Primero dibujamos el diagrama de cuerpo libre:

Ahora aplicando las ecuaciones de equilibrio:

Σ𝑀! = 0 −   𝐹!" cos 30 15 − 𝐹!" sen 30 12 + 𝑃 sen 40 15 + 𝑃 cos 40 30  = 0

  𝐹!" cos 30 15 + sen 30 12 = 𝑃 sen 40 15 + cos 40 30  = 0

18,990 𝐹!" =  32,623   𝑃

𝐹!" =  32,623  18,990 𝑃

𝐹!" =  1,718   𝑃

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 AB

M TΣ = − =

300AB

T∴ =(b) From free-body diagram of lever BCD

( )0: 200 N 0.6 300 N 0x x

F CΣ = + + =

380 N or 380 Nx x

C∴ = − =C

( )0: 0.8 300 N 0y y

F CΣ = + =

N 240or N 240 =−=∴yy

C C

Then ( ) ( )2 22 2380 240 449.44 N

x yC C C= + = + =

and °=⎟⎠

⎞⎜⎝

⎛

−−=⎟⎟

⎠

⎞⎜⎜⎝

⎛= −−

276.32380

240tantan

11

x

y

C

Cθ

or 449 N=C 32.3°▹

𝐴ℎ𝑜𝑟𝑎  𝑐𝑜𝑛  𝑃 = 0,084  𝐾𝑖𝑝𝑠

𝐹!" =  1,718   𝑃 = 1,718   0,084 = 0,144  𝐾𝑖𝑝𝑠

a.

𝐹. 𝑆.=𝐹!"#𝐹!"

=  250,144 = 173,61

b.

𝐹. 𝑆.= 𝐹. 𝑆.     <    1        ,      𝐹𝑎𝑙𝑙𝑎  𝑒𝑙  𝑐𝑎𝑏𝑙𝑒  𝐵𝐷                      𝐹. 𝑆.     >    1        ,      𝐸𝑙  𝑐𝑎𝑏𝑙𝑒  𝐵𝐷  𝑛𝑜  𝑓𝑎𝑙𝑙𝑎𝑟𝑎