1) The document uses the method of joints to analyze a lever system subjected to forces. It draws the free body diagram and applies equations of equilibrium to calculate the reactions at points B and C. 2) It then uses the reactions to determine the internal forces (tension or compression) in members BA, BC, and CA. It finds the forces to be 375 lb of compression in BA, 300 lb of compression in BC, and 780 lb of tension in CA. 3) Angles of 36.87° and 67.38° are also calculated in analyzing the system.

1.
Mediante
el
método
de
los
nodos:
Dibujando
el
diagrama
de
cuerpo
libre:
or C = 449 N
!
⎛ Cy ⎞
− 240 ⎞
⎟ = tan −1 ⎛
⎟ = 32.276°
⎜
⎟
⎝ − 380 ⎠
⎝ Cx ⎠
( 380 )2 + ( 240 )2
= 449.44 N
!
or
ΣFy = 0: C y + 0.8 ( 300 N ) = 0
C y = 240 N
C x = 380 N
!!!
or
Chapter 4, Solution 19.
ΣFx = 0: 200 N + Cx + 0.6 ( 300 N ) = 0
ΣM C = 0: TAB ( 50 mm ) − 200 N ( 75 mm ) = 0
!!!
!!
COSMOS: Complete Online Solutions Manual Organization System
Free-Body Diagram:
945!!"!
32.3° ▹
∴ TAB = 300
!!
!!
!!!
and
θ = tan −1 ⎜
⎜
C =
ΣFy = 0: C y + 0.8 ( 300 N ) = 0
𝐹! = 0 ,
𝑅! + 𝑅! − 945 = 0
∴ C y!= −240 !
N
or
C y = 240 N
2
𝑅!! = C!!+= y = ( 380 )2 + ( 240 )2 = 449.44 N
945 (𝑎)
C + 𝑅x C 2
and
Then
2
2
Cx + C y =
∴ C y = −240 N
∴ C x = −380 N
ΣFx = 0: 0 , N + Cx + = 0
300 N ) = 0
𝐹! = 200
𝑅!! 0.6 (
∴ C x = −380 N
or
C x = 380 N
Then
⎛ Cy ⎞
− 240 ⎞
⎟ = tan −1 ⎛
⎟ = 32.276°
⎜
⎟
⎝ − 380 ⎠
⎝ Cx ⎠
θ = tan −1 ⎜
⎜
or C = 449 N
ram:
tion 19.
nline Solutions Manual Organization System
(b) From free-body diagram of lever BCD
(a) From free-body diagram of lever BCD
(a) From free-body diagram of lever BCD
Aplicando
las
ecuaciones
de
equilibrio
para
calcular
las
(reacciones
en
B
y
C
ΣM C = 0: TAB ( 50 mm ) − 200 N 75 mm ) = 0
obtenemos
que:
∴ TAB = 300
(b) From free-body diagram of lever BCD
32.3° ▹

2. Chapter 4, Solution 19.
Free-Body Diagram:
(a) From free-body diagram of lever BCD
ΣM C = 0: TAB ( 50 mm ) − 200 N ( 75 mm ) = 0
𝑀! = 0:
(b) From free-body diagram of lever BCD
ΣFx 3,75
N C +
−12 945 + 12 += 0: 200𝑅!! + =x 0
0.6 ( 300 N ) = 0
∴ C x = −380 N
𝑅!! =
Ahora
de
(b)
en
(a)
tenemos:
Then
∴ TAB = 300
and
C x = 380 N
or
12 945 = 0: C + 0.8 300 N = 0
ΣFy = 720 𝑙𝑏 (𝑏)
(
)
y
12 + 3,75
∴ C y = −240 N
C =
2
2
Cx + C y =
C y = 240 N
or
( 380 )2 + ( 240 )2
= 449.44 N
𝑅!! + 𝑅!! = ⎛ 945
Cy ⎞
−1
− 240 ⎞
⎟ = tan −1 ⎛
⎟ = 32.276°
⎜
⎟
⎝ − 380 ⎠
⎝ Cx ⎠
θ = tan ⎜
⎜
𝑅!! + 720 = 945
or C = 449 N
32.3° ▹
𝑅!! = 945 − 720 = 225 𝑙𝑏
Aplicando
el
método
de
los
nodos
para
obtener
las
fuerzas
internas
en
los
elementos
BA,
AC
y
BC,
y
así
poder
conocer
si
están
en
tracción
o
compresión
En
el
nodo
B
tenemos:
!!"
36,87°
!!
!!"
225!!"!
𝐶𝑂
9
3
tan 𝛼 =
=
=
𝐶𝐴 12
4
𝛼 = tan!!
3
= 36,87°
4
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.

