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Método de los nodos: Primero dibujamos el diagrama de cuerpo libre:
Ahora aplicando las ecuaciones de equilibrio para calcular las reacciones en B y C tenemos:
𝐹! = 0 , 𝑅!! = 0
𝐹! = 0 , 𝑅!! + 𝑅!! − 945 = 0
𝑅!! + 𝑅!! = 945 (𝐼)
!!!
945!!"! !!
!! !!
!!!
! !
!!!
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 © 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 TΣ = − =
300AB
T∴ =(b) From free-body diagram of lever BCD
( )0: 200 N 0.6 300 N 0x x
F CΣ = + + =
380 N or 380 Nx x
C∴ = − =C
( )0: 0.8 300 N 0y y
F CΣ = + =
N 240or N 240 =−=∴yy
C C
Then ( ) ( )2 22 2380 240 449.44 N
x yC C C= + = + =
and °=⎟⎠
⎞⎜⎝
⎛
−−=⎟⎟
⎠
⎞⎜⎜⎝
⎛= −−
276.32380
240tantan
11
x
y
C
Cθ
or 449 N=C 32.3°▹
CO
SM
OS
: C
om
ple
te O
nli
ne
So
luti
on
s M
anu
al O
rgan
izat
ion
Sy
stem
V
ecto
r M
echa
nic
s fo
r E
ng
inee
rs:
Sta
tics
and
Dyn
am
ics,
8/e
, F
erd
inan
d P
. B
eer,
E.
Ru
ssel
l Jo
hn
sto
n,
Jr.,
Ell
iot
R.
Eis
enb
erg
, W
illi
am E
. C
lau
sen
, D
avid
Maz
ure
k,
Ph
illi
p J
. C
orn
wel
l ©
20
07
Th
e M
cGra
w-H
ill
Co
mp
anie
s.
Ch
ap
ter
4, S
olu
tion
19.
Fre
e-B
od
y D
iag
ram
:
(a)
Fro
m f
ree-
bo
dy
dia
gra
m o
f le
ver
BCD
()
()
0:
50
mm
200
N75 m
m0
CAB
MT
Σ=
−=
300
AB
T∴
=(b
) F
rom
fre
e-b
od
y d
iag
ram
of
lev
er BCD
()
0:
200
N0.6
300
N0
xx
FC
Σ=
++
=
3
80
N
o
r3
80
Nx
xC
∴=
−=
C
()
0:
0.8
30
0 N
0y
yF
CΣ
=+
=
N
240
or
N 240
=−
=∴
yy
CC
Th
en
()
()
22
22
38
02
40
449
.44
Nx
yC
CC
=+
=+
=
and
°
=⎟ ⎠⎞
⎜ ⎝⎛
−−=
⎟⎟ ⎠⎞⎜⎜ ⎝⎛
=−
−276
.32
380
240
tan
tan
11
xy
CCθ
or
4
49
N=
C32.3
°▹
𝑀! = 0:
−12 945 + 12+ 3,75 𝑅!! = 0
𝑅!! = 12 94512+ 3,75 = 720 𝑙𝑏 (𝐼𝐼)
Ahora de (II) en (I) tenemos:
𝑅!! + 𝑅!! = 945
𝑅!! + 720 = 945
𝑅!! = 945− 720 = 225 𝑙𝑏 Se aplica el método de los nodos para determinar las fuerzas internas en los elementos BA, AC y BC, y conocer si están en tracción o compresión Ahora en el nodo B
tan𝛼 =𝐶𝑂𝐶𝐴 =
912 =
34
𝛼 = tan!!34 = 36,87°
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 © 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 TΣ = − =
300AB
T∴ =(b) From free-body diagram of lever BCD
( )0: 200 N 0.6 300 N 0x x
F CΣ = + + =
380 N or 380 Nx x
C∴ = − =C
( )0: 0.8 300 N 0y y
F CΣ = + =
N 240or N 240 =−=∴yy
C C
Then ( ) ( )2 22 2380 240 449.44 N
x yC C C= + = + =
and °=⎟⎠
⎞⎜⎝
⎛
−−=⎟⎟
⎠
⎞⎜⎜⎝
⎛= −−
276.32380
240tantan
11
x
y
C
Cθ
or 449 N=C 32.3°▹
!! !!"
!!"
225!!"!
