N N N N FCA = 833.33 N N N N N N FCD = 333.33 N N N

446

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•6–49. Determine the force in members KJ, KC, and BCof the Howe truss, and state if the members are in tension orcompression.

2 kN3 kN

4 kN5 kN

4 kN

6 kN

5 kN

BA

C D E FG

H

I

J

K

L

2 m

4 m

2 m 2 m 2 m 2 m 2 m

06b Ch06b 446-490.indd 44606b Ch06b 446-490.indd 446 6/12/09 8:51:41 AM6/12/09 8:51:41 AM

447


6–50. Determine the force in each member of the trussand state if the members are in tension or compression. Set

, .P2 = 10 kNP1 = 20 kN

AG F

E

DCB

P2P1

1.5 m 1.5 m 1.5 m 1.5 m

2 m


448



449


6–51. Determine the force in each member of the trussand state if the members are in tension or compression. Set

, .P2 = 20 kNP1 = 40 kN

AG F

E

DCB

P2P1

1.5 m 1.5 m 1.5 m 1.5 m

2 m


450



451


*6–52. Determine the force in members KJ, NJ, ND, andCD of the K truss. Indicate if the members are in tension orcompression. Hint: Use sections aa and bb.

1800 lb

15 ft

15 ft

20 ft 20 ft 20 ft 20 ft 20 ft

A B

I H

20 ft

L

M N O P

GFEDC

1500 lb1200 lb

a b

JKa b

6 kN7.5 kN 9 kN

4.5 m

4.5 m

6 m 6 m 6 m 6 m 6 m 6 m

6 m 6 m 6 m 18 m

9 m

6 m 6 m

9 m

6 m 6 m

7.5 kN

7.5 kN

7.5 kN

6 kN 9 kN

6 kN

6 kN

Ay = 14.5 kN

Ay = 14.5 kN

Support Reactions :

�

+ ΣMG = 0; 6 (30) + 7.5 (24) + 9 (18) – Ay (36) = 0

Ay = 14.5 kN

+→ ΣFx = 0; Ax = 0

Method of Sections : From section a – a, FKJ and FCD can be obtained directly by summing moment about points C and K respectively.

�

+ ΣMC = 0; FKI (9) + 6 (6) – 14.5 (12) = 0

FKJ = 15.33 kN (C) Ans

�

+ ΣMK = 0; FCD (9) + 6 (6) – 14.5 (12) = 0

FCD = 15.33 kN (T) Ans

From sec b – b, summing forces along x and y axes yields

+→ ΣFx = 0; FND

4

5

⎛⎝⎜

⎞⎠⎟

– FNJ 4

5

⎛⎝⎜

⎞⎠⎟

+ 15.33 – 15.33 = 0

FND = FNJ [1]

+↑ΣFy = 0; 14.5 – 6 – 7.5 – FND 3

5

⎛⎝⎜

⎞⎠⎟

– FNJ 3

5

⎛⎝⎜

⎞⎠⎟

= 0

FND + FNJ = 1.6667 kN [2]

Solving Eqs. [1] and [2] yields

FND = 0.833 kN (T)

FNJ = 0.833 kN (C) Ans


452


•6–53. Determine the force in members JI and DE ofthe K truss. Indicate if the members are in tension orcompression.

1800 lb

15 ft

15 ft

20 ft 20 ft 20 ft 20 ft 20 ft

A B

I H

20 ft

L

M N O P

GFEDC

1500 lb1200 lb

a b

JKa b

6 kN7.5 kN 9 kN

4.5 m

4.5 m

6 m 6 m 6 m 6 m 6 m 6 m

6 m 6 m 6 m 18 m

7.5 kN6 kN 9 kN

12 m

9 m

Gy = 8 kN

Support Reactions :

�

+ ΣMA = 0; Gy (36) – 9 (18) – 7.5 (12) – 6 (6) = 0

Gy = 8 kN

Method of Sections :

�

+ ΣMg = 0; 8 (12) – FJI (9) = 0

FJI = 10.67 kN (C) Ans

�

+ ΣMl = 0; 8 (12) – FDE (9) = 0

FDg = 10.67 kN (T) Ans


453


6–54. The space truss supports a force. Determine the force in

each member, and state if the members are in tension orcompression.

