Alguns Problemas do Capítulo 4 -...

Alguns Problemas do Capítulo 4

Prof. Ettore Baldini-Neto

Exemplo Aula Passada: Determine a intensidade do momento produzido pela força F em relação ao segmento OA do encanamento da figura abaixo. 4.5 MOMENT OF A FORCE ABOUT A SPECIFIED AXIS 143

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EXAMPLE 4.9

Determine the magnitude of the moment of force F about segmentOA of the pipe assembly in Fig. 4–24a.

SOLUTIONThe moment of F about the OA axis is determined from

where r is a position vector extending from anypoint on the OA axis to any point on the line of action of F. Asindicated in Fig. 4–24b, either can be used;however, will be considered since it will simplify the calculation.

The unit vector , which specifies the direction of the OA axis, is

and the position vector is

rOD = 50.5i + 0.5k6 mrOD

uOA =rOA

rOA=

50.3i + 0.4j6 m210.3 m22 + 10.4 m22 = 0.6i + 0.8j

uOA

rOD

rOD, rOC, rAD, or rAC

MOA = uOA# (r * F),

The force F expressed as a Cartesian vector is

= {200i - 200j + 100k} N

= (300 N)B 50.4i - 0.4j + 0.2k6m2(0.4 m)2 + (-0.4 m)2 + (0.2 m)2RF = Fa rCD

rCDb

Therefore,

Ans.= 100 N #m= 0.6[011002 - ( 0.5 )1-2002] - 0.8[0.511002 - ( 0.5 )12002] + 0

= 3 0.6 0.8 0

0.5 0 0.5

200 -200 100

3MOA = uOA# 1rOD * F2

0.1 m

0.3 m

0.2 m0.4 m

0.5 m

0.5 m

(a)

x y

C

A

O

D

z

F ! 300 N

B

Fig. 4–24

xy

z

F

(b)

D

A

CO

rOD

rAD

rAC

rOC

uOA

4.5 MOMENT OF A FORCE ABOUT A SPECIFIED AXIS 143

4

EXAMPLE 4.9

Determine the magnitude of the moment of force F about segmentOA of the pipe assembly in Fig. 4–24a.

SOLUTIONThe moment of F about the OA axis is determined from

where r is a position vector extending from anypoint on the OA axis to any point on the line of action of F. Asindicated in Fig. 4–24b, either can be used;however, will be considered since it will simplify the calculation.

The unit vector , which specifies the direction of the OA axis, is

and the position vector is

rOD = 50.5i + 0.5k6 mrOD

uOA =rOA

rOA=

50.3i + 0.4j6 m210.3 m22 + 10.4 m22 = 0.6i + 0.8j

uOA

rOD

rOD, rOC, rAD, or rAC

MOA = uOA# (r * F),

The force F expressed as a Cartesian vector is

= {200i - 200j + 100k} N

= (300 N)B 50.4i - 0.4j + 0.2k6m2(0.4 m)2 + (-0.4 m)2 + (0.2 m)2RF = Fa rCD

rCDb

Therefore,

Ans.= 100 N #m= 0.6[011002 - ( 0.5 )1-2002] - 0.8[0.511002 - ( 0.5 )12002] + 0

= 3 0.6 0.8 0

0.5 0 0.5

200 -200 100

3MOA = uOA# 1rOD * F2

0.1 m

0.3 m

0.2 m0.4 m

0.5 m

0.5 m

(a)

x y

C

A

O

D

z

F ! 300 N

B

Fig. 4–24

xy

z

F

(b)

D

A

CO

rOD

rAD

rAC

rOC

uOA

Exemplo Aula Passada: Determine o momento resultante das três forças mostradas em relação aos eixos x, y e z respectivamente.

4.5 MOMENT OF A FORCE ABOUT A SPECIFIED AXIS 141

4

Important Points

● The moment of a force about a specified axis can be determinedprovided the perpendicular distance from the force line ofaction to the axis can be determined.

● If vector analysis is used, where defines thedirection of the axis and r is extended from any point on the axisto any point on the line of action of the force.

● If is calculated as a negative scalar, then the sense of directionof is opposite to

● The moment expressed as a Cartesian vector is determinedfrom Ma = Maua.

