Problem 6.17 Determine the axial forces in themembers in terms of the weight W.

A

B E

D

C

1 m

1 m

0.8 m 0.8 m 0.8 m

W

Solution: Denote the axial force in a member joining two pointsI, K by IK. The angle between member DE and the positive x axisis ˛ D tan�1 0.8 D 38.66°. The angle formed by member DB with thepositive x axis is 90° C ˛. The angle formed by member AB with thepositive x axis is ˛.

Joint E :

∑Fy D �DE cos ˛�W D 0,

from which DE D �1.28W �C� .

∑Fy D �BE� DE sin ˛ D 0,

from which BE D 0.8W �T�

Joint D :

∑Fx D DE cos ˛C BD cos ˛�CD cos ˛ D 0,

from which BD �CD D �DE.

∑Fy D �BD sin ˛C DE sin ˛�CD sin ˛ D 0,

from which BD CCD D DE.

Solving these two equations in two unknowns:

CD D DE D �1.28W �C� , BD D 0

Joint B :

∑Fx D BE� AB sin ˛� BD sin ˛ D 0,

from which AB D BE

sin ˛D 1.28W�T�

∑Fy D �AB cos ˛� BC D 0,

from which BC D �AB cos ˛ D �W�C�

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401

Problem 6.53 Consider the truss in Problem 6.52. Theweight of the bucket is W D 1000 lb. The cable passesover pulleys at A and D. Determine the axial forces inmembers IK and JL.

Solution: Make a cut through JL, JK, and IK, and consider theupper section. Denote the axial force in a member joining, ˛, ˇ by˛ˇ. The section as a free body: The perpendicular distance from pointJ to the line of action of the weight is L D 6 cos ˛C 3

p2 sin ˇ C

3.5 D 12.593 ft. The sum of the moments about J is MJ D �W�L�CW�3.25�� IK�3� D 0, from which IK D �3114.4 lb�C�.

The sum of the forces:

∑Fx D JL cos ˛� IK cos ˛

�W cos ˛� JK cos ˇ D 0,

and∑

Fy D �JL sin ˛� IK sin ˛

�W sin ˛�W� JK sin ˇ D 0,

from which two simultaneous equations:

0.8192JL C 0.1736JK D �1732

and 0.5736JL C 0.9848JK D 212.75.

Solve: JL D 2360 lb�T� ,

and JK D �1158.5 lb�C� .

WW

3.5 ft

3 ft

3.25 ft

βαJL

JKIK

Problem 6.54 The truss supports loads at N, P, and R.Determine the axial forces in members IL and KM.

2 m

2 m

2 m

2 m

1 m

6 m

2 m 2 m 2 m 2 m 2 m

K

I

M

L

O

N

Q

P RJ

H

F

D

G

E

C

BA

1 kN 2 kN 1 kNSolution: The strategy is to make a cut through KM, IM, andIL, and consider only the outer section. Denote the axial force in amember joining, ˛, ˇ by ˛ˇ.

The section as a free body: The moment about M is

MM D �IL � 2�1�� 4�2�� 6�1� D 0,

from which IL D �16 kN �C� .

The angle of member IM is ˛ D tan�1�0.5� D 26.57°.

The sums of the forces:

∑Fy D �IM sin ˛� 4 D 0,

from which IM D � 4

sin ˛D �8.944 kN (C).

∑Fx D �KM� IM cos ˛� IL D 0,

from which KM D 24 kN�T�

αKM

IM

IL1 kN 2 kN 1 kN

1 m

2 m 2 m 2 m

432

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Problem 6.55 Consider the truss in Problem 6.54.Determine the axial forces in members HJ and GI.

Solution: The strategy is to make a cut through the four membersAJ, HJ, HI, and GI, and consider the upper section. The axial forcein AJ can be found by taking the moment of the structure about B.

The complete structure as a free body: The angle formed by AJ with the

vertical is ˛ D tan�1

(4

8

)D 26.57° . The moment about B is MB D

6AJ cos ˛� 24 D 0, from which AJ D 4.47 kN (T).

The section as a free body: The angles of members HJ and HI relative

to the vertical are ˇ D tan�1

(2

8

)D 14.0°, and � D tan�1

(1.5

2

)D

36.87° respectively. Make a cut through the four members AJ, HJ,HI, and GI, and consider the upper section. The moment aboutthe point I is MI D �24C 2AJ cos ˛C 2HJ cos ˇ D 0. From which

HJ D 8.25 kN �T� . The sums of the forces:

∑Fx D �AJ sin ˛CHJ sin ˇ�HI sin � D 0,

from which HI D AJ sin ˛�HJ sin ˇ

sin �D 2� 2

sin �D 0.

∑FY D �AJ cos ˛�HJ cos ˇ �HI cos � �GI� 4 D 0,

from which GI D �16 kN �C�

AJ HJ

HI GI2 m 2 m 2 m

1 kN 2 kN 1 kN

2 m 2 m

1 m

Iγα β

Problem 6.56 Consider the truss in Problem 6.54. Bydrawing free-body diagrams of sections, explain why theaxial forces in members DE, FG, and HI are zero.