F = 5-500i + 600j + 400k6 lb

A

B

C

D

x

y

z

F

8 ft

6 ft

6 ft

6 ft

6 ft

0.8 m

0.6 m

0.6 m

0.6 m0.6 m

F = {–500i + 600j + 400k} N. Determine the force in


454


400 N

500 N

600 N

FCA = 833.33 N

FCD = 333.33 N

FAD = 353.55 N

N N

NN

N

N

N N

N

N

N


455


6–55. The space truss supports a force. Determine the force in each

member, and state if the members are in tension orcompression.

F = 5600i + 450j - 750k6 lb

A

B

C

D

x

y

z

F

8 ft

6 ft

6 ft

6 ft

6 ft

FCD = 406.25 N

FCA = 1000 N

FAD = 424.26 N

750 N

450 N600 N

0.8 m

0.6 m

0.6 m

0.6 m0.6 m

F = {600i + 450j – 750k} N. Determine the force in each

N N

N NN N

N N

N N

N

N

N

N


456


*6–56. Determine the force in each member of the spacetruss and state if the members are in tension orcompression. The truss is supported by ball-and-socketjoints at A, B, and E. Set . Hint: The supportreaction at E acts along member EC. Why?

F = 5800j6 N

FD

A

z

2 m

x

y

B

C

E

5 m

1 m

2 m1.5 m


457


•6–57. Determine the force in each member of the spacetruss and state if the members are in tension orcompression. The truss is supported by ball-and-socketjoints at A, B, and E. Set . Hint: Thesupport reaction at E acts along member EC. Why?

F = 5-200i + 400j6 N

FD

A

z

2 m

x

y

B

C

E

5 m

1 m

2 m1.5 m


458


6–58. Determine the force in members BE, DF, and BC ofthe space truss and state if the members are in tension orcompression.

2 m

2 m

2 m

E

A

3 m

F

D

C

B

{�2k} kN

{�2k} kN

2 m


459


6–59. Determine the force in members AB, CD, ED, andCF of the space truss and state if the members are in tensionor compression.

2 m

2 m

2 m

E

A

3 m

F

D

C

B

{�2k} kN

{�2k} kN

2 m


460


*6–60. Determine the force in the members AB, AE, BC,BF, BD, and BE of the space truss, and state if the membersare in tension or compression.

F

E

D

x

z

y

C

BA

4 ft

4 ft2 ft

2 ft300 lb

600 lb

400 lb

4 ft1 m

0.5 m1 m

0.5 m

1 m

2 kN

1.5 kN

3 kN

1.5 kN

3 kN

1 m

1 m

0.5 m

1 m

0.5 m

1 m

1 m

0.5 m

1 m

1 m

2 kN

FAB = 1.5 kN

1.5 m

Method of Joints: In this case, there is no need to compute the support reactions. We will begin by analyzing the equilibrium of joint A, and then that of joints C and B.

Joint A: From the free – body diagram in Fig. a,

ΣFz = 0; FAE 1

1.5

⎛⎝⎜

⎞⎠⎟

– 1.5 = 0

FAE = 2.25 kN (T) Ans.

ΣFx = 0; 3 – 2.25 1

1.5

⎛⎝⎜

⎞⎠⎟

– FAD 1

1.25

⎛

⎝⎜

⎞

⎠⎟ = 0

FAD = 1.677 kN (T) Ans.

ΣFy = 0; FAB – 1.677 0.5

1.25

⎛

⎝⎜

⎞

⎠⎟ – 2.25

0.5

1.5

⎛⎝⎜

⎞⎠⎟

= 0

FAB = 1.50 kN (T) Ans.

