(1) The document discusses free-body diagrams for parts (a) and (b) of a problem. (2) For part (a), the free-body diagram of lever BCD is used to solve for the force of 300 N. (3) For part (b), the free-body diagram is again used, solving for forces of 380 N and 240 N, which are then combined to find the total force of 449 N at an angle of 32.3 degrees.

1. Del diagrama de cuerpo tenemos:
Para la parte a
Luego la solución de k es:
Para la parte b
MOS: Complete Online Solutions Manual Organization System
Mechanicsfor Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
7 The McGraw-Hill Companies.
pter 4, Solution 19.
e-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 = + =
N240orN240 =-= yyC C
Then ( ) ( )2 22 2
380 240 449.44 Nx yC C C= + = + =
and °=
-
-
== --
276.32
380
240
tantan 11
x
y
C
C
q
or 449 N=C 32.3°
S: CompleteOnlineSolutionsManual Organization System
er 4, Solution 19.
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 = + =
N240orN240 =-= yyC C
Then ( ) ( )
2 22 2
380 240 449.44 Nx yC C C= + = + =
and °=
-
-
== --
276.32
380
240
tantan 11
x
y
C
C
q
or 449 N=C 32.3°
OSMOS: CompleteOnlineSolutionsManual Organization System
hapter 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 = + =
N240orN240 =-= yyC C
Then ( ) ( )2 22 2
380 240 449.44 Nx yC C C= + = + =
and °=
-
-
== --
276.32
380
240
tantan 11
x
y
C
C
q
or 449 N=C 32.3°