Dibujando el diagrama de cuerpo libre obtenemos:

Aplicando las ecuaciones de equilibrio obtenemos:

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

∴  𝑭𝑩𝑫 =  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°▹

𝑪𝒐𝒍𝒐𝒄𝒂𝒏𝒅𝒐    𝑷 = 𝟎,𝟎𝟖𝟔  𝑲𝒊𝒑𝒔

𝑭𝑩𝑫 =  1,718   𝑃 = 1,718   0,086 = 𝟎,𝟏𝟒𝟕  𝑲𝒊𝒑𝒔

a.

𝑭.𝑺.=𝐹!"#𝐹!"

=  250,148 = 𝟏𝟔𝟖,𝟗𝟐

b.

𝑭.𝑺. 𝑭.𝑺.     <    𝟏        ,        𝑬𝒍  𝒄𝒂𝒃𝒍𝒆  𝑩𝑫  𝒇𝒂𝒍𝒍𝒂                      𝑭.𝑺.     >    𝟏        ,      𝑵𝒐  𝒇𝒂𝒍𝒍𝒂𝒓𝒂  𝒆𝒍  𝒄𝒂𝒃𝒍𝒆  𝑩𝑫