Joint C: From the free – body diagram of the joint in Fig. b, notice that FCD, FCF, and Cy lie in the y – z plane (shown shaded). Thus, if we write the force equation of equilibrium along the x axis, we have

ΣFx = 0; FBC 1

1.25

⎛

⎝⎜

⎞

⎠⎟ = 0

FBC = 0 Ans.

Joint B: From the free – body diagram in Fig. c,

ΣFx = 0; –FBF 1

1.5

⎛⎝⎜

⎞⎠⎟

– FBE 1

4.25

⎛

⎝⎜

⎞

⎠⎟ – FBD

1

3.25

⎛

⎝⎜

⎞

⎠⎟ = 0 (1)

ΣFy = 0; FBF 0.5

1.5

⎛⎝⎜

⎞⎠⎟

– FBE 1.5

4.25

⎛

⎝⎜

⎞

⎠⎟ – FBD

1.5

3.25

⎛

⎝⎜

⎞

⎠⎟ – 1.5 = 0 (2)

ΣFz = 0; FBE 1

4.25

⎛

⎝⎜

⎞

⎠⎟ + FBF

1

1.5

⎛⎝⎜

⎞⎠⎟

– 2 = 0 (3)

Solving Eqs. (1) through (3) yields

FBF = 1.125 kN (T) Ans.

FBE = 2.577 kN (T) Ans.

FBD = –3.606 kN = 3.606 kN (C) Ans.


461


•6–61. Determine the force in the members EF, DF, CF,and CD of the space truss, and state if the members are intension or compression.

F

E

D

x

z

y

C

BA

4 ft

4 ft2 ft

2 ft300 lb

600 lb

400 lb

4 ft1 m

0.5 m1 m

0.5 m

1 m

2 kN

1.5 kN

3 kN

1 m

0.5 m

1 m

0.5 m

1 m2 m

1 m

Cy = 3.25 kN

Fz = 3.5 kN

Fz = 0.75 kN

Fy = 3.25 kN

0.5 m

0.5 m

1 m1 m

1 m3 kN

1.5 kN

2 kN

Support Reactions: In this case, it is easier to compute the support reactions first. From the free – body diagram of the truss, Fig. a, and writing the equations of equilibrium,

ΣMy′ = 0; 2 (1) + 1.5 (1) – 3 (1) – Dx (1) = 0

Dx = 0.5 kN

ΣMx′ = 0; 2 (0.5) + 1.5 (1.5) – Cy (1) = 0

Cy = 3.25 kN

ΣMz′ = 0; 3 (1.5) + 0.5 (2) – Ex (2) = 0

Ex = 2.75 kN

ΣFx = 0; –Fx + 3 + 0.5 – 2.75 = 0

Fx = 0.75 kN

ΣFy = 0; Fy – 3.25 = 0

Fy = 3.25 kN

ΣFz = 0; Fz – 1.5 – 2 = 0

Fz = 3.5 kN

Method of Joints: Using the above results, we will begin by analyzing the equilibrium of joint C, and then proceed to analyzing that of joint F.

Joint C: From the free – body diagram in Fig. b,

ΣFx = 0; FCB 1

1.25

⎛

⎝⎜

⎞

⎠⎟ = 0 FCB = 0

ΣFy = 0; FCD – 3.25 = 0 FCD = 3.25 kN (C) Ans.

ΣFz = 0; FCF = 0 Ans.

Joint F: From the free – body diagram in Fig. c,

ΣFx = 0; FFB 1

1.5

⎛⎝⎜

⎞⎠⎟

– 0.75 = 0 FBF = 1.125 kN (T)

ΣFz = 0; 3.5 – 1.125 1

1.5

⎛⎝⎜

⎞⎠⎟

– FDF 1

5

⎛

⎝⎜

⎞

⎠⎟ = 0

FDF = 6.149 kN = 6.149 kN (T) Ans.

