SOLUTION Continued
(b) Load applied at D.
Free body: Member ACF.
directed along CF.
7 in. 14 lb
yy
F= =F
PROBLEM 6.61
The 48–lb load is removed and a 288–lb · in. clockwise couple is applied
successively at A, D, and E. Determine the components of the reactions at B
and F if the couple is applied (a) at A, (b) at D, (c) at E.
SOLUTION
Free body: Entire frame
The following analysis is valid for (a), (b), and (c), since the point of
application of the couple is immaterial.
0: 288 lb in. (12 in.) 0
Fx
MBΣ = − ⋅− =
24 lb 24 lb
xx
B=−=B
0: 24 lb+ 0
xx
FFΣ= − =
24 lb 24 lb
xx
F= =F
0: 0
y yy
F BFΣ= + =
(1)
(a) Couple applied at A.
Free body: Member CDB
5 in. 7.5 lb
24 lb 16 in.
yy
B= =B
From Eq. (1):
7.5 lb 0
y
F− +=
7.5 lb 7.5 lb
yy
F= =F
Thus, reactions are:
24.0 lb
x=B
,
7.50 lb
y
=B
24.0 lb
x
=
F
,
7.50 lb
y=F
consent of McGraw–Hill Education.
SOLUTION Continued
(b) Couple applied at D.
Free body: Member ACF.
PROBLEM 6.62
Determine all the forces exerted on member AI if the frame is loaded
by a clockwise couple of magnitude 1200 lb ∙ in. applied (a) at Point D,
(b) at Point E.
SOLUTION
Free body: Entire frame:
Location of couple is immaterial.
0: (48 in.) 1200 lb in. 0
H
MIΣ = − ⋅=
25.0 lbI= +
(a) and (b)
25.0 lb=I
We note that AB, BC, and FG are two–force members.
Free body: Member AI:
20 5
tan 22.6
48 12
aa
= = = °
(a) Couple applied at D.
5
0: 25 lb 0
13
y
FAΣ= − + =
65.0 lbA= +
65.0 lb=A
22.6°
12
0: (65 lb)(40 in.) (20 in.) 0
13
G
MCΣ= − =
120 lbC= +
120 lb=C
12
0: (65 lb) 120 lb 0
13
x
FGΣ= − + +=
60.0 lbG= −
60 lb=G
consent of McGraw–Hill Education.
SOLUTION Continued
(b) Couple applied at E.
5
0: 25 lb 0
13
y
FA
Σ= − + =
65.0 lbA= +
65.0 lb=A
22.6°
12
0: (65 lb) (40 in.) (20 in.) 1200 lb in. 0
13
G
MCΣ = + − − ⋅=
60.0 lbC= +
60.0 lb=C
12
0: (65 lb) 60 lb 0
13
x
FGΣ= − + +=
0=G
consent of McGraw–Hill Education.
PROBLEM 6.63
The hydraulic cylinder CF, which partially controls the position of
rod DE, has been locked in the position shown. Knowing that
θ = 60°, determine (a) the force P for which the tension in link AB is
410 N, (b) the corresponding force exerted on member BCD at
point C.
SOLUTION
Free body: Member BCD:
Since AB is a two–force member, the force it exerts at B is directed as shown above.
(a)
40 9
0: (410N)(100cos20 ) (410N)(100sin 20 )
41 41
( cos60 )(175sin 20 ) ( sin60 )(175cos20 ) 0
C
M
PP
Σ= −
− −=
(175)sin(60 20 ) (400cos20 90sin 20 )(100)
200.24N
P
P
+= −
=
P
= 200 N 60°
(b)
F 0: (410 N)+(200.24 N)cos 60 0
10.12N
40
F 0: (410 N) (200.24 N)sin60 0
41
=+573.4 N
xx
x
yy
y
C
C
C
C
+↑
+= =
= −
=−− =
∑
∑
FC
P
= 573 N 89.0°
consent of McGraw–Hill Education.
PROBLEM 6.64
The hydraulic cylinder CF, which partially controls the position of
rod DE, has been locked in the position shown. Knowing that
P = 400 N and θ = 75°, determine (a) the force in link AB (b) the
corresponding force exerted on member BCD at point C.
SOLUTION
Free body: Member BCD:
Since AB is a two–force member, the force it exerts at B is directed as shown above.
(a)
40 9
0: (100cos 20 ) (100sin 20 )
41 41
(400 N)cos75 (175sin 20 ) (400 N)sin75 (175cos20 ) 0
100
(40cos20 9sin 20 ) (400 N)(175)sin(75 20 )
41
C AB AB
AB
MF F
F
Σ= −
−−=
−= +
AA
AA A A
A A AA
41 sin95
(400)(175) = 828.49 N
100 40cos 20 9sin 20
AB AB
FF=−
A
AA
FAB = 828 N T
consent of McGraw–Hill Education.