Ma

ua.Ma

Ma

uaMa = ua# 1r * F2,Ma = Fda.

da

EXAMPLE 4.7

Determine the resultant moment of the three forces in Fig. 4–22 aboutthe x axis, the y axis, and the z axis.

SOLUTIONA force that is parallel to a coordinate axis or has a line of action thatpasses through the axis does not produce any moment or tendency forturning about that axis.Therefore, defining the positive direction of themoment of a force according to the right-hand rule, as shown in thefigure, we have

Ans.

Ans.

Ans.

The negative signs indicate that and act in the and directions, respectively.

-z-yMzMy

Mz = 0 + 0 - (40 lb)(2 ft) = -80 lb # ft

My = 0 - (50 lb)(3 ft) - (40 lb)(2 ft) = -230 lb # ft

Mx = (60 lb)(2 ft) + (50 lb)(2 ft) + 0 = 220 lb # ft2 ft

2 ft2 ft 3 ft

xy

z

B

C

A

O

F3 ! 40 lb

F2 ! 50 lb

F1 ! 60 lb

Fig. 4–22

Problema Fundamental 4.5 (pág 96): Determine o momento da força em relação ao ponto O, desprezando a espessura do braço.

4.4 PRINCIPLE OF MOMENTS 131

4

FUNDAMENTAL PROBLEMS

F4–1. Determine the moment of the force about point O.

5 ft

0.5 ft

600 lb

20!

30!

O

F4–1

5 m

2 m

100 N3

4

5

O

F4–2

30!

45!

F " 300 N

0.4 m

0.3 mO

F4–3

4 ft

3 ft

1 ft

600 lb

O

45!

F4–4

50 N

60!

45!

100 mm

100 mm

200 mmO

F4–5

500 N

3 m

O

45!

F4–6


F4–5. Determine the moment of the force about point O.Neglect the thickness of the member.

F4–6. Determine the moment of the force about point O.F4–3. Determine the moment of the force about point O.


Problema Fundamental 4.12 (pág 97): Se F1=100i-120j+75k e F2=-200i+250j+100k, ambas medidas em libras (lb), determine o momento resultante produzido por estas forças em relação ao ponto O. Expresse-o como um vetor cartesiano.

132 CHAPTER 4 FORCE SYSTEM RESULTANTS

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F4–10. Determine the moment of force F about point O.Express the result as a Cartesian vector.

F4–9. Determine the resultant moment produced by theforces about point O.

F 4–11. Determine the moment of force F about point O.Express the result as a Cartesian vector.

F4–12. If and , determine the resultant moment

produced by these forces about point O. Express the resultas a Cartesian vector.

+ 250j + 100k6 lb F2 = 5-200iF1 = 5100i - 120j + 75k6 lb

O

2 m

2.5 m45!

1 m

600 N

300 N

500 N

F4–7

F1 " 500 N

F2 " 600 N

A

0.25 m

0.3 m0.125 m

60!

435

O

F4–8

O

30!30!

6 ft

6 ft

F2 " 200 lb

F1 " 300 lb

F4–9

x

z

y

O

AB

4 m3 m

F " 500 N

F 4–10

x

z

y

O

A

B

C 2 ft

1 ft

4 ft

4 ft

F " 120 lb

F4–11

z

O

A

x

y

4 ft

3 ft 5 ft

F1

F2

F4–12



Problema 4.5 (pág 97): Dois homens exercem forças de F=400N e P=250N sobre as cordas conforme mostra a figura. Determine o momento de cada força em relação ao ponto A. Em que sentido o poste vai girar? Horário ou anti-horário? Conversão: 1ft=0,305m

4.4 PRINCIPLE OF MOMENTS 133

4

PROBLEMS

A

P

F

B

C

6 ft

45!

12 ft3

4

5

Probs. 4–4/5

3 m

0.45 m

4 kN

A

u

Probs. 4–6/7

F

B

A

18 in.

5 in.

30!

Probs. 4–8/9

4 kN

800 N 800 N

4 kN

Case 1 Case 2

0.4 m

0.05 m

0.05 m

0.4 m

OO

Prob. 4–10

4–6. If , determine the moment produced by the 4-kN force about point A.