ΣFy = 0; FEF + 3.25 – 1.125 0.5

1.5

⎛⎝⎜

⎞⎠⎟

– 6.149 2

5

⎛

⎝⎜

⎞

⎠⎟ = 0

FEF = 2.625 kN (C) Ans.


462


6–62. If the truss supports a force of ,determine the force in each member and state if themembers are in tension or compression.

F = 200 N

y

D

E

F

x

z

C

B

A

200 mm

200 mm

200 mm

200 mm

500 mm

300 mm


463


6–63. If each member of the space truss can support amaximum force of 600 N in compression and 800 N intension, determine the greatest force F the truss cansupport.

y

D

E

F

x

z

C

B

A

200 mm

200 mm

200 mm

200 mm

500 mm

300 mm


464


*6–64. Determine the force developed in each member ofthe space truss and state if the members are in tension orcompression. The crate has a weight of 150 lb.

xy

z

A

B

C

D

6 ft

6 ft6 ft

6 ft1 m

1 m

1 m1 m

compression. The crate has a weight of 750 N.

750 N

F = 612.7 N

FAC = 612.7 N

FCA = FCA –1 + 2 + 2 sin 60°

8

i j k⎡

⎣⎢⎢

⎤

⎦⎥⎥

= –0.35 FCA i + 0.707 FCA j + 0.612 FCA k

FCB = 0.354 FCB i + 0.707 FCB j + 0.612 FCB k

FCD = –FCD j

W = –750 k

ΣFx = 0; –0.354 FCA + 0.354 FCB = 0

ΣFy = 0; 0.707 FCA + 0.707 FCB – FCD = 0

ΣFz = 0; 0.612 FCA + 0.612 FCB – 750 = 0

Solving :

FCA = FCB = 612.7 N = 612.7 N (C) Ans

FCD = 866.4 N (T) Ans

FCA = FBA i

FBD = FBD cos 60° i + FBD sin 60° k

FCB = 612.7 (–0.354i – 0.707j – 0.612k)

= –216.9i – 433.2j – 375.0k

ΣFx = 0; FBA + FBD cos 60° – 216.9 = 0

ΣFz = 0; FBD sin 60° – 375.0 = 0

Solving :

FBD = 433.0 N (T) Ans

FBA = 0 Ans

FAC = 612.7 (0.354 FAC i – 0.708 FAC j – 0.612 FAC k)

ΣFz = 0; FDA cos 30° – 0.612 (612.7) = 0

FDA = 433.0 N (T) Ans


465


•6–65. Determine the force in members FE and ED of thespace truss and state if the members are in tension orcompression. The truss is supported by a ball-and-socketjoint at C and short links at A and B.

z

x

y

{�500k} lb

G {200j} lb

6 ft

6 ft

F

ED

C

4 ft

2 ft3 ft3 ft

A

B

{–500k} N

{200j} N

0.6 m

0.6 m0.4 m

0.2 m0.3 m

0.3 m


466


6–66. Determine the force in members GD, GE, and FDof the space truss and state if the members are in tension orcompression.

z

x

y

{�500k} lb

G {200j} lb

6 ft

6 ft

F

ED

C

4 ft

2 ft3 ft3 ft

A

B

{–500k} N

{200j} N

0.6 m

0.6 m0.4 m

0.2 m0.3 m

0.3 m

0.3 m

0.2 m 0.3 m

0.4 m

0.2 m

1.2 m

1.2 m

1.2 m

500 N

200 N

N N

N

N


467


6–67. Determine the force required to hold the 100-lb weight in equilibrium.

P

PA

B

C

D

100 (9.81) N

TA = 490.5 NTB = 245.25 N

100-kg weight in equilibrium.

Equations of Equilibrium: Applying the force equation of equilibrium along the y axis of pulley A on the free – body diagram, Fig. a,

+↑ΣFy = 0; 2TA – 100 (9.81) = 0 TA = 490.5 N

Applying ΣFy = 0 to the free – body diagram of pulley B, Fig. b,

+↑ΣFy = 0; 2TB – 490.5 = 0 TB = 245.25 N

From the free – body diagram of pulley C, Fig. c,

+↑ΣFy = 0; 2P – 245.25 = 0 P = 122.625 N Ans.


