Solucionario de mecanica vectorial para ingenieros edicion 8 Cap 10 solucionario

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Transcript of Solucionario de mecanica vectorial para ingenieros edicion 8 Cap 10 solucionario

COSMOS: Complete Online Solutions Manual Organization System

Chapter 10, Solution 1.

Link ABC: Assume clockwise Then, for point C

xC = (125 mm ) and for point D

xD = xC = (125 mm ) and for point E

xE = Link DEFG: Thus

250 mm 2 xD = xD 3 375 mm

xD = ( 375 mm )

(125 mm ) = ( 375 mm ) = 1 3

Then

G = 100 2 mm = 3 yG = G cos 45 =

(

)

100

2 mm

100 100 2 mm cos 45 = mm 3 3

Virtual Work:

Assume P acts downward at G

U = 0:

( 9000 N mm ) (180 N )( xE

mm ) + P ( yG mm ) = 0

2 100 = 0 9000 180 125 + P 3 3 or

P = 180.0 N

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 10, Solution 2.

Have

FA = 20 lb at AFD = 30 lb at D

Link ABC:

y A = (16 in.)

Link BF:

yF = yB

yB = (10 in.) Link DEFG: or

yF = ( 6 in.) = (10 in.) = yG = (12 in.) = ( 20 in.) d ED = 5 3

( 5.5 in.)2 + ( 4.8 in.)25

= 7.3 in.

D = ( 7.3 in.) = 7.3 in. 3 xD =Virtual Work: 4.8 4.8 5 D = 7.3 in. = ( 8 in.) 7.3 7.3 3

Assume P acts upward at G

U = 0:or or

FA y A + FD xD + P yG = 0

( 20 lb ) (16 in.) + ( 30 lb ) (8 in.) + P ( 20 in.) = 0 P = 28.0 lbP = 28.0 lb

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 10, Solution 3.

Link ABC: Assume clockwise Then, for point C

xC = (125 mm ) and for point D

xD = xC = (125 mm ) and for point E

xE = Link DEFG: Thus or Virtual Work:

250 mm 2 xD = xD 3 375 mm

xD = ( 375 mm )

(125 mm ) = ( 375 mm ) = Assume M acts clockwise on link DEFG

1 3

U = 0:

( 9000 N mm ) (180 N )( xE

mm ) + M = 0

2 1 9000 180 125 + M = 0 3 3 or

M = 18000 N mm

or

M = 18.00 N m

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 10, Solution 4.

Have

FA = 20 lb at A FD = 30 lb

at D

Link ABC:

y A = (16 in.) yB = (10 in.) yF = ( 6 in.) = (10 in.) or

Link BF:

yF = yB

Link DEFG:

=

5 3

d ED =

( 5.5 in.)2 + ( 4.8 in.)25 3

= 7.3 in.

D = ( 7.3 in.) = 7.3 in. xD =Virtual Work: 4.8 4.8 5 D = 7.3 in. = ( 8 in.) 7.3 7.3 3 on DEFG 5 = 0 3 M = 28.0 lb ft

Assume M acts

U = 0:or or

FA y A + FD xD + M = 0

( 20 lb ) (16 in.) + ( 30 lb ) (8 in.) + M M = 336.0 lb in.

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 10, Solution 5.

Assume

x A = 10 in.

yC = 4 in. yD = yC = 4 in.

=

yD6

=

2 3 2

xG = 15 = 15 = 10 in. 3Virtual Work: Assume that force P is applied at A.

U = 0:

U = P x A + 30 yC + 60 yD + 240 + 80 xG = 02 P (10 in.) + ( 30 lb )( 4 in.) + ( 60 lb )( 4 ) + ( 240 lb in.) 3 + ( 80 lb )(10 in.) = 0 10P + 120 + 240 + 160 + 800 = 0 10P = 1320P = 132.0 lb

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 10, Solution 6.

