y  0
Ay  112 kN  10 kN  0
Ay   122 kN
A y  122.0 kN 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.107 (Continued)
(b)
Force exerted at B on AB.
From Eq. (4):
Bx  (3.75  112  3  140)/4.2  200 kN
B x  200 kN
From Eq. (5):

By  ( 9  112  9.6  140)/33.6   10 kN
B y  10.00 kN 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.108
The axis of the three-hinge arch ABC is a
parabola with the vertex at B. Knowing that
P  140 kN and Q  112 kN, determine (a) the
components of the reaction at A, (b) the
components of the force exerted at B on
segment AB.
SOLUTION
Free body: Segment AB:
M A  0: Bx (3.2 m)  B y (8 m)  P (5 m)  0
(1)
Bx (2.4 m)  B y (6 m)  P (3.75 m)  0
(2)
0.75 (Eq. 1):
Free body: Segment BC:
M C  0: Bx (1.8 m)  B y (6 m)  Q (3 m)  0
Add Eqs. (2) and (3):
(3)
4.2 Bx  3.75 P  3Q  0
Bx  (3.75 P  3Q )/4.2
(3.75 P  3Q )
From Eq. (1):
(4)
3.2
 8B y  5 P  0
4.2
B y  ( 9P  9.6Q )/33.6
(5)
given that P  140 kN and Q  112 kN.
(a)
Reaction at A.
Fx  0: Ax  Bx  0;
Ax  Bx  205 kN
A x  205 kN

Fy  0: Ay  P  B y  0
Ay  140 kN  ( 5.5 kN)  0
Ay  134.5 kN
A y  134.5 kN 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.108 (Continued)
(b)
Force exerted at B on AB.
From Eq. (4):
Bx  (3.75  140  3  112)/4.2  205 kN
B x  205 kN
From Eq. (5):

B y  ( 9  140  9.6  112)/33.6   5.5 kN
B y  5.50 kN 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.109
Neglecting the effect of friction at the horizontal and vertical
surfaces, determine the forces exerted at B and C on member
BCE.
SOLUTION
Free body of Member ACD
M H  0: C x (2 cm)  C y (18 cm)  0 C x  9C y
(1)
Free body of Member BCE
M B  0: C x (6 cm)  C y (6 cm)  (50 N)(12 cm)  0
9C y (6)  C y (6)  600  0
Substitute from (1):
C y   10 N; C x  9 C y  9(10)   90 N
C  90.6 N
Fx  0: Bx  90 N  0
Bx  90 N
Fy  0: B y  10 N  50 N  0
B y  40 N
B  98.5 N
6.34° 
24.0° 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
.
PROBLEM 6.110
Neglecting the effect of friction at the horizontal and
vertical surfaces, determine the forces exerted at B and C
on member BCE.
SOLUTION
Free body of Member ACD
We note that AC is a two-force member.
Cx C y

or C x  3C y
18
6
(1)
On free body of Member BCE
M H  0: C x (6 cm) C y (6 cm)  (50.0 N) (12 cm)  0
Substitute from (1):
3C y (6)  C y (6)  600  0
C y   25.0 N; C x  3C y  3(25.0)   75.0 N
C  79.1 N
Fy  0: B  25.0 N  50.0 N  0
18.43° 
B  25.0 N
B  25.0 N
Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 6.111
Members ABC and CDE are pin-connected at C and supported by
four links. For the loading shown, determine the force in each
link.
SOLUTION
Member FBDs:
I
FBD I:
FBD II:
FBDs combined:
II
aC y  a
1
M D  0: aC y  a
1
M B  0:
M G  0:
aP  a
FAF  0 FAF  2C y
2
2
1
2
FEH  0 FEH  2C y
1
FAF  a
2
FEH  0 P 
Cy 
P
2

FBD I:
Fy  0:
1
2
FAF 
1
2
FBG  P  C y  0
1
2
2C y 
1
2
2C y
so FAF 
2
P C 
2
FEH 
2
P T 
2
P
1
P

FBG  P   0
2
2
2
FBG  0 
FBD II:
Fy  0:  C y 
1
2
FDG 
1
2
FEH  0 
P
1
P

FDG   0
2
2
2
FDG  2 P C 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.112
Members ABC and CDE are pin-connected at C and supported by
four links. For the loading shown, determine the force in each
link.
SOLUTION
Member FBDs:
I
II
FBD I:
M I  0: 2aC y  aC x  aP  0 2C y  C x  P
FBD II:
M J  0: 2aC y  aC x  0 2C y  C x  0
Solving,
FBD I:
Cx 
P
P
; Cy 
as shown.
2
4
Fx  0: 
1
2
FBG  C x  0 FBG  Cx 2
Fy  0: FAF  P 
FBD II:
Fx  0:  Cx 
Fy  0:  C y 
1
2
1  2  P
P   0

2  2  4
FDG  0 FDG  C x 2
1  2 
P P
P
P   FEH  0 FEH    


4 2
4
2 2 
FBG 
P
2
P
4
FAF 
FDG 
P
FEH 
2
P
4
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
C 
C 
C 
T 
PROBLEM 6.113
Members ABC and CDE are pin-connected at C and supported by
four links. For the loading shown, determine the force in each
link.
SOLUTION
Member FBDs:
I
II
From FBD I:
M J  0:
a
3a
a
Cx 
C y  P  0 C x  3C y  P
2
2
2
FBD II:
M K  0:
a
3a
C x  C y  0 C x  3C y  0
2
2
Solving,
Cx 
FBD I:
P
P
; Cy 
as drawn.
2
6
M B  0: aC y  a
Fx  0: 
1
2
1
2
FAG 
FAG  0 FAG  2C y 
1
2
2
P
6
FAG 
2
2
P
P
6
2
FBF  Cx  0 FBF  FAG  Cx 2 
FBF 
FBD II:
M D  0: a
1
2
Fx  0: C x 
FEH  aC y  0 FEH   2C y  
1
2
FDI 
1
2
2
P
6
2
P C 
6
2 2
P C 
3
FEH 
FEH  0 FDI  FEH  Cx 2  
2
P T 
6
2
2
P
P
6
2
FDI 
2
P C 
3
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.114
Members ABC and CDE are pin-connected at C and supported by
the four links AF, BG, DG, and EH. For the loading shown,
determine the force in each link.
SOLUTION
We consider members ABC and CDE:
Free body: CDE:
M J  0: C x (4a )  C y (2a )  0
C y  2C x
(1)
Free body: ABC:
M K  0: C x (2a )  C y (4a )  P (3a )  0
Substituting for Cy from Eq. (1):
C x (2a )  2C x (4a )  P (3a )  0
1
Cx   P
2
Fx  0:
1
2
 1 
C y  2   P    P
 2 
FBG  C x  0
P
 1 
FBG   2C x   2   P   
,
2
 2 
FBG 
P
2
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
T 
PROBLEM 6.114 (Continued)
Fy  0:  FAF 
1
2
FBG  P  C y  0
FAF  
1
2
P
2
PP
P
2
FAF 
P
2
C 
Free body: CDE:
Fy  0:
1
2
FDG  C y  0
FDG  2C y   2 P
Fx  0:  FEH  Cx 
1
2
FDG  2 P T 
FDG  0
P
 1  1
FEH     P  
2P  
2
2
 2 
FEH 
P
2
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
C 
PROBLEM 6.115
Solve Problem 6.112 assuming that the force P is replaced by a
clockwise couple of moment M0 applied to member CDE at D.
PROBLEM 6.112 Members ABC and CDE are pin-connected at
C and supported by four links. For the loading shown, determine
the force in each link.
SOLUTION
Free body: Member ABC:
M J  0: C y (2a )  C x (a )  0
C x   2C y
Free body: Member CDE:
M K  0: C y (2a )  C x (a )  M 0  0
M0
v
4a
M
Cx   0 v
2a
C y (2a )  ( 2C y )(a )  M 0  0
Cy 
C x   2C y :
D
Fx  0:
2
D
 Cx  0;
2

M0
D
M0
0
2a
FDG 
2a
M0
2a
T 
M D  0: E ( a )  C y (a )  M 0  0
M
E (a )   0
 4a

 (a)  M 0  0

E
3 M0
4 a
3 M0
4 a
C 
FBG 
M0
T 
FAF 
M0
4a
FEH 
Return to free body of ABC:
Fx  0:
B
2
 Cx  0;
B
B
2

M0
0
2a
M0
2a
M B  0: A( a )  C y ( a );
A(a ) 
2a
M0
(a)  0
4a
A
M0
4a
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
C 
PROBLEM 6.116
Solve Problem 6.114 assuming that the force P is replaced by a
clockwise couple of moment M0 applied at the same point.
PROBLEM 6.114 Members ABC and CDE are pin-connected at C
and supported by the four links AF, BG, DG, and EH. For the
loading shown, determine the force in each link.
SOLUTION
Free body: CDE: (same as for Problem 6.114)
M J  0: C x (4a )  C y (2a )  0
C y  2C x
(1)
Free body: ABC:
M K  0: C x (2a )  C y (4a )  M 0  0
Substituting for Cy from Eq. (1): C x (2 a )  2C x (4a )  M 0  0
M0
6a
M
 M 
C y  2   0    0
3a
 6a 
Cx  
Fx  0:
1
2
FBG  C x  0
2M 0
6a
FBG   2C x  
Fy  0:  FAF 
1
2
FBG 
2M 0
6a
T 
M0
6a
T 
FBG  C y  0
FAF  
1
2
2M 0 M 0
M

 0
6a
3a
6a
FAF 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.116 (Continued)
Free body: CDE:
(Use F.B. diagram of Problem 6.114.)
Fy  0:
1
2
FDG  C y  0
FDG  2C y  
Fx  0:  FEH  Cx 
1
2
2M 0
3a
FDG 
2M 0
3a
T 
M0
6a
C 
FDG  0
 M   1   2M 0
FEH     0   

 6a   2   3a

M
   0 ,
6a

FEH 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.117
Four beams, each of length 2a, are nailed together at their midpoints to form the support system shown.
Assuming that only vertical forces are exerted at the connections, determine the vertical reactions at A, D,
E, and H.
SOLUTION
Note that, if we assume P is applied to EG, each individual member FBD looks like
2 Fleft  2 Fright  Fmiddle
so
Labeling each interaction force with the letter corresponding to the joint of its application, we see that
B  2 A  2F
C  2B  2D
G  2C  2 H
P  F  2G ( 4C  8 B  16 F )  2 E
From P  F  16F ,

