Ejercicio

Del diagrama de cuerpo tenemos: Para la parte a

Luego la solución de k es:

Para la parte 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 °=-

-==

-- 276.32380

240tantan 11

x

y

C

Cq

or 449 N=C 32.3°




Free-Body Diagram:


( ) ( )0: 50 mm 200 N 75 mm 0C ABM TS = - =

300ABT\ =


( )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°




Free-Body Diagram:


( ) ( )0: 50 mm 200 N 75 mm 0C ABM TS = - =

300ABT\ =


( )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°

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