The document describes solving a mechanics problem by using a free-body diagram of a lever BCD. Two forces acting on the lever are calculated in parts (a) and (b). The results are then combined to find the magnitude and direction of the net force C acting on the lever, which is determined to be 449 N at an angle of 32.3 degrees.

1. Con P = 191 N
Del diagrama de cuerpo libre tenemos:
Para la parte a del diagrama de cuerpo libre tomaremos la parte BCD
Para la parte b del diagrama de cuerpo libre tomaremos la parte BCD
S: Complete Online Solutions Manual Organization System
echanicsfor Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
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°
MOS: CompleteOnlineSolutionsManual Organization System
apter 4, Solution 19.
ee-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°
mpleteOnlineSolutionsManual Organization System
, Solution 19.
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°
COSMOS: CompleteOnlineSolutionsManual Organization System
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 = + =
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°
te Online Solutions Manual Organization System
olution 19.
iagram:
(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 Manual Organization System
(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°