F
P
15
so A 
P

15
D
2P

15
H
4P

15
E
8P

15
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.118
Four beams, each of length 3a, are held together by single nails at A, B, C, and D. Each beam is attached
to a support located at a distance a from an end of the beam as shown. Assuming that only vertical forces
are exerted at the connections, determine the vertical reactions at E, F, G, and H.
SOLUTION
We shall draw the free body of each member. Force P will be applied to member AFB. Starting with
member AED, we shall express all forces in terms of reaction E.
Member AED:
M D  0: A(3a )  E ( a )  0
E
3
M A  0:  D (3a )  E (2a )  0
A
D
2E
3
Member DHC:
 2E 
M C  0:  
 (3a )  H ( a )  0
 3 
H   2E
(1)
 2E 
M H  0:  
 (2a )  C (a )  0
 3 
C
4E
3
Member CGB:
 4E 
M B  0:  
 (3a )  G (a )  0
 3 
G   4E
 4E 
M G  0:  
 (2a )  B ( a )  0
 3 
B
8E
3
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
(2)
PROBLEM 6.118 (Continued)
Member AFB:
Fy  0: F  A  B  P  0
 E   8E 
F    
P 0
 3  3 
F  P  3E
(3)
M A  0: F ( a )  B (3a )  0
 8E 
( P  3E )( a )   
 (3a )  0
 3 
P  3E  8 E  0; E  
P
5
E
P

5
Substitute E   P5 into Eqs. (1), (2), and (3).
 P
H  2 E   2   
 5
H 
2P
5
H
2P

5
 P
G   4E  4   
 5
G
4P
5
G
4P

5
 P
F  P  3E  P  3   
 5
F 
8P
5
F
8P

5

Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.119
Each of the frames shown consists of two L-shaped members connected by two rigid links. For each
frame, determine the reactions at the supports and indicate whether the frame is rigid.
SOLUTION
(a)
Member FBDs:
I
FBD II:
FBD I:
Fy  0: B y  0
II
M B  0: aF2  0
F2  0
M A  0: aF2  2aP  0, but F2  0
so P  0
(b)
not rigid for P  0 
Member FBDs:
Note: Seven unknowns ( Ax , Ay , Bx , B y , F1 , F2 , C ), but only six independent equations
System is statically indeterminate. 

System is, however, rigid. 
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PROBLEM 6.119 (Continued)
(c)
FBD whole:
FBD right:
I
FBD I:
II
M A  0: 5aB y  2aP  0
Fy  0: Ay  P 
2
P0
5
a
5a
Bx 
By  0
2
2
FBD II:
M C  0:
FBD I:
Fx  0: Ax  Bx  0
Bx  5 B y
Ax   Bx
By 
2
P 
5
Ay 
3
P
5
B x  2P
A x  2P
A  2.09P
16.70 
B  2.04P
11.31 
System is rigid. 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.120
Each of the frames shown consists of two L-shaped members connected by two rigid links. For each
frame, determine the reactions at the supports and indicate whether the frame is rigid.
SOLUTION
Note: In all three cases, the right member has only three forces acting, two of which are parallel. Thus,
the third force, at B, must be parallel to the link forces.
(a)
FBD whole:
M A  0:  2aP 
Fx  0: Ax 
Fy  0:
a 4

B  5a
B  0 B  2.06 P
4 17
17
4
17
Ay  P 
B0
1
17
B  2.06P
14.04 
A  2.06P
14.04 
A x  2P
B  0 Ay 
P
2
rigid 
(b)
FBD whole:
Since B passes through A,
no equilibrium if P  0
M A  2aP  0 only if P  0.
not rigid 
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PROBLEM 6.120 (Continued)
(c)
FBD whole:
M A  0: 5a
1
17
Fx  0: Ax 
B
4
17
Fy  0: Ay  P 
3a 4
17
B  2aP  0 B 
P
4 17
4
B0
1
17
B  1.031P
14.04 
Ax   P
B0
Ay  P 
P 3P

4
4
A  1.250P
36.9 
System is rigid. 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.121
Each of the frames shown consists of two L-shaped members connected by two rigid links. For each
frame, determine the reactions at the supports and indicate whether the frame is rigid.
SOLUTION
(a)
Member FBDs:
I
II
FBD I:
M A  0: aF1  2aP  0 F1  2 P;
FBD II:
M B  0:  aF2  0 F2  0
FBD I:
Fy  0:
Ay  P  0 A y  P
Fx  0: Bx  F1  0, Bx   F1  2 P
B x  2P
Fy  0: B y  0
so B  2P
Fx  0:  Ax  F1  F2  0
Ax  F2  F1  0  2 P

A x  2P
so A  2.24P
26.6 
Frame is rigid. 
(b)
FBD left:
FBD whole:
I
FBD I:
M E  0:
FBD II:
M B  0:
II
a
a
5a
P  Ax 
Ay  0 Ax  5 Ay   P
2
2
2
3aP  aAx  5aAy  0 Ax  5 Ay  3P
This is impossible unless P  0.
not rigid 
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PROBLEM 6.121 (Continued)
(c)
Member FBDs:
I
FBD I:
II
A  P 
Fy  0: A  P  0
M D  0: aF1  2aA  0
Fx  0: F2  F1  0
FBD II:
F1  2 P
F2  2 P 
M B  0: 2aC  aF1  0 C 
Fx  0: F1  F2  Bx  0
F1
P
2
C  P 
Bx  P  P  0 
Fx  0: B y  C  0 B y   C   P
BP 
Frame is rigid. 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.122
The shear shown is used to cut and trim electronic circuit board
laminates. For the position shown, determine (a) the vertical
component of the force exerted on the shearing blade at D, (b) the
reaction at C.
SOLUTION
We note that BD is a two-force member.
Free body: Member ABC: We have the components:
Px  (400 N)sin 30  200 N
Py  (400 N) cos 30  346.41 N
(FBD ) x 
25
FBD
65
(FBD ) y 
60
FBD
65
M C  0: ( FBD ) x (45)  ( FBD ) y (30)  Px (45  300sin 30)
 Py (30  300cos30)  0
 25

 60

 65 FBD  (45)   65 FBD  (30)  (200)(195)  (346.41)(289.81)




45 FBD  139.39  103
FBD  3097.6 N
(a)
Vertical component of force exerted on shearing blade at D.
( FBD ) y 
60
60
FBD 
(3097.6 N)  2859.3 N
65
65
(FBD ) y  2860 N 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.122 (Continued)
(b)
Returning to FB diagram of member ABC,
Fx  0: ( FBD ) x  Px  Cx  0
C x  ( FBD ) x  Px 
25
FBD  Px
65
25
(3097.6)  200
65
C x  991.39

Cx  991.39 N
Fy  0: ( FBD ) y  Py  C y  0
C y  ( FBD ) y  Py 
C y  2512.9 N
C  2295 N
60
60
FBD  Py 
(3097.6)  346.41
65
65
C y  2512.9
C  2700 N
68.5° 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.123
A 100-N force directed vertically downward is applied to the
toggle vise at C. Knowing that link BD is 6 cm long and that
a = 4 cm, determine the horizontal force exerted on block E.
SOLUTION
Free body: Entire Frame
BF  4 cm sin15  1.03528 cm
We have
sin  =
1.03528 cm
6 cm
  9.9359
AD  AG  FD  4cos15  6cos9.9359  9.7737 cm
 
M A  0: (100 N)(10 cm) cos 15   (FBD ) sin (9.7737 cm)  0
FBD  572.77 N
Horizontal force exerted on block.

( FBD ) x  FBD cos   (572.77 N) cos 9.9359 

(FBD ) x  564 N
Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 6.124
A 100-N force directed vertically downward is applied to the
toggle vise at C. Knowing that link BD is 6 cm long and that
a = 8 cm, determine the horizontal force exerted on block E.
SOLUTION
Free body: Entire Frame
BF  8 cm sin15  2.0706 cm
We have
sin  =
2.0706 cm
6 cm
  20.188
AF  8 cm cos15  7.7274 cm
FD  6 cm cos 20.188  5.6314 cm
AD AF  FD 13.359 cm
 
M A  0: (100 N)(14 cm) cos 15  ( FBD )sin (13.359 cm)  0
FBD  293.33 N
Horizontal force exerted on block.