SOLUTION Continued
(b)
9
F 0: (828.49)+400 cos 75 0
41
78.34 N
40
F 0: (828.49) 400 sin 75 0
41
=+1194.7 N
xx
x
yy
y
C
C
C
C
+↑
+= − =
=
=−+=
∑
∑
A
A
FC
C
= 1197 N
∡
86.2°
consent of McGraw–Hill Education.
PROBLEM 6.65
Two 9–in.–diameter pipes (pipe 1 and pipe 2) are supported every 7.5 ft
by a small frame like that shown. Knowing that the combined weight of
each pipe and its contents is 30 lb/ft and assuming frictionless surfaces,
determine the components of the reactions at A and G.
SOLUTION
Free-body: Pipe 2
(30 lb/ft)(7.5 ft) 225 lbW= =
225 lb
8 17 15
FD
= =
120 lb=F
255 lb=D
Geometry of pipe 2
4.5 in.r=
By symmetry:
CF CD=
(1)
Equate horizontal distance:
8 15
17 17
25 15
17 17
25 5
15 3
r r CD
r CD
CD r r
+=
=
= =
From Eq. (1):
55
(4.5 in.)
33
7.5 in.
CF r
CF
= =
=
Free-body: Member CFG
0: (120 lb)(7.5 in.) (16 in.) 0
Cx
MGΣ= − =
56.25 lb
x
G=
56.3 lb
x
=G
consent of McGraw–Hill Education.
SOLUTION Continued
Free body: Frame and pipes
Note: Pipe 2 is similar to pipe 1.
7.5 in.
120 lb
AE CF
= =
= =EF
0: (15 in.) (56.25 lb)(24 in.) (225 lb)(4.5 in.)
Ay
MGΣ= − −
(225 lb)(19.5 in.) (120 lb)(7.5 in.) 0− −=
510 lb
y
G=
510 lb
y=G
0: 120 lb 56.25 lb 0
xx
FAΣ= + + =
176.25 lb
x
A=
176.3 lb
x=A
0: 510 lb 225 lb 225 lb 0
yy
FAΣ= +−−=
60 lb
y
A= −
60.0 lb
y=A
consent of McGraw–Hill Education.
PROBLEM 6.61
The 48–lb load is removed and a 288–lb · in. clockwise couple is applied
successively at A, D, and E. Determine the components of the reactions at B
and F if the couple is applied (a) at A, (b) at D, (c) at E.
SOLUTION
Free body: Entire frame
The following analysis is valid for (a), (b), and (c), since the point of
application of the couple is immaterial.
0: 288 lb in. (12 in.) 0
Fx
MBΣ = − ⋅− =
24 lb 24 lb
xx
B=−=B
0: 24 lb+ 0
xx
FFΣ= − =
24 lb 24 lb
xx
F= =F
0: 0
y yy
F BFΣ= + =
(1)
(a) Couple applied at A.
Free body: Member CDB
5 in. 7.5 lb
24 lb 16 in.
yy
B= =B
From Eq. (1):
7.5 lb 0
y
F− +=
7.5 lb 7.5 lb
yy
F= =F
Thus, reactions are:
24.0 lb
x=B
,
7.50 lb
y
=B
24.0 lb
x
=
F
,
7.50 lb
y=F
consent of McGraw–Hill Education.
SOLUTION Continued
(b) Couple applied at D.
Free body: Member ACF.
PROBLEM 6.62
Determine all the forces exerted on member AI if the frame is loaded
by a clockwise couple of magnitude 1200 lb ∙ in. applied (a) at Point D,
(b) at Point E.
SOLUTION
Free body: Entire frame:
Location of couple is immaterial.
0: (48 in.) 1200 lb in. 0
H
MIΣ = − ⋅=
25.0 lbI= +
(a) and (b)
25.0 lb=I
We note that AB, BC, and FG are two–force members.
Free body: Member AI:
20 5
tan 22.6
48 12
aa
= = = °
(a) Couple applied at D.
5
0: 25 lb 0
13
y
FAΣ= − + =
65.0 lbA= +
65.0 lb=A
22.6°
12
0: (65 lb)(40 in.) (20 in.) 0
13
G
MCΣ= − =
120 lbC= +
120 lb=C
12
0: (65 lb) 120 lb 0
13
x
FGΣ= − + +=
60.0 lbG= −
60 lb=G
consent of McGraw–Hill Education.
SOLUTION Continued
(b) Couple applied at E.
