Ejerccio uno
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Transcript of Ejerccio uno
1.
En el nodo B
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 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 = + =
N 240or N 240 =-=\ yyC C
Then ( ) ( )2 22 2 380 240 449.44 Nx yC C C= + = + =
and °=-
-== -- 276.32
380
240tantan 11
x
y
C
Cq
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 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 = + =
N 240or N 240 =-=\ yyC C
Then ( ) ( )2 22 2 380 240 449.44 Nx yC C C= + = + =
and °=-
-==
-- 276.32380
240tantan 11
x
y
C
Cq
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 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 = + =
N 240or N 240 =-=\ yyC C
Then ( ) ( )2 22 2 380 240 449.44 Nx yC C C= + = + =
and °=-
-==
-- 276.32380
240tantan 11
x
y
C
Cq
or 449 N=C 32.3°
Ahora hallamos
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En el nodo C
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