( FBD ) x  FBD cos   (293.33 N) cos 20.188

(FBD ) x  275 N
Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 6.125
The control rod CE passes through a
horizontal hole in the body of the toggle
system shown. Knowing that link BD is 250
mm long, determine the force Q required to
hold the system in equilibrium when   20.
SOLUTION
We note that BD is a two-force member.
Free body: Member ABC:
Dimensions in mm
Since BD  250,
  sin 1
33.404
;   7.679
250
 M C  0: ( FBD sin  )187.94  ( FBD cos )68.404  (100 N)328.89  0
FBD [187.94sin 7.679  68.404cos 7.679]  32,889
FBD  770.6 N
Fx  0: (770.6 N)cos7.679  Cx  0
Cx   763.7 N
Member CQ:
Fx  0: Q  Cx  763.7 N
Q  764 N
Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 6.126
Solve Problem 6.125 when (a)   0,
(b)   6.
PROBLEM 6.125 The control rod CE passes
through a horizontal hole in the body of the
toggle system shown. Knowing that link BD
is 250 mm long, determine the force Q
required to hold the system in equilibrium
when   20.
SOLUTION
We note that BD is a two-force member.
  0:
(a)
Free body: Member ABC
BD  250 mm, sin 
Since
35 mm
;   8.048
250 mm
M C  0: (100 N)(350 mm)  FBD sin  (200 mm)  0
FBD  1250 N
Fx  0: FBD cos  Cx  0
(1250 N)(cos 8.048)  Cx  0
Cx  1237.7 N
Member CE:
Fx  0: (1237.7 N)  Q  0
Q  1237.7 N
  6
(b)
Q  1238 N
Free body: Member ABC
Dimensions in mm
Since
14.094 mm
250 mm
  3.232
BD  250 mm,   sin 1
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
PROBLEM 6.126 (Continued)
M C  0: ( FBD sin  )198.90  ( FBD cos )20.906  (100 N)348.08  0
FBD [198.90sin 3.232  20.906cos3.232]  34808
FBD  1084.8 N
Fx  0: FBD cos  Cx  0
(1084.8 N)cos3.232  Cx  0
Cx  1083.1 N
Member DE:
Fx  0: Q  Cx
Q  1083.1 N
Q  1083 N
Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 6.127
The press shown is used to emboss a small seal at E.
Knowing that P  250 N, determine (a) the vertical
component of the force exerted on the seal, (b) the reaction
at A.
SOLUTION
FBD Stamp D:
Fy  0: E  FBD cos 20  0, E  FBD cos 20
FBD ABC:
M A  0: (0.2 m)(sin 30)( FBD cos 20)  (0.2 m)(cos30)( FBD sin 20)
 [(0.2 m)sin 30  (0.4 m) cos15](250 N)  0
FBD  793.64 N C
and, from above,
E  (793.64 N) cos 20
(a)
E  746 N 
Fx  0: Ax  (793.64 N)sin 20  0
A x  271.44 N
Fy  0: Ay  (793.64 N) cos 20  250 N  0
A y  495.78 N
(b)
A  565 N
61.3° 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.128
The press shown is used to emboss a small seal at E.
Knowing that the vertical component of the force exerted on
the seal must be 900 N, determine (a) the required vertical
force P, (b) the corresponding reaction at A.
SOLUTION
FBD Stamp D:
Fy  0: 900 N  FBD cos 20  0, FBD  957.76 N C
(a)
FBD ABC:
M A  0: [(0.2 m)(sin 30)](957.76 N)cos 20  [(0.2 m)(cos30)](957.76 N)sin 20
 [(0.2 m)sin 30  (0.4 m) cos15]P  0
P  301.70 N,
(b)
P  302 N 
Fx  0: Ax  (957.76 N)sin 20  0
A x  327.57 N
Fy  0:  Ay  (957.76 N) cos 20  301.70 N  0
A y  598.30 N
A  682 N
61.3° 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
50 N
A
PROBLEM 6.129
The pin at B is attached to member ABC and can slide freely along
the slot cut in the fixed plate. Neglecting the effect of friction,
determine the couple M required to hold the system in equilibrium
when θ = 30°.
10 cm
6 cm
B
C
M
8 cm
D
SOLUTION
Free body: Member ABC
ΣM C = 0: (50 N)(13.856 cm) − B (3 cm) = 0
B = +230.9 N
ΣFy = 0: − 50 N + C y = 0
C y = +50 N
ΣFx = 0:
230.9 N − Cx = 0
Cx = +230.9 N
Free body: Member CD
β = sin −1
5.196
; β = 40.504°
8
CD cos β = (8 cm) cos 40.504° = 6.083 cm
ΣM D = 0: M − (50 N)(5.196 cm) − (230.9 N)(6.083 cm) = 0
M = +1664 N ⋅ cm
M = 16.6 N ⋅ m
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932
PROBLEM 6.13 0
50 N
A
The pin at B is attached to member ABC and can slide freely along
the slot cut in the fixed plate. Neglecting the effect of friction,
determine the couple M required to hold the system in equilibrium
when θ = 60°.
10 cm
6 cm
B
C
M
8 cm
D
SOLUTION
Free body: Member ABC
ΣM C = 0: (50 N)(8 cm) − B (5.196 cm) = 0
B = +77 N
ΣFx = 0: 77 N − Cx = 0
Cx = +77 N
ΣFy = 0: − 50 N + C y = 0
C y = +50 N
Free body: Member CD
3
8
β = sin −1 ; β = 22.024°
CD cos β = (8 cm) cos 22.024° = 7.416 cm
ΣM D = 0: M − (50 N)(3 cm) − (77 N)(7.416 cm) = 0
M = +721 N ⋅ cm
M = 7.2 N ⋅ m
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
933
PROBLEM 6.131
Arm ABC is connected by pins to a collar at B and to crank CD at
C. Neglecting the effect of friction, determine the couple M
required to hold the system in equilibrium when   0.
SOLUTION
Free body: Member ABC
Fx  0: Cx  240 N  0
Cx  240 N
M C  0: (240 N)(500 mm)  B(160 mm)  0
B  750 N
Fy  0: C y  750 N  0
C y  750 N
Free body: Member CD
M D  0: M  (750 N)(300 mm)  (240 N)(125 mm)  0
M  195  103 N  mm
M  195.0 kN  m
Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 6.132
Arm ABC is connected by pins to a collar at B and to crank CD at
C. Neglecting the effect of friction, determine the couple M
required to hold the system in equilibrium when   90.
SOLUTION
Free body: Member ABC
Fx  0: Cx  0
M B  0: C y (160 mm)  (240 N)(90 mm)  0
C y  135 N
Free body: Member CD
M D  0: M  (135 N)(300 mm)  0
M  40.5  103 N  mm
M  40.5 kN  m
Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 6.133
The Whitworth mechanism shown is used to produce a quick-return
motion of Point D. The block at B is pinned to the crank AB and is free to
slide in a slot cut in member CD. Determine the couple M that must be
applied to the crank AB to hold the mechanism in equilibrium when
(a) α  0, (b) α  30°.
SOLUTION
(a)
Free body: Member CD:
M C  0: B(0.5 m)  (1200 N)(0.7 m)  0
B  1680 N
Free body: Crank AB:
M A  0: M  (1680 N)(0.1 m)  0
M  168.0 N  m
(b)
M  168.0 N  m
Geometry:
AB  100 mm,   30°
tan  
50
  5.87
486.6
BC  (50) 2  (486.6) 2
 489.16 mm
Free body: Member CD:
M C  0: B(0.48916)  (1200 N)(0.7)cos5.87  0
B  1708.2 N
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
PROBLEM 6.133 (Continued)
Free body: Crank AB:
M A  0: M  ( B sin 65.87)(0.1 m)  0
M  (1708.2 N)(0.1 m)sin 65.87
M  155.90 N  m
M  155.9 N  m
Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 6.134
Solve Problem 6.133 when (a) α  60, (b) α  90.
PROBLEM 6.133 The Whitworth mechanism shown is used to produce
a quick-return motion of Point D. The block at B is pinned to the crank
AB and is free to slide in a slot cut in member CD. Determine the couple
M that must be applied to the crank AB to hold the mechanism in
equilibrium when (a) α  0, (b) α  30°.
SOLUTION
(a)
Geometry:
AB  100 mm,   60°
BE  100 cos 60  86.60
CE  400  100sin 60  450
tan  
86.60
  10.89
450
BC  (86.60) 2  (450)2  458.26 mm
Free body: Member CD:
M C  0: B(0.45826 m)  (1200 N)(0.7 m)cos10.89  0
B  1800.0 N
Free body: Crank AB:
M A  0: M  ( B sin 40.89)(0.1 m)  0
M  (1800.0 N)(0.1 m)sin 40.89
M  117.83 N  m
M  117.8 N  m
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
PROBLEM 6.134 (Continued)
(b)
Free body: Member CD:
0.1 m
0.4 m
  14.04
tan  
BC  (0.1)2  (0.4)2  0.41231 m
M C  0: B(0.41231)  (1200 N)(0.7cos14.04)  0
B  1976.4 N
Free body: Crank AB:
M A  0: M  ( B sin14.04)(0.1 m)  0
M  (1976.4 N)(0.1 m)sin14.04
M  47.948 N  m
M  47.9 N  m
Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 6.135
Two rods are connected by a frictionless collar B. Knowing that the magnitude
of the couple MA is 5 N.m., determine (a) the couple MC required for
equilibrium, (b) the corresponding components of the reaction at C.
SOLUTION
cm
cm
5 N.m
cm
(a)
cm
Free body: Rod AB & collar:
ΣM A = 0: ( B cos α )(6 cm) + (B sinα )(8 cm) −M A = 0
B (6cos 21.8° + 8sin 21.8°) − 500 = 0
B = 58.535 N
Free body: Rod BC:
+ΣM C = 0: M C − B = 0
M C = B = (58.535 N)(0.21541 m) = 12.609 N. m
(b)
MC = 12.61 N⋅m

C x = 54.3 N

ΣFx = 0: Cx + B cos α = 0
Cx = − B cos α = −(58.535 N) cos 21.8 ° = −54.3 N
ΣFy = 0: C y − B sin α = 0
C y = B sin α = (58.535 N) sin 21.8 ° = +21.7 N
C y = 21.7 N 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.136
Two rods are connected by a frictionless collar B. Knowing that the magnitude
of the couple MA is 5 N·m ., determine (a) the couple MC required for
equilibrium, (b) the corresponding components of the reaction at C.
SOLUTION
cm
cm
cm
cm
5 N.m
cm
(a)
Free body: Rod AB:
ΣM A = 0: B(10 cm) − 500 N⋅ cm = 0
B = 50.0 N
Free body: Rod BC & collar:
ΣM C = 0: M C − (0.6 B)(0.2 m) − (0.8 B)(0.08 m) = 0
M C − (30 N)(0.2 m) − (40 N)(0.08 m) = 0
M C = 9.2 N ⋅ m
(b)
M C = 9.2 N ⋅ m