468


*6–68. Determine the force required to hold the 150-kg crate in equilibrium.

P

P

A

B

C


469


•6–69. Determine the force required to hold the 50-kgmass in equilibrium.

P

P

A

B

C


470


6–70. Determine the force needed to hold the 20-lb blockin equilibrium.

P

C

B

A

P

6–70. Determine the force P needed to hold the 10-kg block in equilibrium.

Pulley C :

+↑ΣFy = 0; T – 2P = 0

Pulley A :

+↑ΣFy = 0; 2P + T – 10 (9.81) = 0

P = 24.525 N Ans10 (9.81) N


471


6–71. Determine the force needed to support the 100-lbweight. Each pulley has a weight of 10 lb. Also, what are thecord reactions at A and B?

P

P

2 in.

2 in.

2 in.

CA

B

50 (9.81) N

50 N50 N

50 mm

50 mm

50 mm

6–71. Determine the force P needed to support the 50-kg weight. Each pulley has a weight of 50 N. Also, what are the cord reactions at A and B?

Equations of Equilibrium : From FBD (a),

+↑ΣFy = 0; P′ – 2P – 50 = 0 [1]

From FBD (b),

+↑ΣFy = 0; 2P + P′ – 50 (9.81) – 50 = 0 [2]

Solving Eqs. [1] and [2] yields,

P = 122.625 N Ans

P′ = 295.25 N

The cord reactions at A and B are

FA = P = 122.625 N

FB = P′ = 295.25 N Ans


472


*6–72. The cable and pulleys are used to lift the 600-lbstone. Determine the force that must be exerted on the cableat A and the corresponding magnitude of the resultant forcethe pulley at C exerts on pin B when the cables are in theposition shown.

P

A

C

B

D

30�

300 (9.81) N

T = 1471.5 N

T = 1471.5 N

*6–72. The cable and pulleys are used to lift the 300-kg

Pulley D :

+↑ΣFy = 0; 2T – 300 (9.81) = 0

T = 1471.5 N = 1.47 kN Ans

Pulley B :

+↑ΣFx = 0; Bx – 1471.5 sin 30° = 0

Bx = 735.75 N = 0.74 kN

+↑ΣFy = 0; By – 1471.5 – 1471.5 cos 30° = 0

By = 2745.86 N

FB = (735.75) + (2745.86)2 2

= 2842.7 N = 2.84 kN Ans


473


•6–73. If the peg at B is smooth, determine thecomponents of reaction at the pin A and fixed support C.

A

B C

600 mm

800 mm

900 N�m

600 mm

500 N

45�


474


6–74. Determine the horizontal and vertical componentsof reaction at pins A and C.

BA

C

2 ft3 ft

150 lb

100 lb

2 ft

45�

By = 450 N

500 N0.6 m

0.6 m0.9 m

750 N

0.6 m0.9 m

750 N

500 N

0.6 m

Equations of Equilibrium: From the free – body diagram of member AB in Fig. a, we have

�

+ ΣMA = 0; By (1.5) – 750 (0.9) = 0 By = 450 N

�

+ ΣMB = 0; 750 (0.6) – Ay (1.5) = 0 Ay = 300 N Ans.

+↑ΣFx = 0; Ax – Bx = 0 (1)

From the free – body diagram of the member BC in Fig. b and using the result for By, we can write

�

+ ΣMC = 0; 450 (0.6) + 500 sin 45° (0.6) – Bx (0.6) = 0

Bx = 803.55 N

+↑ΣFy = 0; Cy – 450 – 500 sin 45° = 0

Cy = 803.55 N Ans.

+↑ΣFx = 0; 803.55 – 500 cos 45° – Cx = 0

Cx = 450 N Ans.

