1. 2.
Para las reacciones A e I tenemos:
Por simetría tenemos que
Vector Mechanicsfor Engineers: Staticsand Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 TheMcGraw-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 ABM TS = - =
300ABT =
(b) From free-body diagram of lever BCD
( )0: 200 N 0.6 300 N 0x xF CS = + + =
380 N or 380 Nx xC = - =C
( )0: 0.8 300 N 0y yF CS = + =
N240orN240 =-= yyC C
Then ( ) ( )
2 22 2
380 240 449.44 Nx yC C C= + = + =
and °=
-
-
== --
276.32
380
240
tantan 11
x
y
C
C
q
or 449 N=C 32.3°
COSMOS: Complete Online Solutions Manual Organization System
Chapter 4, Solution 19.
Free-Body Diagram:
(a) From free-body diagram of lever BCD
( ) ( )0: 50 mm 200 N 75 mm 0C ABM TS = - =
300ABT =
(b) From free-body diagram of lever BCD
( )0: 200 N 0.6 300 N 0x xF CS = + + =
380 N or 380 Nx xC = - =C
( )0: 0.8 300 N 0y yF CS = + =
N240orN240 =-= yyC C
Then ( ) ( )2 22 2
380 240 449.44 Nx yC C C= + = + =
and °=
-
-
== --
276.32
380
240
tantan 11
x
y
C
C
q
or 449 N=C 32.3°
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 13.
FBD Truss:
Joint FBDs:
Joint A:
Joint B:
Joint C:
0: 0x xFS = =A
0: (8 m) (4 m)(4.2 kN) (2m)(2.8 kN) 0AM GyS = - - =
2.80 kNy =G
0: 2.8 kN 4.2 kN + 2.8 kN = 0y yF AS = - -
4.2 kNy =A
5 4
0: 0
529
x AC ABF F FS = - =
2 3
0: 4.2 kN = 0
529
y AC ABF F FS = - +
15.00 kN CABF = !
2.4 29ACF = 12.92 kN TACF = !
( )
4 1
0: 15.00 kN 0
5 2
x BD BCF F FS = - - =
( )
3 1
0: 15.00 kN 2.8 kN 0
5 2
y BD BCF F FS = - + - =
13.00 kN CBDF = !
1.6 2 kN,BCF = 2.26 kN CBCF = !
( )4 2 1
0: 2.4 29 kN (1.6 2 kN) 0
5 29 2
y CDF FS = - - =
8.00kN TCDF = !
( )
3 5
0: 8.00 kN (2.4 29 kN)
5 29
x CFF FS = + -
1
(1.6 2 kN) 0+ =
COSMOS: Complete Online Solutions Manual Organization System
Chapter 6, Solution 13.
FBD Truss:
Joint FBDs:
Joint A:
Joint B:
0: 0x xFS = =A
0: (8 m) (4 m)(4.2 kN) (2m)(2.8 kN) 0AM GyS = - - =
2.80 kNy =G
0: 2.8 kN 4.2 kN + 2.8 kN = 0y yF AS = - -
4.2 kNy =A
5 4
0: 0
529
x AC ABF F FS = - =
2 3
0: 4.2 kN = 0
529
y AC ABF F FS = - +
15.00 kN CABF = !
2.4 29ACF = 12.92 kN TACF = !
( )
4 1
0: 15.00 kN 0
5 2
x BD BCF F FS = - - =
( )
3 1
0: 15.00 kN 2.8 kN 0
5 2
y BD BCF F FS = - + - =

2. Vector Mechanicsfor Engineers: Staticsand 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.
( ) ( )0: 50 mm 200 N 75 mm 0C ABM TS = - =
300ABT =
(b) From free-body diagram of lever BCD
( )0: 200 N 0.6 300 N 0x xF CS = + + =
380 N or 380 Nx xC = - =C
( )0: 0.8 300 N 0y yF CS = + =
N240orN240 =-= yyC C
Then ( ) ( )2 22 2
380 240 449.44 Nx yC C C= + = + =
and °=
-
-
== --
276.32
380
240
tantan 11
x
y
C
C
q
or 449 N=C 32.3°