Post on 06-May-2015
description
Dibujamos el diagrama de cuerpo libre:
Llevamos las medidas de mm a metros:
280 ππ = 0,28 π 180 = 0,18 π 100 = 0,10 π
Aplicando las ecuaciones de equilibrio obtenemos:
π! = 0: β π΄ 0,18 + 150 sin 30 0,10 + 150 cos 30 0,28 = 0
0,28!!
0,18!!
0,10!!
30Β°
!
!
150!!
!!
!!
!
!
!
!!
COSMOS: Complete Online Solutions Manual Organization System
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 Β© 2007 The McGraw-Hill Companies.
Chapter 4, Solution 19.
Free-Body Diagram:
(a) From free-body diagram of lever BCD
( ) ( )0: 50 mm 200 N 75 mm 0C AB
M TΞ£ = β =
300AB
Tβ΄ =(b) From free-body diagram of lever BCD
( )0: 200 N 0.6 300 N 0x x
F CΞ£ = + + =
380 N or 380 Nx x
Cβ΄ = β =C
( )0: 0.8 300 N 0y y
F CΞ£ = + =
N 240or N 240 =β=β΄yy
C C
Then ( ) ( )2 22 2380 240 449.44 N
x yC C C= + = + =
and Β°=ββ
βββ
β
ββ=ββ
β
ββββ
β= ββ
276.32380
240tantan
11
x
y
C
CΞΈ
or 449 N=C 32.3Β°βΉ
π΄ = 150 sin 30 0,10 + 150 cos 30 0,28
0,18 = πππ,ππ π΅
π π΄ = 244 π β
πΉ! = 0: 243,74+ 150 sin 30+ π·! = 0
π·! = β243,74β 150 sin 30 = βπππ,ππ π΅
πΉ! = 0: π·! β 150 cos 30 = 0
π·! = 150 cos 30 = πππ,πππ π΅
β΄ π· = π·!! + π·!! = β318,74 ! + 129,904 ! = πππ,ππ π΅
π¦ π = tan!!π·!π·!
= tan!!129,904β318,74 = βππ,πππΒ°
π π· = 344 π π = 22,2Β°
COSMOS: Complete Online Solutions Manual Organization System
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 Β© 2007 The McGraw-Hill Companies.
Chapter 4, Solution 19.
Free-Body Diagram:
(a) From free-body diagram of lever BCD
( ) ( )0: 50 mm 200 N 75 mm 0C AB
M TΞ£ = β =
300AB
Tβ΄ =(b) From free-body diagram of lever BCD
( )0: 200 N 0.6 300 N 0x x
F CΞ£ = + + =
380 N or 380 Nx x
Cβ΄ = β =C
( )0: 0.8 300 N 0y y
F CΞ£ = + =
N 240or N 240 =β=β΄yy
C C
Then ( ) ( )2 22 2380 240 449.44 N
x yC C C= + = + =
and Β°=ββ
βββ
β
ββ=ββ
β
ββββ
β= ββ
276.32380
240tantan
11
x
y
C
CΞΈ
or 449 N=C 32.3Β°βΉ
COSMOS: Complete Online Solutions Manual Organization System
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 Β© 2007 The McGraw-Hill Companies.
Chapter 4, Solution 19.
Free-Body Diagram:
(a) From free-body diagram of lever BCD
( ) ( )0: 50 mm 200 N 75 mm 0C AB
M TΞ£ = β =
300AB
Tβ΄ =(b) From free-body diagram of lever BCD
( )0: 200 N 0.6 300 N 0x x
F CΞ£ = + + =
380 N or 380 Nx x
Cβ΄ = β =C
( )0: 0.8 300 N 0y y
F CΞ£ = + =
N 240or N 240 =β=β΄yy
C C
Then ( ) ( )2 22 2380 240 449.44 N
x yC C C= + = + =
and Β°=ββ
βββ
β
ββ=ββ
β
ββββ
β= ββ
276.32380
240tantan
11
x
y
C
CΞΈ
or 449 N=C 32.3Β°βΉ
COSMOS: Complete Online Solutions Manual Organization System
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 Β© 2007 The McGraw-Hill Companies.
Chapter 4, Solution 21.
