Solucionario de mecanica vectorial para ingenieros edicion 8 Cap 10 solucionario
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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