36,87°
Ahora aplicando las ecuaciones de equilibrio:
𝐹! = 0 , − 𝐹!" + 𝐹!" cos 36,87 = 0 (𝐼)
𝐹! = 0 , 225 + 𝐹!" sin 36,87 = 0
𝐹!" sin 36,87 = −225
𝐹!" =−225
sin 36,87 = −375 𝑙𝑏,
𝐹!" = 375 𝑙𝑏 𝐼𝐼 , 𝑪 𝑒𝑙 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑜 𝐵𝐴 𝑒𝑠𝑡𝑎 𝑒𝑛 𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑖ó𝑛
Ahora de (II) en (I) tenemos:
−𝐹!" + 𝐹!" cos 36,87 = 0
−𝐹!" + 375 cos 36,87 = 0
𝐹!" = 375 cos 36,87 = 300 𝑙𝑏
𝐹!" = 300 𝑙𝑏 , 𝑪 𝑒𝑙 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑜 𝐵𝐶 𝑒𝑠𝑡𝑎 𝑒𝑛 𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑖ó𝑛 Ahora en el nodo C
tan𝛽 =
𝐶𝑂𝐶𝐴 =
93,75 =
125
𝛽 = tan!!125 = 67,38°
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 © 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 TΣ = − =
300AB
T∴ =(b) From free-body diagram of lever BCD
( )0: 200 N 0.6 300 N 0x x
F CΣ = + + =
380 N or 380 Nx x
C∴ = − =C
( )0: 0.8 300 N 0y y
F CΣ = + =
N 240or N 240 =−=∴yy
C C
Then ( ) ( )2 22 2380 240 449.44 N
x yC C C= + = + =
and °=⎟⎠
⎞⎜⎝
⎛
−−=⎟⎟
⎠
⎞⎜⎜⎝
⎛= −−
276.32380
240tantan
11
x
y
C
Cθ
or 449 N=C 32.3°▹
CO
SM
OS
: C
om
ple
te O
nli
ne
So
luti
on
s M
anu
al O
rgan
izat
ion
Sy
stem
V
ecto
r M
echa
nic
s fo
r E
ng
inee
rs:
Sta
tics
and
Dyn
am
ics,
8/e
, F
erd
inan
d P
. B
eer,
E.
Ru
ssel
l Jo
hn
sto
n,
Jr.,
Ell
iot
R.
Eis
enb
erg
, W
illi
am E
. C
lau
sen
, D
avid
Maz
ure
k,
Ph
illi
p J
. C
orn
wel
l ©
20
07
Th
e M
cGra
w-H
ill
Co
mp
anie
s.
Ch
ap
ter
4, S
olu
tion
19.
Fre
e-B
od
y D
iag
ram
:
(a)
Fro
m f
ree-
bo
dy
dia
gra
m o
f le
ver
BCD
()
()
0:
50
mm
200
N75 m
m0
CAB
MT
Σ=
−=
300
AB
T∴
=(b
) F
rom
fre
e-b
od
y d
iag
ram
of
lev
er BCD
()
0:
200
N0.6
300
N0
xx
FC
Σ=
++
=
3
80
N
o
r3
80
Nx
xC
∴=
−=
C
()
0:
0.8
30
0 N
0y
yF
CΣ
=+
=
N
240
or
N 240
=−
=∴
yy
CC
Th
en
()
()
22
22
38
02
40
449
.44
Nx
yC
CC
=+
=+
=
and
°
=⎟ ⎠⎞
⎜ ⎝⎛
−−=
⎟⎟ ⎠⎞⎜⎜ ⎝⎛
=−
−276
.32
380
240
tan
tan
11
xy
CCθ
or
4
49
N=
C32.3
°▹
!!
!!"
!!"
720!!"!
67,38°
Ahora aplicando las ecuaciones de equilibrio:
𝐹! = 0 , 𝐹!" − 𝐹!" cos 67,38 = 0
𝐹!" cos 67,38 = 𝐹!"
𝐹!" = 𝐹!"
cos 67,38 = 300
cos 67,38 = 780 𝑙𝑏
𝐹!" = 780 𝑙𝑏 , 𝑻 𝑒𝑙 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑜 𝐶𝐴 𝑒𝑠𝑡𝑎 𝑒𝑛 𝑡𝑟𝑎𝑐𝑐𝑖ó𝑛
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 © 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 TΣ = − =
300AB
T∴ =(b) From free-body diagram of lever BCD
( )0: 200 N 0.6 300 N 0x x
F CΣ = + + =
380 N or 380 Nx x
C∴ = − =C
( )0: 0.8 300 N 0y y
F CΣ = + =
N 240or N 240 =−=∴yy
C C
Then ( ) ( )2 22 2380 240 449.44 N
x yC C C= + = + =
and °=⎟⎠
⎞⎜⎝
⎛
−−=⎟⎟
⎠
⎞⎜⎜⎝
⎛= −−
276.32380
240tantan
11
x
y
C
Cθ
or 449 N=C 32.3°▹
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