Solution: Define ˛, ˇ to be the interior angles BAJ and ABJrespectively. The sum of the forces in the x-direction at the base yieldsAX C BX D 0, from which Ax D �Bx . Make a cut through AJ, BD andBC, from which the sum of forces in the x-direction, Ax � BD sin ˇ D0. Since Ax D AJ sin ˛, then AJ sin ˛� BD sin ˇ D 0. A repeat of thesolution to Problem 6.55 shows that this result holds for each section,where BD is to be replaced by the member parallel to BD. For example:make a cut through AJ, FD, DE, and CE. Eliminate the axial forcein member AJ as an unknown by taking the moment about A. Repeatthe solution process in Problem 6.55, obtaining the result that

DE D AJ sin ˛� DF sin ˇ

cos �DED 0

where �DE is the angle of the member DE with the vertical. Similarly,a cut through AJ, FH, FG, and EG leads to

FG D AJ sin ˛� FH sin ˇ

cos �FGD 0,

and so on. Thus the explanation is that each member BD, DF, FH andHJ has equal tension, and that this tension balances the x-componentin member AJ

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433

Problem 6.99 Figure a is a diagram of the bones andbiceps muscle of a person’s arm supporting a mass.Tension in the biceps muscle holds the forearm in thehorizontal position, as illustrated in the simple mechan-ical model in Fig. b. The weight of the forearm is 9 N,and the mass m D 2 kg.

(a) Determine the tension in the biceps muscle AB.(b) Determine the magnitude of the force exerted on

the upper arm by the forearm at the elbow joint C.

(a)

A

B

C

290 mm

50 mm

150 mm

9 N

200 mm

m

(b)

Solution: Make a cut through AB and BC just above the elbowjoint C. The angle formed by the biceps muscle with respect to the

forearm is ˛ D tan�1

(290

50

)D 80.2°. The weight of the mass is W D

2�9.81� D 19.62 N.

The section as a free body: The sum of the moments about C is

∑MC D �50T sin ˛C 150�9� C 350W D 0,

from which T D 166.76 N is the tension exerted by the biceps muscleAB. The sum of the forces on the section is

∑FX D Cx C T cos ˛ D 0,

from which Cx D �28.33 N.

∑FY D Cy C T sin ˛� 9�W D 0,

from which Cy D �135.72. The magnitude of the force exerted by theforearm on the upper arm at joint C is

F D√

C2x CC2

y D 138.65 N

W

T

200 mm

150 mm

50 mm

9N

Cy

Cx

α

476

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Problem 6.106 The woman exerts 20-N forces on thehandles of the shears. Determine the magnitude of theforces exerted on the branch at A.

BAC

D

E

25 mm25 mm36 mm

65 mm 20 N

20 N

Solution: Assume that the shears are symmetrical.

Consider the 2 pieces CD and CE

∑Fx D 0) Dx D Ex

∑Fy D 0) Dy D Ey

∑MC D 0) Dx D Ex D 0

20 N

20 NEy

Ex

C

Dx

Dy

Now examine CD by itself

∑MC D ��20 N��90 mm�C Dy�25 mm� D 0) Dy D 72 N

20 N

Cy

Cx

Dx = 0

Dy

Finally examine DBA

∑MB : A�36 mm�� Dy�50 mm� D 0

Bx

Dx = 0By

Dy

A

Solving we find A D 100 N

482

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Problem 7.12 Determine the coordinates of the cen-troid of the area.

x

y

y � � x2 � 4x � 714

Solution: Use a vertical strip. We first need to find the x intercepts.

y D � 1

4x2 C 4x � 7 D 0 ) x D 2, 14

x D

∫ 14

2x�y dx�∫ 14

2y dx

D

∫ 14

2x

(� 1

4x2 C 4x � 7

)dx∫ 14

2

(� 1

4x2 C 4x � 7

)dx

D 8

y D

∫ 14

2

1

2y�y dx�∫ 14

2y dx

D

∫ 14

2

1

2

(� 1

4x2 C 4x � 7

)2

dx∫ 14

2

(� 1

4x2 C 4x � 7

)dx

D 18

5

x D 8y D 3.6

Problem 7.13 Determine the coordinates of the cen-troid of the area.

x

y

y � 5

y � � x2 � 4x � 714

Solution: Use a vertical strip. We first need to find the x intercepts.

y D � 1

4x2 C 4x � 7 D 5 ) x D 4, 12

x D

∫ 12

4x�y dx�∫ 12

4y dx

D

∫ 12

4x

[(� 1

4x2 C 4x � 7

)� 5

]dx∫ 12

4

[(� 1

4x2 C 4x � 7

)� 5

]dx

D 8

y D

∫ 12

4yc�y dx�∫ 12

4y dx

D

∫ 12

4

1

2

[(� 1

4x2 C 4x � 7

)C 5

] [(� 1

4x2 C 4x � 7

)� 5

]dx∫ 12

4

[(� 1

4x2 C 4x � 7

)� 5

]dx

D 33

5

x D 8, y D 6.6

516

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Problem 7.14 Determine the x coordinate of the cent-roid of the area.

x

y

y = x3

y = x

Solution: Work this problem like Example 7.2

x D

∫ 1

0x dA∫ 1

0dA

D

∫ 1

0x�x � x3� dx∫ 1

0�x � x3� dx

x D

[x3

3� x5

5

]1

0[x2

2� x4

4

]1

0

D

(1

3� 1

5

)(

1

2� 1

4

) D2

151

4

D 0.533

x D 0.533

x

y

y = x3

dA

y = x

Problem 7.15 Determine the y coordinate of the cent-roid of the area shown in Problem 7.14.

Solution: Solve this problem like example 7.2.

y D

∫A

y dA∫A

dAD

∫ 1

0

[1

2�x C x3�

]�x � x3� dx∫ 1

0�x � x3� dx

y D 1

2

∫ 1

0�x2 � x6� dx∫ 1

0�x � x3� dx

D

[x3

3� x7

7

]1

0

2

[x2

2� x4

4

]1

0

y D

[1

3� 1

7

]

2

[1

2� 1

4

] D

(4

21

)2

D 8

21D 0.381

y D 0.381

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517