Eligheor
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Transcript of Eligheor
Dibujamos el diagrama de cuerpo libre:
Llevamos las medidas de mm a metros:
280 𝑚𝑚 = 0,28 𝑚 180 = 0,18 𝑚 100 = 0,10 𝑚
Aplicando las ecuaciones de equilibrio obtenemos:
𝑀! = 0: − 𝐴 0,18 + 150 sin 30 0,10 + 150 cos 30 0,28 = 0
0,28!!
0,18!!
0,10!!
30°
!
!
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 © 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°▹
𝐴 = 150 sin 30 0,10 + 150 cos 30 0,28
0,18 = 𝟐𝟒𝟑,𝟕𝟒 𝑵
𝑜 𝐴 = 244 𝑁 →
𝐹! = 0: 243,74+ 150 sin 30+ 𝐷! = 0
𝐷! = −243,74− 150 sin 30 = −𝟑𝟏𝟖,𝟕𝟒 𝑵
𝐹! = 0: 𝐷! − 150 cos 30 = 0
𝐷! = 150 cos 30 = 𝟏𝟐𝟗,𝟗𝟎𝟒 𝑵
∴ 𝐷 = 𝐷!! + 𝐷!! = −318,74 ! + 129,904 ! = 𝟑𝟒𝟒,𝟐𝟎 𝑵
𝑦 𝜃 = tan!!𝐷!𝐷!
= tan!!129,904−318,74 = −𝟐𝟐,𝟏𝟕𝟒°
𝑜 𝐷 = 344 𝑁 𝜃 = 22,2°
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°▹
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°▹
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 21.
Free-Body Diagram:
(a)
( ) 0in.9.0cos
in.2.4:0 =−⎟
⎠
⎞⎜⎝
⎛−=Σ spΒx FAΜα
or ( )8lb 1.2 in.
cos30sp
F kx k= = =°
Solving for k:
7.69800 lb/in.k = 7.70 lb/in.k = ▹
(b)
( ) 8 lb0: 3 lb sin30 0
cos30x x
F B⎛ ⎞Σ = ° + + =⎜ ⎟°⎝ ⎠
or 10.7376 lbx
B = −
( )0: 3 lb cos30 0y y
F BΣ = − ° + =
or 2.5981 lby
B =
( ) ( )2 210.7376 2.5981 11.0475 lb,B = − + = and
1 2.5981tan 13.6020
10.7376θ −= = °
Therefore: 11.05 lb=B 13.60° ▹
Dibujamos el diagrama de cuerpo libre:
Aplicando las ecuaciones de equilibrio obtenemos:
𝑀! = 0: 𝑇 2𝑎 + 𝑎 cos𝜃 − 𝑇𝑎 + 𝑃𝑎 = 0
𝑇 =𝑷
𝟏+ 𝐜𝐨𝐬𝜽 (𝑎)
!! !
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 © 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°▹
𝐹! = 0: 𝐶! − 𝑇 sin𝜃 = 0
𝐶! = 𝑻 𝐬𝐢𝐧𝜽 (𝑏) De la ecuación (a) en la ecuación (b) se tiene que:
𝐶! = 𝑷 𝐬𝐢𝐧𝜽𝟏+ 𝐜𝐨𝐬𝜽 (𝑐)
𝐹! = 0: 𝐶! + 𝑇 + 𝑇 cos𝜃 − 𝑃 = 0
𝐶! = 𝑷− 𝑻 𝟏+ 𝐜𝐨𝐬𝜽 (𝑑) De la ecuación (a) en la ecuación (d) se tiene que:
𝐶! = 𝑃 −𝑃 1+ cos𝜃1+ cos𝜃 = 0
𝐶! = 0 , 𝐶 = 𝐶!
𝐶 = 𝑷 𝐬𝐢𝐧𝜽𝟏+ 𝐜𝐨𝐬𝜽 (𝑒)
𝑃𝑎𝑟𝑎 𝜃 = 60° 𝑎 𝑡𝑟𝑎𝑣𝑒𝑠 𝑑𝑒𝑙 𝑒𝑛𝑢𝑛𝑐𝑖𝑎𝑑𝑜 De la ecuación (a) se tiene que:
𝑇 =𝑃
1+ cos𝜃 = 𝑃
1+ cos 60 = 𝑃
1+ 12=
𝟐𝟑 𝑷
De la ecuación (e) se tiene que:
𝐶 = 𝑃 sin𝜃1+ cos𝜃 =
𝑃 sin 601+ cos 60 =
𝑃 0,87
1+ 12= 𝟎,𝟓𝟖 𝑷
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°▹
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°▹