3. ⎛ Cy ⎞
− 240 ⎞
⎟ = tan −1 ⎛
⎟=3
⎜
Cx ⎟
⎝ − 380 ⎠
⎠
⎝
( 380 )2 + ( 24
or
ΣFy = 0: C y + 0.8 ( 300 N ) = 0
or
ΣFx = 0: 200 N + Cx + 0.6 ( 300
(b) From free-body diagram of lever BCD
ΣM C = 0: TAB ( 50 mm ) − 200 N (
(a) From free-body diagram of lever BCD
∴ C x = −380 N
ΣM C = 0: TAB ( 50 mm ) − 200 N ( 75 mm ) = 0
2
2
Cx + C y =
∴ C y = −240 N
(a) From free-body diagram of lever BCD
-Body Diagram:
Aplicando
las
ecuaciones
de
equilibrio
obtenemos
que:
(b) From free-body diagram of lever BCD
θ = tan −1 ⎜
⎜
C =
ΣFy = 0: C y + 0.8 ( 300 N ) = 0
𝐹! = 0 ,
225 + 𝐹!" sin 36,87 = 0
∴ C y = −240 N
C =
C y = 240 N
or
2
2
C x + C y𝐹!" sin 36,87( 240 −225
= ( 380 ) + = ) = 449.44 N
2
and
Then
ΣFx = 0: 0 , − Cx + 0.6 ( 𝐹 cos = 0
𝐹! = 200 N + 𝐹!" + 300 N ) 36,87 = 0 (𝑎)
!"
∴ C x = −380 N
or
C x = 380 N
Then
2
⎛ Cy ⎞
−1 ⎛ − 240
⎟ = tan−225 ⎞ = 32.276°
⎟
⎜
⎟=
!"
⎝ − 380 = −375 𝑙𝑏,
⎝ C x ⎠ sin 36,87 ⎠
θ = tan −1 ⎜ 𝐹
⎜
and
or C = 449 N
Free-Body Diagram:
Chapter 4, Solution 19.
COSMOS: Complete Online Solutions Manual Organization System
∴ TAB = 300
32.3° ▹
𝐹!" = 375 𝑙𝑏 𝑏 , 𝑪 𝑒𝑙 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑜 𝐵𝐴 𝑒𝑠𝑡𝑎 𝑒𝑛 𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑖ó𝑛
Ahora
de
(b)
en
(a)
tenemos:
−𝐹!" + 𝐹!" cos 36,87 = 0
−𝐹!" + 375 cos 36,87 = 0
𝐹!" = 375 cos 36,87 = 300 𝑙𝑏
𝐹!! = 300 𝑙𝑏 , 𝑪 𝑒𝑙 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑜 𝐵𝐶 𝑒𝑠𝑡𝑎 𝑒𝑛 𝑐𝑜𝑚𝑝𝑟𝑒𝑠𝑖ó𝑛
En
el
nodo
C
tenemos
que:
!!"
67,38°
!!
!!"
720!!"!
𝐶𝑂
9
12
tan 𝛽 =
=
=
𝐶𝐴
5
Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr., 3,75
Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
The McGraw-Hill Companies.
12
𝛽 = tan!!
= 67,38°
5

4. Free-Body Diagram:
(a) From free-body diagram of lever BCD
ΣM C = 0: TAB ( 50 mm ) − 200 N ( 75 mm ) = 0
Aplicando
las
ecuaciones
de
equilibrio
tenemos
que:
(b) From free-body diagram of lever BCD
ΣFx = 0: 0 , 𝐹 C x− 0.6 ( 300 N ) = 0 = 0
𝐹! = 200 N + + 𝐹!" cos 67,38
!"
∴ C x = −380 N
or
C x = 380 N
ΣFy = 0:𝐹!" cos 0.8 ( 300 N ) = 𝐹!"
C y + 67,38 = 0
or
C y = 240 N
300
= 780 𝑙𝑏
cos2 + ( 240 2
2
2
C = cos 67,38 ( 380 ) 67,38 ) = 449.44 N
Cx + C y =
𝐹!" =
∴ TAB = 300
Then
∴ 𝐹C y = −240 N
!"
=
⎛ Cy ⎞
− 240 ⎞
⎟ = tan −1 ⎛
⎟ = 32.276°
⎜
⎟
𝐹!" = 780 𝑙𝑏 , 𝑻 𝑒𝑙 𝑒𝑙𝑒𝑚𝑒𝑛𝑡𝑜 𝐶𝐴 𝑒𝑠𝑡𝑎 𝑒𝑛 𝑡𝑟𝑎𝑐𝑐𝑖ó𝑛
⎝ − 380 ⎠
⎝ Cx ⎠
and
θ = tan −1 ⎜
⎜
ector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
lliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
or C = 449 N
32.3° ▹