Note: xE = 2 x D xG = 3xD x H = 4 xD xI = 5 x D (a) Virtual Work:

xE = 2 xD xG = 3 xD xH = 4 xD

x I = 5 xD

U = 0: FG xG FSP xI = 0

( 90 N )( 3 xD ) FSP ( 5 xD ) = 0or Now FSP = k xI 54 N = ( 720 N/m ) xI xI = 0.075 m and FSP = 54.0 N

xD = x H = x IxH = 4 4 xI = ( 0.075 m ) = 0.06 m 5 5 or xH = 60.0 mm

1 4

1 5

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

(b)

Virtual Work:

U = 0: FG xG + FH xH FSP ( xI ) = 0

( 90 N )( 3 xD ) + ( 90 N )( 4 xD ) FSP ( 5 xD ) = 0or NowFSP = k xI FSP = 126.0 N

126.0 N = ( 720 N/m ) xI xI = 0.175 m From Part (a) xH = 4 4 xI = ( 0.175 m ) = 0.140 m 5 5 or xH = 140.0 mm

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 10, Solution 7.

Note:xE = 2 x D xG = 3xD xH = 4 xD

xE = 2 xD xG = 3 xD xH = 4 xD

xI = 5 xD(a) Virtual Work:

xI = 5 x D

U = 0: FE xE FSP xI = 0

( 90 N )( 2 xD ) FSP ( 5 xD ) = 0or NowFSP = k xI

FSP = 36.0 N

36 N = ( 720 N/m ) xI xI = 0.050 mand

xD = xH = xIxH = 4 4 xI = ( 0.050 m ) = 0.04 m 5 5or

1 4

1 5

xH = 40.0 mm

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

(b)

Virtual Work:

U = 0: FD xD + FE xE FSP xI = 0

( 90 N ) xD + ( 90 N )( 2 xD ) FSP ( 5 xD ) = 0or NowFSP = 54.0 N

FSP = k xI

54 N = ( 720 N/m ) xIxI = 0.075 m

From Part (a)

xH =

4 4 xI = ( 0.075 ) = 0.06 m 5 5orxH = 60.0 mm

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 10, Solution 8.

Assume y A

yA16 in.

=

yC8 in.

;

yC = y A

1 2

Bar CFDE moves in translation

yE = yF = yC = y AVirtual Work:

1 2

U = 0: P ( y A in.) + (100 lb )( yE in.) + (150 lb )( yF in.) = 01 1 P y A + 100 y A + 150 y A = 0 2 2 P = 125 lb

P = 125 lb

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 10, Solution 9.

Have

y A = 2l cos ; CD = 2l sin ; 2

y A = 2l sin ( CD ) = l cos 2

Virtual Work:

U = 0: P y A Q ( CD ) = 0

P ( 2l sin ) Q l cos = 0 2 Q = 2P sin cos 2

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 10, Solution 10.

Virtual Work: Have

x A = 2l sin

x A = 2l cos andyF = 3l cos

yF = 3l sin Virtual Work:

U = 0: Q x A + P yF = 0Q ( 2l cos ) + P ( 3l sin ) = 0

Q=

3 P tan 2

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 10, Solution 11.

Virtual Work: We note that the virtual work of Ax , Ay and C is zero, since A is fixed and C is to xC .

U = 0:xD = 3l cos yD = l sin

P xD + Q yD = 0

xD = 3l sin yD = l cos

Thus:

P ( 3l sin ) + Q ( l cos ) = 0

3Pl sin + Ql cos = 0Q=

3P sin = 3P tan cos

Q = 3P tan

Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell 2007 The McGraw-Hill Companies.

COSMOS: Complete Online Solutions Manual Organization System

Chapter 10, Solution 12.

x A = ( a + b ) cos yG = a sin

x A = ( a + b ) sin

yG = a cos

Virtual Work: The reactions at A and B are perpendicular to the displacements of A and B hence do no work.

U = 0:

T x A + W yG = 0 T ( a + b ) sin + W ( a cos ) = 0

T ( a + b ) sin + Wa cos = 0T = a W cot a+b T = a W cot a+b

Vector Mechanics for Engineer