C x = 30.0 N

ΣFx = 0: Cx + 0.6 B = 0
Cx = −0.6 B = −0.6(50.0 N) = −30.0 N
ΣFy = 0: C y − 0.8B = 0
C y = 0.8 B = 0.8(50.0 N) = + 40.0 N
C y = 40.0 N 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.137
Rod CD is attached to the collar D and passes through a collar
welded to end B of lever AB. Neglecting the effect of friction,
determine the couple M required to hold the system in equilibrium
when   30.
SOLUTION
FBD DC:
Fx  0:
D y sin 30  (150 N) cos 30  0
D y  (150 N) ctn 30  259.81 N
FBD machine:
M A  0: (0.100 m)(150 N)  d (259.81 N)  M  0
d  b  0.040 m
b
so
0.030718 m
tan 30
b  0.053210 m
d  0.0132100 m
M  18.4321 N  m
M  18.43 N  m
Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 6.138
Rod CD is attached to the collar D and passes through a
collar welded to end B of lever AB. Neglecting the effect
of friction, determine the couple M required to hold the
system in equilibrium when   30.
SOLUTION
B  CD
Note:
FBD DC:
Fx  0: D y sin 30  (300 N) cos 30  0
Dy 
300 N
 519.62 N
tan 30
FBD machine:
M A  0:
0.200 m
519.62 N  M  0
sin 30
M  207.85 N  m
M  208 N  m
Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 6.139
Two hydraulic cylinders control the position
of the robotic arm ABC. Knowing that in the
position shown the cylinders are parallel,
determine the force exerted by each cylinder
when P  160 N and Q  80 N.
SOLUTION
Free body: Member ABC:
M B  0:
4
FAE (150 mm)  (160 N)(600 mm)  0
5
FAE  800 N
FAE  800 N T 
3
Fx  0:  (800 N)  Bx  80 N  0
5
Bx  560 N
Fy  0: 
4
(800 N)  B y  160 N  0
5
By  800 N
Free body: Member BDF:
M F  0: (560 N)(400 mm)  (800 N)(300 mm) 
FDG  100 N
4
FDG (200 mm)  0
5
FDG  100.0 N C 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.140
Two hydraulic cylinders control the position of the robotic arm ABC. In the position shown, the cylinders
are parallel and both are in tension. Knowing the FAE  600 N and FDG  50 N, determine the forces P and
Q applied at C to arm ABC.
SOLUTION
Free body: Member ABC:
M B  0:
4
(600 N)(150 mm)  P (600 mm)  0
5
P  120 N
M C  0:
P  120.0 N 
4
(600)(750 mm)  B y (600 mm)  0
5
By  600 N
Free body: Member BDF:
M F  0: Bx (400 mm)  (600 N)(300 mm)
4
 (50 N)(200 mm)  0
5
Bx  470 N
Return to free body: Member ABC:
3
Fx  0:  (600 N)  470 N  Q  0
5
Q  110 N
Q  110.0 N
Copyright © McGraw-Hill Education. Permission required for reproduction or display.

240 mm
240 mm
PROBLEM 6.141
A 12-m length of railroad rail of weight 660 N/m is lifted by the tongs
shown. Determine the forces exerted at D and F on tong BDF.
A
150 mm
B
C
300 mm
D
200 mm
E
F
20 mm
20 mm
SOLUTION
Free body: Rail
W = (12 m)(660 N/m) = 7920 N
Free body: Upper link
7920 N
By symmetry
1
E y = Fy = W = 3960 N
2
By symmetry,
1
( FAB ) y = ( FAC ) y = W = 3960 N
2
Since AB is a two-force member,
( FAB ) x ( FAB ) y
=
240
150
Free Body: Tong BDF
240 mm
150 mm
( FAB ) x =
240
(3960) = 6336 N
150
ΣM D = 0: (Attach FAB at A)
Fx (0.2 m) − (FAB)x (0.45 m) − Fy (0.02 m) = 0
Fx (0.2 m) − (6336 N)(0.45 m) − (3960 N)(0.02 m) = 0
300 mm
Fx = +14652 N
200 mm
F = 15.2 kN
15.1°
ΣFx = 0: − Dx + ( FAB ) x + Fx = 0
20 mm
Dx = ( FAB ) x + Fx = 6336 + 14652 = 20988 N
ΣFy = 0: D y + ( FAB ) y − Fy = 0
Dy = 0
D = 21 kN
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
943
PROBLEM 6.142
A log weighing 4 kN is lifted by a pair of tongs as shown.
Determine the forces exerted at E and F on tong DEF.
SOLUTION
FBD AB:
By symmetry:
and
Ay  B y  4 kN
Ax  Bx
6
(4 kN)
5
 4.8 kN

Note:
so
FBD DEF:
D  B
Dx  4.8 kN
D y  4 kN
M F  (0.252 m)(4 kN)  (0.372 m)(4.8 kN )  ( 0.288 m) Ex 0
Ex  9.7 kN
E  9.7 kN

Fx  0:  4.8 kN  9.7 kN  Fx  0
Fx  4.9 kN
Fy  0: 4 kN  Fy  0
Fy  4 kN
F  6.33 kN
39.2 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.143
The tongs shown are used to apply a total upward force of 45 kN on a pipe cap.
Determine the forces exerted at D and F on tong ADF.
SOLUTION
FBD whole:
By symmetry,
FBD ADF:
A  B  22.5 kN
M F  0: (75 mm) D  (100 mm)(22.5 kN)  0
D  30.0 kN

Fx  0: Fx  D  0
Fx  D  30 kN
Fy  0: 22.5 kN  Fy  0
Fy  22.5 kN
so
F  37.5 kN
36.9 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.144
If the toggle shown is added to the tongs of Prob. 6.141 and a single
vertical force is applied at G, determine the forces exerted at D and F on
tong ADF.
SOLUTION
Free body: Toggle:
By symmetry,
Ay 
1
(45 kN)  22.5 kN
2
AG is a two-force member.
Ax
22.5 kN

22 mm 55 mm
Ax  56.25 kN
Free body: Tong ADF:
Fy  0: 22.5 kN  Fy  0
Fy  22.5 kN
M F  0 : D(75 mm)  (22.5 kN)(100 mm)  (56.25 kN)(160 mm)  0
D  150 kN
D  150.0 kN

Fx  0: 56.25 kN  150 kN  Fx  0
Fx  93.75 kN
F  96.4 kN
13.50 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
30 mm
240 mm
A
PROBLEM 6.14 5
250 N
B
The pliers shown are used to grip a 8-mm-diameter rod.
Knowing that two 250-N forces are applied to the handles,
determine (a) the magnitude of the forces exerted on the rod,
(b) the force exerted by the pin at A on portion AB of the
pliers.
30°
C
250 N
SOLUTION
Free body: Portion AB
(a)
ΣM A = 0: Q(30 mm) − (250 N)(240 mm) = 0
Q = 2 kN
250 N
30 mm
240 mm
(b)
ΣFx = 0: Q(sin 30°) + Ax = 0
(2000 N)(sin30°) + Ax = 0
Ax = −1000 N
Ax = 1 kN
ΣFy = 0: − Q (cos 30°) + Ay − 250 N = 0
−(2000 N)(cos30°) + Ay − 250 N = 0
Ay = +1982 N
Ay = 1.982 kN
A = 2.2 kN
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
946
63.2°
a
18 mm
200 N
108 mm
12 mm
Determine the magnitude of the gripping forces exerted along
line aa on the nut when two 200-N forces are applied to the
handles as shown. Assume that pins A and D slide freely in
slots cut in the jaws.
B
A
C
D
PROBLEM 6.146
E
a
200 N
SOLUTION
FBD jaw AB:
ΣFx = 0: Bx = 0
ΣM B = 0: (12 mm) Q − (36 mm)A = 0
36 mm
12 mm
A=
Q
3
ΣFy = 0: A + Q − B y = 0
By = A + Q =
FBD handle ACE:
4Q
3
By symmetry and FBD jaw DE:
Q
3
E x = Bx = 0
D= A=
E y = By =
4Q
3
ΣM C = 0: (126 mm)(200 N) + (18 mm)
Q
4Q
− (18 mm)
=0
3
3
Q = 1.4 kN
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
948
PROBLEM 6.147
In using the bolt cutter shown, a worker applies two 300-N
forces to the handles. Determine the magnitude of the
forces exerted by the cutter on the bolt.
SOLUTION
FBD cutter AB:
I
FBD I:
FBD handle BC:
II
Dimensions in mm
Fx  0: Bx  0
FBD II:
M C  0: (12 mm)B y  (448 mm)300 N  0
B y  11, 200.0 N
Then
FBD I:
M A  0: (96 mm) B y  (24 mm) F  0
F  4 By
F  44,800 N  44.8 kN 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.148
Determine the magnitude of the gripping forces produced
when two 300-N forces are applied as shown.
SOLUTION
We note that AC is a two-force member.
FBD handle CD:
M D  0:  (126 mm)(300 N)  (6 mm)
2.8
8.84
A
 1

 (30 mm) 
A  0
 8.84 
A  2863.6 8.84 N
Dimensions in mm
FBD handle AB:
M B  0: (132 mm)(300 N)  (120 mm)
1
8.84
(2863.6 8.84 N)
 (36 mm) F  0
F  8.45 kN 
Dimensions in mm
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.149
Determine the force P that must be applied to the toggle CDE to
maintain bracket ABC in the position shown.
SOLUTION
We note that CD and DE are two-force members.
Free body: Joint D:
( FCD ) x ( FCD ) y

30
150
Similarly,
( FCD ) y  5( FCD ) x
Dimensions in mm
( FDE ) y  5( FDE ) x
Fy  0: ( FDE ) y  ( FCD ) y
It follows that
( FDE ) x  ( FCD ) x
Fx  0: P  ( FDE ) x  ( FCD ) x  0
( FDE ) x  ( FCD ) x 
1
P
2
1 
( FDE ) y  ( FCD ) y  5  P   2.5P
2 
Also,
Free body: Member ABC:
1 
M A  0: (2.5P)(300)   P  (450)  (910 N)(150)  0
2 
(750  225) P  (910 N)(150)
P  140.0 N 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.150
Determine the force P that must be applied to the toggle CDE to maintain
bracket ABC in the position shown.
SOLUTION
We note that CD and DE are two-force members.
Free body: Joint D:
( FCD ) x ( FCD ) y