Free-Body Diagram:
(a)
( ) 0in.9.0cos
in.2.4:0 =ββ
β
βββ
ββ=Ξ£ spΞx FAΞΞ±
or ( )8lb 1.2 in.
cos30sp
F kx k= = =Β°
Solving for k:
7.69800 lb/in.k = 7.70 lb/in.k = βΉ
(b)
( ) 8 lb0: 3 lb sin30 0
cos30x x
F Bβ βΞ£ = Β° + + =β βΒ°β β
or 10.7376 lbx
B = β
( )0: 3 lb cos30 0y y
F BΞ£ = β Β° + =
or 2.5981 lby
B =
( ) ( )2 210.7376 2.5981 11.0475 lb,B = β + = and
1 2.5981tan 13.6020
10.7376ΞΈ β= = Β°
Therefore: 11.05 lb=B 13.60Β° βΉ
Dibujamos el diagrama de cuerpo libre:
Aplicando las ecuaciones de equilibrio obtenemos:
π! = 0: π 2π + π cosπ β ππ + ππ = 0
π =π·
π+ ππ¨π¬π½ (π)
!! !
2!+ !
cos!
!
! !
!!
!!
!!
!!
!! !!
!!
COSMOS: Complete Online Solutions Manual Organization System
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 Β© 2007 The McGraw-Hill Companies.
Chapter 4, Solution 19.
Free-Body Diagram:
(a) From free-body diagram of lever BCD
( ) ( )0: 50 mm 200 N 75 mm 0C AB
M TΞ£ = β =
300AB
Tβ΄ =(b) From free-body diagram of lever BCD
( )0: 200 N 0.6 300 N 0x x
F CΞ£ = + + =
380 N or 380 Nx x
Cβ΄ = β =C
( )0: 0.8 300 N 0y y
F CΞ£ = + =
N 240or N 240 =β=β΄yy
C C
Then ( ) ( )2 22 2380 240 449.44 N
x yC C C= + = + =
and Β°=ββ
βββ
β
ββ=ββ
β
ββββ
β= ββ
276.32380
240tantan
11
x
y
C
CΞΈ
or 449 N=C 32.3Β°βΉ
πΉ! = 0: πΆ! β π sinπ = 0
πΆ! = π» π¬π’π§π½ (π) De la ecuaciΓ³n (a) en la ecuaciΓ³n (b) se tiene que:
πΆ! = π· π¬π’π§π½π+ ππ¨π¬π½ (π)
πΉ! = 0: πΆ! + π + π cosπ β π = 0
πΆ! = π·β π» π+ ππ¨π¬π½ (π) De la ecuaciΓ³n (a) en la ecuaciΓ³n (d) se tiene que:
πΆ! = π βπ 1+ cosπ1+ cosπ = 0
πΆ! = 0 , πΆ = πΆ!
πΆ = π· π¬π’π§π½π+ ππ¨π¬π½ (π)
ππππ π = 60Β° π π‘πππ£ππ πππ πππ’ππππππ De la ecuaciΓ³n (a) se tiene que:
π =π
1+ cosπ = π
1+ cos 60 = π
1+ 12=
ππ π·
De la ecuaciΓ³n (e) se tiene que:
πΆ = π sinπ1+ cosπ =
π sin 601+ cos 60 =
π 0,87
1+ 12= π,ππ π·
COSMOS: Complete Online Solutions Manual Organization System
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 Β© 2007 The McGraw-Hill Companies.
Chapter 4, Solution 19.
Free-Body Diagram:
(a) From free-body diagram of lever BCD
( ) ( )0: 50 mm 200 N 75 mm 0C AB
M TΞ£ = β =
300AB
Tβ΄ =(b) From free-body diagram of lever BCD
( )0: 200 N 0.6 300 N 0x x
F CΞ£ = + + =
380 N or 380 Nx x
Cβ΄ = β =C
( )0: 0.8 300 N 0y y
F CΞ£ = + =
N 240or N 240 =β=β΄yy
C C
Then ( ) ( )2 22 2380 240 449.44 N
x yC C C= + = + =
and Β°=ββ
βββ
β
ββ=ββ
β
ββββ
β= ββ
276.32380
240tantan
11
x
y
C
CΞΈ
or 449 N=C 32.3Β°βΉ
COSMOS: Complete Online Solutions Manual Organization System
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 Β© 2007 The McGraw-Hill Companies.
Chapter 4, Solution 19.
Free-Body Diagram:
(a) From free-body diagram of lever BCD
( ) ( )0: 50 mm 200 N 75 mm 0C AB
M TΞ£ = β =
300AB
Tβ΄ =(b) From free-body diagram of lever BCD
( )0: 200 N 0.6 300 N 0x x
F CΞ£ = + + =
380 N or 380 Nx x
Cβ΄ = β =C
( )0: 0.8 300 N 0y y
F CΞ£ = + =
N 240or N 240 =β=β΄yy
C C
Then ( ) ( )2 22 2380 240 449.44 N
x yC C C= + = + =
and Β°=ββ
βββ
β
ββ=ββ
β
ββββ
β= ββ
276.32380
240tantan
11
x
y
C
CΞΈ
or 449 N=C 32.3Β°βΉ