Dibujando el diagrama de cuerpo libre obtenemos:
Aplicando las ecuaciones de equilibrio obtenemos:
Ξ£π! = 0 β πΉ!" cos 30 15 β πΉ!" sen 30 12 + π sen 40 15 + π cos 40 30 = 0
πΉ!" cos 30 15 + sen 30 12 = π sen 40 15 + cos 40 30 = 0
18,990 πΉ!" = 32,623 π
πΉ!" = 32,623 18,990 π
β΄ ππ©π« = π,πππ π·
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 Β© 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 TΞ£ = β =
300AB
Tβ΄ =(b) From free-body diagram of lever BCD
( )0: 200 N 0.6 300 N 0x x
F CΞ£ = + + =
380 N or 380 Nx x
Cβ΄ = β =C
( )0: 0.8 300 N 0y y
F CΞ£ = + =
N 240or N 240 =β=β΄yy
C C
Then ( ) ( )2 22 2380 240 449.44 N
x yC C C= + = + =
and Β°=ββ
βββ
β
ββ=ββ
β
ββββ
β= ββ
276.32380
240tantan
11
x
y
C
CΞΈ
or 449 N=C 32.3Β°βΉ
πͺπππππππ π π· = π,πππ π²πππ
ππ©π« = 1,718 π = 1,718 0,086 = π,πππ π²πππ
a.
π.πΊ.=πΉ!"#πΉ!"
= 250,148 = πππ,ππ
b.
π.πΊ. π.πΊ. < π , π¬π πππππ π©π« πππππ π.πΊ. > π , π΅π πππππππ ππ πππππ π©π«
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