30
150
Similarly,
( FCD ) y  5( FCD ) x
( FDE ) y  5( FDE ) x
Dimensions
in mm
Fy  0: ( FDE ) y  ( FCD ) y
It follows that
( FDE ) x  ( FCD ) x
 Fx  0: ( FDE ) x  ( FCD ) x  P  0
( FDE ) x  ( FCD ) x 
1
P
2
1 
( FDE ) y  ( FCD ) y  5  P   2.5P
2 
Also,
Free body: Member ABC:
1 
M A  0: (2.5P)(300)   P  (450)  (910 N)(150)  0
2 
(750  225) P  (910 N)(150)
P  260 N 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.151
Since the brace shown must remain in position even when the magnitude of P is
very small, a single safety spring is attached at D and E. The spring DE has a
constant of 50 N/cm and an unstretched length of 7 cm. Knowing that l = 10 cm and
that the magnitude of P is 800 N, determine the force Q required to release the
brace.
SOLUTION
M A  0 : Q 15 cm  Cx 35 cm  0
Free body:
7
Q  C x (1)
3
Fy  0 : C y  800 N  0 or C y  800 N
Spring force.
Unstretched length:
Stretched length:
 0  7 cm
  10 cm
k  50 N/cm
F  kx  k (   0 )
 50(10  7)  150 N
Free body: Portion BC
M B  0 :
150 N 2 cm  800 N  3 cm  Cx 20 cm  0
C x  135.0 N
Substituting into equation (1) gives Q  315 N 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
B
15 mm
PROBLEM 6.152
A
C
9 mm
27 mm
The specialized plumbing wrench shown is used in confined areas (e.g., under a
basin or sink). It consists essentially of a jaw BC pinned at B to a long rod.
Knowing that the forces exerted on the nut are equivalent to a clockwise (when
viewed from above) couple of magnitude 14 N ⋅ m, determine (a) the magnitude
of the force exerted by pin B on jaw BC, (b) the couple M0 that is applied to the
wrench.
M0
SOLUTION
Free body: Jaw BC
This is a two-force member
Cy
36 mm
Free body: Nut
=
Cx
15 mm
C y = 2.4 C x
ΣFx = 0: Bx = C x
(1)
ΣFy = 0: By = C y = 2.4 Cx
(2)
ΣFx = 0: Cx = Dx
ΣM = 14 N ⋅ m
Cx (0.027 m) = 14 N ⋅ m
Cx = 518.5 N
(a)
Eq. (1):
Bx = Cx = 518.5 N
Eq. (2):
By = Cy = 2.4(518.5 N) = 1244.4 N
(
B = Bx2 + By2
(b)
)
1/ 2
= (518.52 + 1244.42)1/ 2
B = 1.35 kN
Free body: Rod
Σ
= 0:
0.015 m
0.009 m
−M0 + (1244.4)(0.015) − (518.5)(0.009) = 0
M 0 = 14 N ⋅ m
,
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
985
0.4 m
0.24 m
1.5 m
B
0.4 m
0.48 m
PROBLEM 6.153
0.56 m
D
20 kN
A
C
E
0.24 m
0.24 m
0.36 m
F
G
H
0.44 m
J
0.2 m
The motion of the bucket of the front-end loader shown is
controlled by two arms and a linkage that are pin-connected
at D. The arms are located symmetrically with respect to
the central, vertical, and longitudinal plane of the loader;
one arm AFJ and its control cylinder EF are shown. The
single linkage GHDB and its control cylinder BC are located
in the plane of symmetry. For the position and loading
shown, determine the force exerted (a) by cylinder BC,
(b) by cylinder EF.
SOLUTION
0.4 m
20 kN
Free body: Bucket
0.44 m
ΣM J = 0: (20 kN)(0.4 m) − FGH (0.44 m) = 0
FGH = 18.18 kN
0.4 m
Free body: Arm BDH
ΣM D = 0: −(18.18 kN)(0.48 m) − FBC (0.4 m) = 0
0.48 m
FBC = −21.82 kN
FGH = 18.18 kN
FBC = 21.82 kN C
Free body: Entire mechanism
(Two arms and cylinders)
2.46 m
0.4 + 0.56 + 1.5 = 2.46
0.24 m
0.24 m
0.36 m
20 kN
1.3 m
Note: Two arms thus 2 FEF
0.36 m
1.3 m
β = 15.48°
tan β =
ΣM A = 0: (20 kN)(2.46 m) + FBC (0.24 m) + 2FEF cos b (0.48 m) = 0
(20 kN)(2.46 m) − (21.82 kN)(0.24 m) + 2FEF cos 15.48°(0.48 m) = 0
FEF = −47.52 kN
FEF = 47.52 kN C
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
956
0.3
0.4
0.48 0.12
0.32
PROBLEM 6.154
16 kN
A
0.32
D
0.3
The bucket of the front-end loader shown carries a 16-kN
load. The motion of the bucket is controlled by two
identical mechanisms, only one of which is shown.
Knowing that the mechanism shown supports one-half
of the 16-kN load, determine the force exerted (a) by
cylinder CD, (b) by cylinder FH.
B
0.16
C
E
0.3
F
Dimensions in meters
0.24
0.12
G
0.48
H
SOLUTION
16
kN
2
Free body: Bucket (One mechanism)
0.32
ΣM D = 0: (8 kN)(0.3 m) − FAB (0.32 m) = 0
0.3
FAB = 7.5 kN
Note: There are 2 identical support mechanisms.
= 7.5 kN
Free body: One arm BCE
0.16
0.3
0.16
tan β =
0.4
β = 21.8°
0.3
0.1
ΣM E = 0: (7.5 kN)(0.46 m) + FCD cos 21.8°(0.3 m) − FCD sin 21.8°(0.1 m) = 0
FCD = −14.29 kN
FCD = 14.3 kN C
Free body: Arm DFG
0.3 + 0.4 + 0.32 + 0.48 = 1.5 m
16
kN
2
0.48 m
0.12 m
0.6 m
0.6 m
ΣM G = 0: (8 kN)(1.5 m) + FFH sin 45°(0.48 m) − FFH cos 45°(0.12 m) = 0
FFH = −47.14 kN
FFH = 47.1 kN C
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.155
The telescoping arm ABC is used to provide an elevated
platform for construction workers. The workers and the
platform together have a mass of 200 kg and have a
combined center of gravity located directly above C. For the
position when   20, determine (a) the force exerted at B
by the single hydraulic cylinder BD, (b) the force exerted on
the supporting carriage at A.
SOLUTION
a  (5 m) cos 20  4.6985 m
Geometry:
b  (2.4 m) cos 20  2.2553 m
c  (2.4 m)sin 20  0.8208 m
d  b  0.5  1.7553 m
e  c  0.9  1.7208 m
tan  
e 1.7208

;   44.43
d 1.7553
Free body: Arm ABC:
We note that BD is a two-force member.
W  (200 kg)(9.81 m/s 2 )  1.962 kN
(a)
M A  0: (1.962 kN)(4.6985 m)  FBD sin 44.43(2.2553 m)  FBD cos 44.43(0.8208 m)  0
9.2185  FBD (0.9927)  0: FBD  9.2867 kN
FBD  9.29 kN
(b)
44.4 
Fx  0: Ax  FBD cos   0
Ax  (9.2867 kN)cos 44.43  6.632 kN
A x  6.632 kN
Fy  0: Ay  1.962 kN  FBD sin   0
Ay  1.962 kN  (9.2867 kN) sin 44.43  4.539 kN
A y  4.539 kN
A  8.04 kN
34.4 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.156
The telescoping arm ABC of Prob. 6.155 can be lowered
until end C is close to the ground, so that workers can easily
board the platform. For the position when  = 20,
determine (a) the force exerted at B by the single hydraulic
cylinder BD, (b) the force exerted on the supporting carriage
at A.
SOLUTION
a  (5 m) cos 20  4.6985 m
Geometry:
b  (2.4 m) cos 20  2.2552 m
c  (2.4 m)sin 20  0.8208 m
d  b  0.5  1.7553 m
e  0.9  c  0.0792 m
tan  
e 0.0792