Substituting Bx = 803.55 N into Eq. (1) yields

Ax = 803.55 N Ans.


475


6–75. The compound beam is fixed at A and supported byrockers at B and C. There are hinges (pins) at D and E.Determine the components of reaction at the supports.

6 m2 m

6 m

30 kN � m

2 m 2 m

15 kN

A D B EC


476


*6–76. The compound beam is pin-supported at C andsupported by rollers at A and B. There is a hinge (pin) at D.Determine the components of reaction at the supports.Neglect the thickness of the beam. A D B C

8 ft

3

45

8 ft

12 kip

15 kip � ft

4 kip30�

8 kip

8 ft4 ft 2 ft

6 ft3 m 4 m 4 m 4 m2 m 1 m

40 kN 60 kN

25 kN · m

4 m 4 m 4 m

Dy = 9.35 kN

60 kN

25 kN · mDx = 10 kN

40 kN

20 kN

3 m 2 m1 m

20 kN

Equations of Equilibrium : From FBD (a),

�

+ ΣMD = 0; 20 cos 30° (6) + 40 (1) – Ay (3) = 0

Ay = 47.97 kN Ans

+↑ΣFy = 0; Dy + 47.97 – 20 cos 30° – 40 = 0

Dy = 9.35 kN

+→ ΣFx = 0; Dx – 20 sin 30° = 0

Dx = 10 kN

From FBD (b),

�

+ ΣMC = 0; 9.35 (12) + 25 + 60 4

5

⎛⎝⎜

⎞⎠⎟

(4) – By (8) = 0

By = 41.15 kN Ans

+↑ΣFy = 0; Cy + 41.15 – 9.35 – 60 4

5

⎛⎝⎜

⎞⎠⎟

= 0

Cy = 16.2 kN Ans

+→ ΣFx = 0; Cx – 10 – 60

3

5

⎛⎝⎜

⎞⎠⎟

= 0

Cx = 46.0 kN Ans


477


•6–77. The compound beam is supported by a rocker at Band is fixed to the wall at A. If it is hinged (pinned) togetherat C, determine the components of reaction at the supports.Neglect the thickness of the beam.

4 ft 4 ft

500 lb200 lb

4000 lb � ft

4 ft8 ft

A C B

1213

5 60�

1 m

2.5 kN1 kN

6 kN · m

1 m 2 m 1 m

1 kN

6 kN · m

2 m 1 m

1 m 1 m

Cx = 0.5 kN

Cy = 1.711 kN

2.5 kN

Member CB :

+→ ΣFx = 0; –Cx + 1 cos 60° = 0

Cx = 0.5 kN

�

+ ΣMC = 0; –1 sin 60° (2) + By (3) – 6 = 0

By = 2.577 kN Ans

+↑ΣFy = 0; Cy – 1 sin 60° + 2.577 = 0

Cy = –1.711 kN

Member AC :

+→ ΣFx = 0; Ax – 2.5

5

13

⎛⎝⎜

⎞⎠⎟

+ 0.5 = 0

Ax = 0.462 kN Ans

+↑ΣFy = 0; Ay – 2.5 12

13

⎛⎝⎜

⎞⎠⎟

+ 1.711 = 0

Ay = 0.597 kN Ans

�

+ ΣMA = 0; –MA – 2.5 12

13

⎛⎝⎜

⎞⎠⎟

(1) + 1.711 (2) = 0

MA = 1.114 kN · m Ans


478


6–78. Determine the horizontal and vertical componentsof reaction at pins A and C of the two-member frame.

3 m

3 m

200 N/m

A

B

C


479


6–79. If a force of acts on the rope, determinethe cutting force on the smooth tree limb at D and thehorizontal and vertical components of force acting on pin A.The rope passes through a small pulley at C and a smoothring at E.

F = 50 N

F � 50 N

BC

E

30 mm

100 mm

A

D


480


*6–80. Two beams are connected together by the shortlink BC. Determine the components of reaction at the fixedsupport A and at pin D.