4–7. If the moment produced by the 4-kN force aboutpoint A is clockwise, determine the angle , where

.0° … u … 90°u10 kN # m

u = 45°

•4–1. If A, B, and D are given vectors, prove thedistributive law for the vector cross product, i.e.,

.

4–2. Prove the triple scalar product identity.

4–3. Given the three nonzero vectors A, B, and C, showthat if , the three vectors must lie in thesame plane.

*4–4. Two men exert forces of and onthe ropes. Determine the moment of each force about A.Which way will the pole rotate, clockwise or counterclockwise?

•4–5. If the man at B exerts a force of on hisrope, determine the magnitude of the force F the man at Cmust exert to prevent the pole from rotating, i.e., so theresultant moment about A of both forces is zero.

P = 30 lb

P = 50 lbF = 80 lb

A # (B : C) = 0

A # B : C = A : B # C

A : (B + D) = (A : B) + (A : D)

*4–8. The handle of the hammer is subjected to the forceof Determine the moment of this force about thepoint A.

•4–9. In order to pull out the nail at B, the force F exertedon the handle of the hammer must produce a clockwisemoment of about point A. Determine therequired magnitude of force F.

500 lb # in.

F = 20 lb.

4–10. The hub of the wheel can be attached to the axleeither with negative offset (left) or with positive offset(right). If the tire is subjected to both a normal and radialload as shown, determine the resultant moment of theseloads about point O on the axle for both cases.

Problema Fundamental 4.13 (pág 105): Determine a intensidade do momento da força F=300i-200j+150k medida em Newtons (N) em relação ao eixo x, expressando-o como um vetor cartesiano.

Problema Fundamental 4.15 (pág 105): Determine a intensidade do momento da força de 200N em relação ao eixo x.


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F4–18. Determine the moment of force F about the x, they, and the z axes. Use a scalar analysis.

F4–15. Determine the magnitude of the moment of the200-N force about the x axis.

F4–13. Determine the magnitude of the moment of theforce about the x axis.Express the result as a Cartesian vector.

F4–14. Determine the magnitude of the moment of theforce about the OA axis.Express the result as a Cartesian vector.

F = 5300i - 200j + 150k6N

F = 5300i - 200j + 150k6 N

z

O

A

BF

xy0.4 m

0.2 m

0.3 m

F4–13/14

x

O

A

45!

120!60!

F " 200 N

z

y

0.25 m

0.3 m

F4–15

2 m

F " {30i # 20j $ 50k} N

4 m

z

x

y

A

3 m

F4–16

2 ft

4 ft3 ftx y

z

B

C

A

F

F4–17

z

A

O

y

x

F " 500 N

3 m

2 m2 m

3

3

4

4

5

5

F4–18

F4–16. Determine the magnitude of the moment of theforce about the y axis.

F4–17. Determine the moment of the forceabout the AB axis. Express the

result as a Cartesian vector.F = 550i - 40j + 20k6 lb


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F = 5300i - 200j + 150k6N

F = 5300i - 200j + 150k6 N

z

O

A

BF

xy0.4 m

0.2 m

0.3 m

F4–13/14

x

O

A

45!

120!60!

F " 200 N

z

y

0.25 m

0.3 m

F4–15

2 m

F " {30i # 20j $ 50k} N

4 m

z

x

y

A

3 m

F4–16

2 ft

4 ft3 ftx y

z

B

C

A

F

F4–17

z

A

O

y

x

F " 500 N

3 m

2 m2 m

3

3

4

4

5

5

F4–18




Problema Fundamental 4.16 (pág 105): Determine a intensidade do momento da força em relação ao eixo y.


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F = 5300i - 200j + 150k6N

F = 5300i - 200j + 150k6 N

z

O

A

BF

xy0.4 m

0.2 m

0.3 m

F4–13/14

x

O

A

45!

120!60!

F " 200 N

z

y

0.25 m

0.3 m

F4–15

2 m

F " {30i # 20j $ 50k} N

4 m

z

x

y

A

3 m

F4–16

2 ft

4 ft3 ftx y

z

B

C

A

F

F4–17

z

A

O

y

x

F " 500 N

3 m

2 m2 m

3

3

4

4

5

5

F4–18




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