;   2.584
d 1.7552
Free body: Arm ABC:
We note that BD is a two-force member.
W  (200 kg)(9.81 m/s2 )
W  1962 N  1.962 kN
(a)
M A  0: (1.962 kN)(4.6985 m)  FBD sin 2.584(2.2553 m)  FBD cos 2.584(0.8208 m)  0
9.2185  FBD (0.9216)  0 FBD  10.003 kN
FBD  10.00 kN
(b)
2.58 
Fx  0: Ax  FBD cos   0
Ax  (10.003 kN)cos 2.583  9.993 kN
A x  9.993 kN
Fy  0: Ay  1.962 kN  FBD sin   0
Ay  1.962 kN  (10.003 kN) sin 2.583  1.5112 kN
A y  1.5112 kN
A  10.11 kN
8.60 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
0.4 m
0.2 m
0.96 m
PROBLEM 6.157
0.3 m
0.2 m
0.7 m
A
0.24 m
C
B
0.72 m
D
E
0.32 m
1.2 m
0.32 m
0.36 m F
I
G
H
P
0.8 m
The motion of the backhoe bucket shown is
controlled by the hydraulic cylinders AD, CG, and
EF. As a result of an attempt to dislodge a portion
of a slab, a 10 kN force P is exerted on the bucket
teeth at J. Knowing that θ = 45°, determine the
force exerted by each cylinder.
0.2 m
J
0.32 m
0.16 m
SOLUTION
0.9
0.2
Free body: Bucket
ΣM H = 0
(Dimensions in meters)
1.2
0.2
0.32
4
3
FCG (0.2) + FCG (0.2) + P cos q(0.32) + P sin q (0.16) = 0
5
5
0.16
P
FCG = −
(0.32 cos q + 0.16 sin q )
0.28
(1)
Free body: Arm ABH and bucket
(Dimensions in meters)
0.96
0.2
0.24
0.72
1.72
0.84
4
3
ΣM B = 0: FAD (0.24) + FAD (0.2) + P cos q(1.72) − P sin q (0.84) = 0
5
5
FAD = −
P
(1.72 cos q − 0.84 sin q )
0.312
(2)
Free body: Bucket and arms IEB + ABH
Geometry of cylinder EF
0.32 m
0.8 m
β = 21.8°
0.32 m
tan β =
0.8 m
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958
PROBLEM 6.157 (Continued)
0.36 m
0.56 m
2.4 m
ΣM I = 0:
FEF cos b(0.36 m) + P cos q(0.56 m) − P sin q(2.4 m) = 0
FEF =
=
P (2.4 sin q − 0.56 cos q)
cos 21.8°(0.36)
P
(2.4 sin q − 0.56 cos q)
0.334
(3)
For P = 10 kN, q = 45°
Eq. (1):
FCG = −
10
(0.32 cos 45° + 0.16 sin 45°) = −12.12 kN
0.28
FCG = 12.12 kN C
Eq. (2):
FAD = −
10
(1.72 cos 45° – 0.84 sin 45°) = −19.94 kN
0.312
FAD = 19.94 kN C
Eq. (3):
FEF =
10
(2.4 sin 45° − 0.56 cos 45°) = +38.95 kN
0.334
FEF = 38.95 kN T
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959
0.4 m
0.2 m
0.96 m
PROBLEM 6.158
0.3 m
0.2 m
0.7 m
A
0.24 m
C
B
0.72 m
D
E
0.32 m
1.2 m
0.32 m
0.36 m F
I
0.8 m
G
H
P
0.2 m
J
0.32 m
Solve Problem 6.157 assuming that the 10 kN force P
acts horizontally to the right (θ = 0).
PROBLEM 6.157 The motion of the backhoe bucket
shown is controlled by the hydraulic cylinders AD,
CG, and EF. As a result of an attempt to dislodge a
portion of a slab, a 10 kN force P is exerted on the
bucket teeth at J. Knowing that θ = 45°, determine the
force exerted by each cylinder.
0.16 m
SOLUTION
0.9
0.2
Free body: Bucket
ΣM H = 0
(Dimensions in meters)
1.2
0.2
0.32
4
3
FCG (0.2) + FCG (0.2) + P cos q(0.32) + P sin q (0.16) = 0
5
5
0.16
P
FCG = −
(0.32 cos q + 0.16 sin q )
0.28
(1)
Free body: Arm ABH and bucket
(Dimensions in meters)
0.96
0.2
0.24
0.72
1.72
0.84
4
3
ΣM B = 0: FAD (0.24) + FAD (0.2) + P cos q(1.72) − P sin q (0.84) = 0
5
5
FAD = −
P
(1.72 cos q − 0.84 sin q )
0.312
(2)
Free body: Bucket and arms IEB + ABH
Geometry of cylinder EF
0.32 m
0.8 m
β = 21.8°
0.32 m
tan β =
0.8 m
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96
PROBLEM 6.158 (Continued)
0.36 m
0.56 m
2.4 m
ΣM I = 0:
FEF cos b(0.36 m) + P cos q(0.56 m) − P sin q(2.4 m) = 0
FEF =
=
P (2.4 sin q − 0.56 cos q)
cos 21.8°(0.36)
P
(2.4 sin q − 0.56 cos q)
0.334
(3)
For P = 10 kN, q = 0
Eq. (1):
FCG = −
10
(0.32 cos 0 + 0.16 sin 0) = −11.43 kN
0.28
FCG = 11.43 kN C
Eq. (2):
FAD = −
10
(1.72 cos 0 − 0.84 sin 0) = −55.13 kN
0.312
FAD = 55.13 kN C
Eq. (3):
FEF =
10
(2.4 sin 0 − 0.56 cos 0) = −16.77 kN
0.334
FEF = 16.77 kN C
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961
PROBLEM 6.159
The gears D and G are rigidly attached
to shafts that are held by frictionless
bearings. If rD  90 mm and rG  30
mm, determine (a) the couple M0 that
must be applied for equilibrium, (b) the
reactions at A and B.
SOLUTION
(a)
Projections on yz plane.
Free body: Gear G:
M G  0: 30 N  m  J (0.03 m)  0; J  1000 N
Free body: Gear D:
M D  0: M 0  (1000 N)(0.09 m)  0
M 0  90 N  m
(b)
M0  (90.0 N  m)i 
Gear G and axle FH:
M F  0: H (0.3 m)  (1000 N)(0.18 m)  0
H  600 N
Fy  0: F  600  1000  0
F  400 N
Gear D and axle CE:
M C  0: (1000 N)(0.18 m)  E (0.3 m)  0
E  600 N
Fy  0: 1000  C  600  0
C  400 N
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PROBLEM 6.159 (Continued)
Free body: Bracket AE:
Fy  0: A  400  400  0
A  0 
M A  0: M A  (400 N)(0.32 m)  (400 N)(0.2 m)  0
M A  48 N  m
M A  (48.0 N  m)i 
Free body: Bracket BH:
Fy  0: B  600  600  0
B0 
M B  0: M B  (600 N)(0.32 m)  (600 N)(0.2 m)  0
M B  72 N  m
M B  (72.0 N  m)i 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.160
In the planetary gear system shown, the radius of the central gear A is
a  18 mm, the radius of each planetary gear is b, and the radius of the
outer gear E is (a  2b). A clockwise couple of magnitude MA  10 N  m
is applied to the central gear A and a counterclockwise couple of
magnitude MS  50 N  m is applied to the spider BCD. If the system is
to be in equilibrium, determine (a) the required radius b of the
planetary gears, (b) the magnitude ME of the couple that must be
applied to the outer gear E.
SOLUTION
FBD Central Gear:
By symmetry,
F1  F2  F3  F
M A  0: 3(rA F )  10 N  m  0,
M C  0: rB ( F  F4 )  0,
FBD Gear C:
F
10
Nm
3rA
F4  F
Fx  0: Cx  0
Fy  0: C y  2 F  0,
C y  2 F
Gears B and D are analogous, each having a central force of 2F.
M A  0: 50 N  m  3(rA  rB )2F  0
FBD Spider:
50 N  m  3(rA  rB )
20
Nm  0
rA
rA  rB
r
 2.5  1  B ,
rA
rA
rB  1.5rA
Since rA  18 mm,
FBD Outer Gear:
rB  27.0 mm 
(a)
M A  0: 3(rA  2rB ) F  M E  0
3(18 mm  54 mm)
(b)
10 N  m
 ME  0
54 mm
M E  40.0 N  m
Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 6.161*
Two shafts AC and CF, which lie in the vertical xy plane, are connected by a universal joint at C. The bearings
at B and D do not exert any axial force. A couple of magnitude 50 N ⋅ m (clockwise when viewed from the
positive x axis) is applied to shaft CF at F. At a time when the arm of the crosspiece attached to shaft CF is
horizontal, determine (a) the magnitude of the couple that must be applied to shaft AC at A to maintain
equilibrium, (b) the reactions at B, D, and E. (Hint: The sum of the couples exerted on the crosspiece must be
zero.)
y
0.4 m
0.6 m
30°
z
C
50 N.m
A
D
B
F
0.5 m
E
x
SOLUTION
We recall from Figure 4.10, that a universal joint exerts on members it connects a force of unknown direction
and a couple about an axis perpendicular to the crosspiece.
Free body: Shaft DF
ΣM x= 0: MC cos 30° − 50 N ⋅ m = 0
MC = 57.735 N ⋅ m
Free body: Shaft BC
We use here x′, y ′, z with x′ along BC
Σ
= 0:
A
57.735 N ⋅ m
0.5 m
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965
PROBLEM 6.161* (Continued)
Equate coefficients of unit vectors to zero:
i:
MA − 57.735 N ⋅ m = 0
MA = 57.735 N ⋅ m
j:
Bz = 0
MA = 57.7 N ⋅ m
k:
By = 0
B=0
ΣF = 0:
B + C = 0,
since B = 0,
B=0
C=0
Return to free body of shaft DF
(Note that C = 0 and MC = 57.735 N ⋅ m)
ΣM D = 0
(57.735 N ⋅ m)(cos30°i + sin30°j) − (50 N ⋅ m)i
+(0.6 m)i × (Exi + Eyj + Ezk) = 0
(50 N ⋅ m)i + (28.868 N ⋅ m)j − (50 N ⋅ m)i
+(0.6 m)Eyk − (0.6 m)Ezj = 0
Equate coefficients of unit vectors to zero:
j:
28.868 N ⋅ m − (0.6 m)Ez = 0 Ez = 48.1 N
k:
Ey = 0
ΣF = 0: C + D + E = 0
0 + Dy j + Dz k + Ex i + (48.1 N)k = 0
i:
Ex = 0
j:
Dy = 0
k:
Dz + 48.1 N = 0 Dz = − 48.1 N
B=0
Reactions are:
D = − (48.1 N)k
E = (48.1 N)k
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966
PROBLEM 6.162*
Solve Problem 6.161 assuming that the arm of the crosspiece attached to shaft CF is vertical.
PROBLEM 6.161 Two shafts AC and CF, which lie in the vertical xy plane, are connected by a universal
joint at C. The bearings at B and D do not exert any axial force. A couple of magnitude 50 N · m (clockwise
when viewed from the positive x axis) is applied to shaft CF at F. At a time when the arm of the crosspiece
attached to shaft CF is horizontal, determine (a) the magnitude of the couple that must be applied to shaft AC
at A to maintain equilibrium, (b) the reactions at B, D, and E. (Hint: The sum of the couples exerted on the
crosspiece must be zero.)
y
0.4 m
0.6 m
C
30°
z
50 N . m
A
B
D
0.5 m
F
E
x
SOLUTION
Free body: Shaft DF
ΣM x = 0: MC − 50 N ⋅ m = 0
MC = 50 N ⋅ m
Free body: Shaft BC
We resolve −(50 N ⋅ m)i into components along x′ and y′ axes:
−MC = −(50 N ⋅ m)(cos30°i′ + sin30°j′)
43.3 N ⋅ m
25 N ⋅ m
0.5 m
0.5 m
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967
PROBLEM 6.162* (Continued)
Equate to zero coefficients of unit vectors:
i ′: MA − 43.3 N ⋅ m = 0
j′: −25 N ⋅ m − (0.5 m)Bz = 0
MA = 43.3 N ⋅ m
Bz = −50 N
k : B y′ = 0
B = −(50 N)k
Reactions at B:
ΣF = 0: B − C = 0
−(50 N)k − C = 0
C = −(50 N)k
Return to free body of shaft DF:
ΣMiD = 0: (0.6 m)i × (Exi + Eyj + Ezk) − (0.4 m)i × (−50 N)k
−(50 N ⋅ m)i + (50 N ⋅ m)i = 0
(0.6 m)Eyk − (0.6 m)Ezj − (20 N ⋅ m)j = 0
k : Ey = 0
j: − (0.6 m)Ez − 20 N ⋅ m = 0
Ez = −33.3 N
ΣF = 0: C + D + E = 0
−(50 N)k + Dyj + Dzk + Exi − (33.3 N)k = 0
i : Ex = 0
k : −50 N − 33.3 N + Dz = 0
Dz = 83.3 N
B = −(50 N)k
Reactions are:
D = (83.3 N)k
E = −(33.3 N)k
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968
PROBLEM 6.163*
The large mechanical tongs shown are used to grab and lift a
thick 7500-kg steel slab HJ. Knowing that slipping does not
occur between the tong grips and the slab at H and J, determine
the components of all forces acting on member EFH. (Hint:
Consider the symmetry of the tongs to establish relationships
between the components of the force acting at E on EFH and the
components of the force acting at D on DGJ.)
SOLUTION
Free body: Pin A:
T  W  mg  (7500 kg)(9.81 m/s 2 )  73.575 kN
Fx  0: ( FAB ) x  ( FAC ) x
1
Fy  0: ( FAB ) y  ( FAC ) y  W
2
( FAC ) x  2( FAC ) y  W
Also,
Free body: Member CDF:
1
 M D  0: W (0.3)  W (2.3)  Fx (1.8)  Fy (0.5 m)  0
2
or
1.8 Fx  0.5 Fy  1.45W
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(1)
PROBLEM 6.163* (Continued)
Fx  0: Dx  Fx  W  0
Ex  Fx  W
or
(2)
1
Fy  0: Fy  D y  W  0
2
1
E y  Fy  W
2
or
(3)
Free body: Member EFH:
1
M E  0: Fx (1.8)  Fy (1.5)  H x (2.3)  W (1.8 m)  0
2
1.8 Fx  1.5 Fy  2.3H x  0.9W
or
(4)
Fx  0: Ex  Fx  H x  0
Ex  Fx  H x
or
2Fx  H x  W
Subtract Eq. (2) from Eq. (5):
Subtract Eq. (4) from 3  (1):
3.6Fx  5.25W  2.3H x
Add Eq. (7) to 2.3  Eq. (6):
8.2Fx  2.95W
Fx  0.35976W
(5)
(6)
(7)
(8)
Substitute from Eq. (8) into Eq. (1):
(1.8)(0.35976W )  0.5 Fy  1.45W
0.5 Fy  1.45W  0.64756W  0.80244W
Fy  1.6049W
(9)
Substitute from Eq. (8) into Eq. (2):
Ex  0.35976W  W ; Ex  1.35976W
Substitute from Eq. (9) into Eq. (3):
1
E y  1.6049W  W
2
From Eq. (5):
H x  Ex  Fx  1.35976W  0.35976W  1.71952W
Recall that
1
Hy  W
2
E y  2.1049W
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PROBLEM 6.163* (Continued)
Since all expressions obtained are positive, all forces are directed as shown on the free-body diagrams.
Substitute
W  73.575 kN
Ex  100.0 kN
E y  154.9 kN 
Fx  26.5 kN
Fy  118.1 kN 
H x  126.5 kN
H y  36.8 kN 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.164
Using the method of joints, determine the force in each member of
the truss shown. State whether each member is in tension or
compression.
SOLUTION
Reactions:
M C  0: A x  16 kN
Fy  0: A y  9 kN
Fx  0:
C  16 kN
Joint E:
FBE FDE 3 kN


