Ejercicio dos

2. Para las reacciones A e I tenemos:

Por simetría tenemos que

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°




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 °=-

-==

-- 276.32380

240tantan 11

x

y

C

Cq

or 449 N=C 32.3°


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.


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: 0529

x AC ABF F FS = - =

2 3

0: 4.2 kN = 0529

y AC ABF F FS = - +

15.00 kN CABF = !

2.4 29ACF = 12.92 kN TACF = !

( )4 1

0: 15.00 kN 05 2

x BD BCF F FS = - - =

( )3 1

0: 15.00 kN 2.8 kN 05 2

y BD BCF F FS = - + - =

13.00 kN CBDF = !

1.6 2 kN,BCF = 2.26 kN CBCF = !

( )4 2 10: 2.4 29 kN (1.6 2 kN) 0

5 29 2y 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) 02

+ =

5.60 kN TCFF = !

continued


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.


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: 0529

x AC ABF F FS = - =

2 3

0: 4.2 kN = 0529

y AC ABF F FS = - +

15.00 kN CABF = !

2.4 29ACF = 12.92 kN TACF = !

( )4 1

0: 15.00 kN 05 2

x BD BCF F FS = - - =

( )3 1

0: 15.00 kN 2.8 kN 05 2

y BD BCF F FS = - + - =

13.00 kN CBDF = !

1.6 2 kN,BCF = 2.26 kN CBCF = !

( )4 2 10: 2.4 29 kN (1.6 2 kN) 0

5 29 2y 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) 02

+ =

5.60 kN TCFF = !

continued




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°

Ejercicio dos

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