5
0: 25 lb 0
13
y
FA
Σ= − + =
65.0 lbA= +
65.0 lb=A
22.6°
12
0: (65 lb) (40 in.) (20 in.) 1200 lb in. 0
13
G
MCΣ = + − − ⋅=
60.0 lbC= +
60.0 lb=C
12
0: (65 lb) 60 lb 0
13
x
FGΣ= − + +=
0=G
consent of McGraw–Hill Education.
PROBLEM 6.63
The hydraulic cylinder CF, which partially controls the position of
rod DE, has been locked in the position shown. Knowing that
θ = 60°, determine (a) the force P for which the tension in link AB is
410 N, (b) the corresponding force exerted on member BCD at
point C.
SOLUTION
Free body: Member BCD:
Since AB is a two–force member, the force it exerts at B is directed as shown above.
(a)
40 9
0: (410N)(100cos20 ) (410N)(100sin 20 )
41 41
( cos60 )(175sin 20 ) ( sin60 )(175cos20 ) 0
C
M
PP
Σ= −
− −=
(175)sin(60 20 ) (400cos20 90sin 20 )(100)
200.24N
P
P
+= −
=
P
= 200 N 60°
(b)
F 0: (410 N)+(200.24 N)cos 60 0
10.12N
40
F 0: (410 N) (200.24 N)sin60 0
41
=+573.4 N
xx
x
yy
y
C
C
C
C
+↑
+= =
= −
=−− =
∑
∑
FC
P
= 573 N 89.0°
consent of McGraw–Hill Education.
PROBLEM 6.64
The hydraulic cylinder CF, which partially controls the position of
rod DE, has been locked in the position shown. Knowing that
P = 400 N and θ = 75°, determine (a) the force in link AB (b) the
corresponding force exerted on member BCD at point C.
SOLUTION
Free body: Member BCD:
Since AB is a two–force member, the force it exerts at B is directed as shown above.
(a)
40 9
0: (100cos 20 ) (100sin 20 )
41 41
(400 N)cos75 (175sin 20 ) (400 N)sin75 (175cos20 ) 0
100
(40cos20 9sin 20 ) (400 N)(175)sin(75 20 )
41
C AB AB
AB
MF F
F
Σ= −
−−=
−= +
AA
AA A A
A A AA
41 sin95
(400)(175) = 828.49 N
100 40cos 20 9sin 20
AB AB
FF=−
A
AA
FAB = 828 N T
consent of McGraw–Hill Education.
SOLUTION Continued
(b)
9
F 0: (828.49)+400 cos 75 0
41
78.34 N
40
F 0: (828.49) 400 sin 75 0
41
=+1194.7 N
xx
x
yy
y
C
C
C
C
+↑
+= − =
=
=−+=
∑
∑
A
A
FC
C
= 1197 N
∡
86.2°
consent of McGraw–Hill Education.
PROBLEM 6.65
Two 9–in.–diameter pipes (pipe 1 and pipe 2) are supported every 7.5 ft
by a small frame like that shown. Knowing that the combined weight of
each pipe and its contents is 30 lb/ft and assuming frictionless surfaces,
determine the components of the reactions at A and G.
SOLUTION
Free-body: Pipe 2
(30 lb/ft)(7.5 ft) 225 lbW= =
225 lb
8 17 15
FD
= =
120 lb=F
255 lb=D
Geometry of pipe 2
4.5 in.r=
By symmetry:
CF CD=
(1)
Equate horizontal distance:
8 15
17 17
25 15
17 17
25 5
15 3
r r CD
r CD
CD r r
+=
=
= =
From Eq. (1):
55
(4.5 in.)
33
7.5 in.
CF r
CF
= =
=
Free-body: Member CFG
0: (120 lb)(7.5 in.) (16 in.) 0
Cx
MGΣ= − =
56.25 lb
x
G=
56.3 lb
x
=G
consent of McGraw–Hill Education.
SOLUTION Continued
Free body: Frame and pipes
Note: Pipe 2 is similar to pipe 1.
7.5 in.
120 lb
AE CF
= =
= =EF
0: (15 in.) (56.25 lb)(24 in.) (225 lb)(4.5 in.)
Ay
MGΣ= − −
(225 lb)(19.5 in.) (120 lb)(7.5 in.) 0− −=
510 lb
y
G=
510 lb
y=G
0: 120 lb 56.25 lb 0
xx
FAΣ= + + =
176.25 lb
x
A=
176.3 lb
x=A
0: 510 lb 225 lb 225 lb 0
yy
FAΣ= +−−=
60 lb
y
A= −
60.0 lb
y=A
consent of McGraw–Hill Education.