5
4
3
FBE  5.00 kN T 
FDE  4.00 kN C 
Joint B:
Fx  0:
4
(5 kN)  FAB  0
5
FAB  4 kN
FAB  4.00 kN T 
3
Fy  0:  6 kN  (5 kN)  FBD  0
5
FBD  9 kN
FBD  9.00 kN C 
Joint D:
Fy  0:  9 kN 
3
FAD  0
5
FAD  15 kN FAD  15.00 kN T 
Fx  0:  4 kN 
4
(15 kN)  FCD  0
5
FCD  16 kN FCD  16.00 kN C 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.165
Using the method of joints, determine the force in each
member of the double-pitch roof truss shown. State whether
each member is in tension or compression.
SOLUTION
Free body: Truss:
M A  0: H (18 m)  (2 kN)(4 m)  (2 kN)(8 m)  (1.75 kN)(12 m)
 (1.5 kN)(15 m)  (0.75 kN)(18 m)  0
H  4.50 kN
Fx  0: Ax  0
Fy  0: Ay  H  9  0
Ay  9  4.50
A y  4.50 kN
Free body: Joint A:
FAC 3.50 kN

2
1
5
FAB  7.8262 kN C
FAB

FAB  7.83 kN C 
FAC  7.00 kN T 

Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.165 (Continued)
Free body: Joint B:
Fx  0:
2
5
FBD 
2
5
(7.8262 kN) 
1
2
FBC  0
FBD  0.79057 FBC  7.8262 kN
or
Fy  0:
1
5
FBD 
1
5
(7.8262 kN) 
1
2
(1)
FBC  2 kN  0
FBD  1.58114FBC  3.3541
or
(2)
Multiply Eq. (1) by 2 and add Eq. (2):
3FBD  19.0065
FBD  6.34 kN C 
FBD  6.3355 kN
Subtract Eq. (2) from Eq. (1):
2.37111FBC  4.4721
FBC  1.886 kN C 
FBC  1.8861 kN
Free body: Joint C:
2
Fy  0:
5
FCD 
1
2
(1.8861 kN)  0
FCD  1.4911 kN
Fx  0: FCE  7.00 kN 
1
2
FCD  1.491 kN T 
(1.8861 kN) 
1
5
(1.4911 kN)  0
FCE  5.000 kN
FCE  5.00 kN T 
Free body: Joint D:
Fx  0:
2
5
FDF 
1
2
FDE 
2
5
(6.3355 kN) 
1
5
(1.4911 kN)  0
FDF  0.79057 FDE  5.5900 kN
or
Fy  0:
1
5
FDF 
1
2
FDE 
1
5
(6.3355 kN) 
(1)
2
5
(1.4911 kN)  2 kN  0
FDF  0.79057 FDE  1.1188 kN
or
Add Eqs. (1) and (2):
2FDF  6.7088 kN
FDF  3.3544 kN
Subtract Eq. (2) from Eq. (1):
(2)
FDF  3.35 kN C 
1.58114FDE  4.4712 kN
FDE  2.8278 kN
FDE  2.83 kN C 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.165 (Continued)
Free body: Joint F:
Fx  0:
1
FFG 
2
2
(3.3544 kN)  0
5
FFG  4.243 kN
Fy  0: FEF  1.75 kN 
1
5
FFG  4.24 kN C 
(3.3544 kN) 
FEF  2.750 kN
1
2
(4.243 kN)  0
FEF  2.75 kN T 
Free body: Joint G:
Fx  0:
1
FGH 
2
1
2
FEG 
1
2
(4.243 kN)  0
FGH  FEG  4.243 kN
or
Fy  0: 
1
2
FGH 
1
2
FEG 
1
2
(1)
(4.243 kN)  1.5 kN  0
FGH  FEG  6.364 kN
or
Add Eqs. (1) and (2):
2FGH  10.607
FGH  5.303
Subtract Eq. (1) from Eq. (2):
(2)
FGH  5.30 kN C 
2FEG  2.121 kN
FEG  1.0605 kN
FEG  1.061 kN C 
Free body: Joint H:
FEH 3.75 kN

1
1
We can also write
FGH
2

3.75 kN
1
FEH  3.75 kN T 
FGH  5.30 kN C (Checks)
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
0.9 kN
1.8 kN
A
PROBLEM 6.166
1.8 kN
0.9 kN
B
C
2.25 m
E
3m
I
8m
K
2m
D
H
G
F
A stadium roof truss is loaded as shown. Determine the force
in members AB, AG, and FG.
3.5 m
3.5 m
J
L
2m
SOLUTION
We pass a section through members AB, AG, and FG, and use the free body shown.
ΣMG = 0:
10
FAB (1.575 m) − (1.8 kN)(3.5 m) − (0.9 kN)(7 m) = 0
10.25
FAB = +8.20 kN
ΣMD = 0:
−
FAB = 8.20 kN
T
FAG = 4.50 kN
T
FFG = 11.60 kN
C
2.25
FAG (7 m) + (1.8 kN)(7 m) + (1.8 kN)(3.5 m) = 0
3.75
FAG = +4.50 kN
MA = 0: −FFG (2.25 m) − (1.8 kN)(3 m) − (1.8 kN)(6.5 m) − (0.9 kN)(10 m) = 0
FFG = −11.60 kN
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828
0.9 kN
PROBLEM 6.167
1.8 kN
1.8 kN
A
0.9 kN
B
C
2.25 m
E
3m
I
8m
K
2m
D
H
G
F
A stadium roof truss is loaded as shown. Determine the force
in members AE, EF, and FJ.
3.5 m
3.5 m
J
L
2m
SOLUTION
We pass a section through members AE, EF, and FJ, and use the free body shown.
ΣMF = 0:
2
2 + 2.252
2
FAE (2.25) − (1.8 kN)(3 m) − (1.8 kN)(6.5 m) − (0.9 kN)(10 m) = 0
FAE = +17.46 kN
ΣMA = 0:
FAE = 17.46 kN
T
−FEF (2.25 m) − (1.8 kN)(3 m) − (1.8 kN)(6.5 m) − (0.9 kN)(10 m) = 0
FEF = −11.60 kN
FEF = 11.60 kN
C
ΣME = 0: −FFJ (2 m) − (0.9 kN)(2 m) − (1.8 kN)(5 m) − (1.8 kN)(8.5 m) − (0.9 kN)(12 m) = 0
FFJ = 18.45 kN
FFJ = −18.45 kN
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829
C
PROBLEM 6.168
Determine the components of all forces acting on member ABD
of the frame shown.
SOLUTION
Free body: Entire frame:
3m
3 kN
4.5 kN
4m
2m
ΣM D = 0: (3 kN) ( 6 m) + (4.5 kN)(2 m) − E (3 m) = 0
E = +9 kN
E = 9 kN

D x = 9 kN

ΣFx = 0: Dx − 9 kN = 0
ΣFy = 0: Dy − 3 kN − 4.5 kN = 0
D y = 7.5 kN 
Free body: Member ABD:
We note that BC is a two-force member and that B is directed along BC.
ΣM A = 0: (7.5 kN)(8 m) − (9 kN)(3 m) − B(4 m) = 0
7.5 kN
4m
B = +8.25 kN
B = 8.25 kN 
4m
9 kN
3m
ΣFx = 0: Ax + 9 kN = 0
= −9 kN
EA
x
A x = 9 kN

ΣFy = 0: Ay + 7.5 kN − 8.25 kN = 0
Ay = 0.75 kN
A y = 0.75 kN 
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PROBLEM 6.169
Determine the components of the reactions at A and E if the frame is loaded
by a clockwise couple of magnitude 36 N · m applied (a) at B, (b) at D.
SOLUTION
Free body: Entire frame:
The following analysis is valid for both parts (a) and (b) since the point of application of the couple is
immaterial.
M E  0:  36 N  m  Ax (0.4 m)  0
Ax   90 N
A x  90.0 N
Fx  0:  90 +E x  0
E x  90 N
E x  90.0 N
Fy  0: Ay  E y  0
(a)
(1)
Couple applied at B.
Free body: Member CE:
AC is a two-force member. Thus, the reaction at E must be directed along EC.
Ey
90 N
From Eq. (1):