A B

C

D

10 kN12 kN

3 m1.5 m1 m 1.5 m


481


•6–81. The bridge frame consists of three segments whichcan be considered pinned at A, D, and E, rocker supportedat C and F, and roller supported at B. Determine thehorizontal and vertical components of reaction at all thesesupports due to the loading shown. 15 ft

20 ft

5 ft 5 ft

15 ft

2 kip/ft

30 ft

A

B

C F

DE

30 kN/m

9 m

6 m4.5 m

1.5 m 1.5 m

4.5 m

4.5 m4.5 m

1.5 m

2.25 m 2.25 m

1.5 m

2.25 m2.25 m

30 (4.5) kN

30 (4.5) kN By = 135 kN

Dy = 135 kN

30 (9) kN

For segment BD :

�

+ ΣMD = 0; 30 (9) (4.5) – By (9) = 0

By = 135 kN Ans

+→ ΣFx = 0; Dx = 0 Ans

+↑ΣFy = 0; Dy + 135 – 30 (9) = 0

Dy = 135 kN Ans

For segment ABC :

�

+ ΣMA = 0; Cy (1.5) – 30 (4.5) (2.25) – 135 (4.5) = 0

Cy = 607.5 kN Ans

+→ ΣFx = 0; Ax = 0 Ans

+↑ΣFy = 0; –Ay + 607.5 – 30 (4.5) – 135 = 0

Ay = 337.5 kN Ans

For segment DEF :

�

+ ΣMg = 0; –Fy (1.5) – 30 (4.5) (2.25) + 135 (4.5) = 0

Fy = 607.5 kN Ans

+→ ΣFx = 0; Ex = 0 Ans

+↑ΣFy = 0; –Ey + 607.5 – 30 (4.5) – 135 = 0

Ey = 337.5 kN Ans


482


6–82. If the 300-kg drum has a center of mass at point G,determine the horizontal and vertical components of forceacting at pin A and the reactions on the smooth pads Cand D. The grip at B on member DAB resists bothhorizontal and vertical components of force at the rim ofthe drum.

P

390 mm

100 mm

60 mm60 mm

600 mm

30�

B

A

C

D G

E


483


6–83. Determine the horizontal and vertical componentsof reaction that pins A and C exert on the two-member arch.

1 m1.5 m

2 kN

1.5 kN

0.5 m

A

B

C


484


*6–84. The truck and the tanker have weights of 8000 lband 20 000 lb respectively. Their respective centers ofgravity are located at points and . If the truck is atrest, determine the reactions on both wheels at A, at B, andat C. The tanker is connected to the truck at the turntableD which acts as a pin.

G2G1G1

15 ft 10 ft 9 ft5 ft

A B

D

C

G2

Dy = 60 kN100 kN 40 kN

1.5 m

2.7 m4.5 m 3 m

4.5 m 3 m 2.7 m1.5 m

*6–84. The truck and the tanker have weights of 40 kg and 100 kN respectively. Their respective centers of gravity are located at points G1 and G2 . If the truck is at rest, determine the reactions on both wheels at A, at B, and at C. The tanker is connected to the truck at the turntable D which acts as a pin.

Equations of Equilibrium: First, we will consider the free – body diagram of the tanker in Fig. a.

�

+ ΣMD = 0; 100 (3) – NA (7.5) = 0

NA = 40 kN Ans.

+→ ΣFx = 0; Dx = 0

+↑ΣFy = 0; Dy + 40 – 100 = 0

Dy = 60 kN

Using the results of Dx and Dy obtained above and considering the free – body diagram of the truck in Fig. b,

�

+ ΣMD = 0; NC (4.2) – 40 (2.7) = 0

NC = 25.71 kN Ans.

+→ ΣFy = 0; NB + 25.71 – 40 – 60 = 0

NB = 74.29 kN Ans.


