1. Solución al ejercicio 1
Ahora en el nodo B
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.
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
Vector Mechanicsfor Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
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°

2. Hallemos
Hallemos
Ahora en el nodo C
Por ultimo hallemos