0.24 m
; E y  45.0 N
0.48 m
Ay  45 N  0
Ay   45 N A y  45.0 N
Thus, reactions are
(b)
A x  90.0 N
, A y  45.0 N 
E x  90.0 N
,
E y  45.0 N 
Couple applied at D.
Free body: Member AC:
AC is a two-force member. Thus, the reaction at A must be directed along AC.
Ay
90 N

0.16 m
; A y  30 N
0.48 m
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PROBLEM 6.169 (Continued)
From Eq. (1):
Ay  E y  0
30 N  E y  0
E y   30 N E y  30 N
Thus, reactions are
A x  90.0 N
E x   90.0 N
,
A y  30.0 N 
, E y  30.0 N 
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PROBLEM 6.170
Knowing that the pulley has a radius of 50 mm, determine the
components of the reactions at B and E.
SOLUTION
Free body: Entire assembly:
M E  0 :  (300 N)(350 mm)  Bx (150 mm)  0
Bx  700 N
B x  700 N
Fx  0:  700 N  E x  0
E x  700 N
E x  700 N
Fy  0: By  E y  300 N  0
(1)
Free body: Member ACE:
M C  0: (700 N)(150 mm)  (300 N)(50 mm)  E y (180 mm)  0
E y  500 N
From Eq. (1):
E y  500 N
By  500 N  300 N  0
By  200 N
B y  200 N
Thus, reactions are
B x  700 N
, B y  200 N 
E x  700 N
, E y  500 N 
Copyright © McGraw-Hill Education. Permission required for reproduction or display.
PROBLEM 6.171
A
6 cm
40 N
C
B
For the frame and loading shown, determine the components of the
forces acting on member CFE at C and F.
4 cm
D
E
F
5 cm
4 cm
4 cm
SOLUTION
Free body: Entire frame
ΣM D = 0: (40 N)(13 cm) + Ax (10 cm) = 0
Ax = −52 N,
A x = 52 N
Free body: Member ABF
ΣM B = 0: −(52 N)(6 cm) + Fx (4 cm) = 0
Fx = +78 N
Free body: Member CFE
Fx = 78.0 N
From above:
ΣM C = 0: (40 N)(9 cm) − (78 N)(4 cm) − Fy (4 cm) = 0
Fy = +12 N
Fy = 12.00 N
ΣFx = 0: Cx − 78 N = 0
Cx = +78 N
ΣFy = 0: − 40 N + 12 N + C y = 0; C y = +28 N
C x = 78.0 N
C y = 28.0 N
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885
8 cm
PROBLEM 6.172
8 cm
A
B
6 cm
For the frame and loading shown, determine the reactions at A, B,
D, and E. Assume that the surface at each support is frictionless.
C
D
E
6 cm
30°
1 kN
SOLUTION
Free body: Entire frame
ΣFx = 0: A − B + (1000 N)sin30° = 0
A − B + 500 = 0
(1)
ΣFy = 0: D + E − (1000 N)cos30° = 0
D + E − 866 = 0
(2)
Free body: Member ACE
ΣM C = 0: −A(6 cm) + E(8 cm) = 0
E=
3
A
4
(3)
Free body: Member BCD
ΣM C = 0: −D(8 cm) + B(6 cm) = 0
D=
3
B
4
(4)
Substitute E and D from (3) and (4) into (2):
3
3
A + B −866 = 0
4
4
A + B − 1154.7 = 0
(5)
(1)
A − B + 500 = 0
(5) + (6)
2A − 654.7 = 0
A = 327.4 N
A = 327 N
(5) − (6)
2B − 1654.7 = 0
B = 827.4 N
B = 827 N
(4)
D=
3
(827.4)
4
D = 620.6 N
D = 621 N
(3)
E=
3
(327.4)
4
E = 245.6 N
E = 246 N
(6)
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981
PROBLEM 6.173
Water pressure in the supply system exerts a downward force of 135 N
on the vertical plug at A. Determine the tension in the fusible link DE
and the force exerted on member BCE at B.
SOLUTION
Free body: Entire linkage:
 Fy  0: C  135  0
C   135 N
Free body: Member BCE:
Fx  0: Bx  0
M B  0: (135 N)(6 mm)  TDE (10 mm)  0
TDE  81.0 N 
Fy  0: 135  81  By  0
By   216 N
B  216 N 
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PROBLEM 6.174
A couple M of magnitude 1.5 kN  m is applied to the crank of the engine system shown. For each of the
two positions shown, determine the force P required to hold the system in equilibrium.
SOLUTION
(a)
FBDs:
Note:
tan  

FBD whole:
FBD piston:
50 mm
175 mm
Dimensions in mm
2
7
M A  0: (0.250 m)C y  1.5 kN  m  0 C y  6.00 kN
Fy  0: C y  FBC sin   0 FBC 
Cy
sin 

6.00 kN
sin
Fx  0: FBC cos  P  0
P  FBC cos  
6.00 kN
 7 kips
tan 
P  21.0 kN
Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 6.174 (Continued)
(b)
FBDs:
Dimensions in mm
2
as above
7
Note:
tan  
FBD whole:
M A  0: (0.100 m)C y  1.5 kN  m  0 C y  15 kN
Fy  0: C y  FBC sin   0 FBC 
Fx  0:
Cy
sin 
FBC cos  P  0
P  FBC cos  
Cy
tan 

15 kN
P  52.5 kN
2/7
Copyright © McGraw-Hill Education. Permission required for reproduction or display.

PROBLEM 6.175
The compound-lever pruning shears shown can be adjusted
by placing pin A at various ratchet positions on blade ACE.
Knowing that 1.2 kN vertical forces are required to complete
the pruning of a small branch, determine the magnitude P of
the forces that must be applied to the handles when the shears
are adjusted as shown.
SOLUTION
We note that AB is a two-force member.
( FAB ) x ( FAB ) y
=
13 mm 11 mm
11
( FAB ) y = ( FAB ) x
13
11 mm
13 mm
(1)
Free body: Blade ACE:
28 mm
10 mm
1200 N
32 mm
ΣM C = 0: (1200 N)(0.032 m) − (FAB )x (0.01 m) − (FAB )y (0.028 m) = 0
Use Eq. (1):
( FAB ) x (0.01 m) +
11
(FAB )x (0.028 m) = 38.4 N⋅m
13
0.03369( FAB )x = 38.4
( FAB ) y =
( FAB ) x = 1139.8 N
11
(1139.8 N)
13
( FAB ) y = 964.4 N
Free body: Lower handle:
964.4 N
1139.8 N
5 mm
15 mm
70 mm
ΣM D = 0: (964.4 N)(0.015 m) − (1139.8 N)(0.005 m) − P (0.07 m) = 0
P = 125.2 N 
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PROBLEM 6.F1
For the frame and loading shown, draw the free-body diagram(s) needed
to determine the force in member BD and the components of the reaction
at C.
SOLUTION
We note that BD is a two-force member.
Free body: ABC


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PROBLEM 6.F2
For the frame and loading shown, draw the free-body diagram(s)
needed to determine the components of all forces acting on
member ABC.
SOLUTION
We note that BE is a two-force member.
Free body: Frame
Note:
Sum forces in y about A to determine Ay.
Sum moments about E to determine Ax.
Free body: ABC

Note:
Sum moments about C to determine FBE.
Sum forces in x to determine Cx.
Sum forces in y to determine Cy.
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PROBLEM 6.F3
Draw the free-body diagram(s) needed to determine all the forces
exerted on member AI if the frame is loaded by a clockwise
couple of magnitude 600 N · m applied at point D.
SOLUTION
We note that AB, BC, and FG are two-force members.
Free body: Frame
Note:
Sum moments about H to determine Iy.
Free body: AI
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PROBLEM 6.F3 (Continued)
Note: Sum forces in y to determine FAB
Sum moments about G to determine FBC
Sum forces in x to determine FFG

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PROBLEM 6.F4
Knowing that the pulley has a radius of 0.5 m, draw the
free-body diagram(s) needed to determine the components
of the reactions at A and E.
SOLUTION
Free body: Frame
Note: Sum moments about E to determine Ay.
Sum forces in y to determine Ey.
Sum forces in x to determine Ax and Ex.
Free body: ABC
Note: Sum moments about C to determine Ax.
Use previously obtained relation between Ax and Ex to determine Ex.

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PROBLEM 6.F5
An 84-N force is applied to the toggle vise at C. Knowing that  = 90,
draw the free-body diagram(s) needed to determine the vertical force
exerted on the block at D.
SOLUTION
We note that BD is a two-force member.
Free body: ABC
Note: Sum moments about A to determine FBD.
Free body: D
Note: Sum forces in y to determine Dy.
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PROBLEM 6.F6
For the system and loading shown, draw the free-body diagram(s)
needed to determine the force P required for equilibrium.
SOLUTION
We note that AB and BD are two-force members.
Free body: CDE
Note: Sum moments about C to determine FBD.
Free body: Pin B
Note: Sum forces in x to determine P.

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PROBLEM 6.F7
A small barrel weighing 300 N is lifted by a pair of tongs as shown.
Knowing that a = 125 mm, draw the free-body diagram(s) needed
to determine the forces exerted at B and D on tong ABD.
SOLUTION
We note that BC is a two-force member.
Free body: ABD
Note: Sum moments about D to determine FBC.
Sum forces in x to determine Dx .
Sum forces in y to determine Dy.

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PROBLEM 6.F8
The position of member ABC is controlled by the
hydraulic cylinder CD. Knowing that θ  30°,
draw the free-body diagram(s) needed to
determine the force exerted by the hydraulic
cylinder on pin C, and the reaction at B.
SOLUTION
We note that CD is a two-force member.
Free body: ABC
Note: To find , consider geometry of triangle BCD:
Law of cosines:
(CD)2  (0.5)2  (1.5)2  2(0.5)(1.5) cos 60
CD  1.32288 m
Law of sines:
sin q
sin 60

0.5 m 1.32288 m
  19.1066 
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