485


•6–85. The platform scale consists of a combination ofthird and first class levers so that the load on one leverbecomes the effort that moves the next lever. Through thisarrangement, a small weight can balance a massive object.If , determine the required mass of thecounterweight S required to balance a 90-kg load, L.x = 450 mm

350 mm150 mm

150 mm100 mm250 mm

BA

C D

E F

H

G

x

L

S


486


6–86. The platform scale consists of a combination ofthird and first class levers so that the load on one leverbecomes the effort that moves the next lever. Through thisarrangement, a small weight can balance a massive object. If

and, the mass of the counterweight S is 2 kg,determine the mass of the load L required to maintain thebalance.

x = 450 mm

350 mm150 mm

150 mm100 mm250 mm

BA

C D

E F

H

G

x

L

S


487


6–87. The hoist supports the 125-kg engine. Determinethe force the load creates in member DB and in memberFB, which contains the hydraulic cylinder H.

C

D

EFG

H

2 m

1 m

1 m

2 m1 m

2 m

A B


488


*6–88. The frame is used to support the 100-kg cylinder E.Determine the horizontal and vertical components ofreaction at A and D.

A

CD

E

0.6 m

1.2 m

r � 0.1 m


489


•6–89. Determine the horizontal and vertical componentsof reaction which the pins exert on member AB of the frame.

A

E

B

CD

500 lb

300 lb

3 ft 3 ft

4 ft

60�

1.5 kN

0.9 m0.9 m

0.9 m 0.9 m

2.5 kN

1.5 kN

1.2 m

Member AB :

�

+ ΣMA = 0; –1.5 sin 60° (0.9) + 4

5 FBD (1.8) = 0

FBD = 0.8119 kN

Thus,

Bx = 3

5 (0.8119) = 0.4871 kN Ans

By = 4

5 (0.8119) = 0.6495 kN Ans

+→ ΣFx = 0; –1.5 cos 60° +

3

5 (0.8119) + Ax = 0

Ax = 0.2629 kN Ans

+↑ΣFy = 0; Ay – 1.5 sin 60° + 4

5 (0.8119) = 0

Ay = 0.6495 kN Ans


490


6–90. Determine the horizontal and vertical components ofreaction which the pins exert on member EDC of the frame.

A

E

B

CD

500 lb

300 lb

3 ft 3 ft

4 ft

60�

1.5 kN

0.9 m0.9 m

2.5 kN

1.2 m

0.9 m 0.9 m

1.5 kN

0.9 m 0.9 m

2.5 kN

FBD = 0.8119 kN

FBD = 0.8119 kNFAD = 7.0619 kN

Member AB :

�

+ ΣMA = 0; –1.5 sin 60° (0.9) + 4

5 FBD (1.8) = 0

FBD = 0.8119 kN

Member EDC :

�

+ ΣMg = 0; –2.5 (1.8) – 4

5 (0.8119) (0.9)

+ 4

5 FAD (0.9) = 0

FAD = 7.0619 kN

+→ ΣFx = 0; Ex – 0.8119

3

5

⎛⎝⎜

⎞⎠⎟

– 7.0619 3

5

⎛⎝⎜

⎞⎠⎟

= 0

Ex = 4.7243 kN Ans

+↑ΣFy = 0; –Ey + 7.0619 4

5

⎛⎝⎜

⎞⎠⎟

– 0.8119 4

5

⎛⎝⎜

⎞⎠⎟

– 2.5 = 0

Ey = 2.5 kN Ans

Pin D :

+→ ΣFx = 0; Dx –

3

5 (0.8119) –

3

5 (7.0619) = 0

Dx = 4.7243 kN Ans

+↑ΣFy = 0; –Dy – 4

5 (0.8119) +

4

5 (7.0619) = 0

Dy = 5 kN Ans


N N N N FCA = 833.33 N N N N N N FCD = 333.33 N N N

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Transcript of N N N N FCA = 833.33 N N N N N N FCD = 333.33 N N N