PROBLEM 6.106
A 3–ft–diameter pipe is supported every 16 ft by a small frame like
that shown. Knowing that the combined weight of the pipe and its
contents is 500 lb/ft and assuming frictionless surfaces, determine
the components (a) of the reaction at E, (b) of the force exerted at C
on member CDE.
consent of McGraw–Hill Education.
SOLUTION Continued
Free body: Member CDE:
From above, we have
(b)
4.00 kips
x=C
0: (10 kips)(7 ft) (4 kips)(6 ft) (8 ft) 0
Ey
MCΣ= − − =
5.75 kips,
y
C= +
5.75 kips
y=
C
(a)
3
0: 4 kips (10 kips) 0
5
xx
FEΣ= − + + =
2 kips,
x
E= −
2.00 kips
x=E
4
0: 5.75 kips (10 kips) 0,
5
yy
FEΣ= − + =
2.25 kips
y=E
consent of McGraw–Hill Education.
PROBLEM 6.107
For the bevel–gear system shown, determine the required value of α
if the ration of MB to MA is to be 3.
Gear A:
A
r
Gear B:
tan
AB
rr
a
=
0: 0
A AA
M M Jr=−==
∑
(1)
AA
M Jr=
0: 0
B BB
M M Jr= −=
∑
(2)
BB
M Jr=
Ratio:
3 : 3
B BB BA
A AA
M Jr r rr
M Jr r
= = = =
3 tan
BB
rr
a
=
1
tan =18.4
3
aa
=
consent of McGraw–Hill Education.
PROBLEM 6.108
A 400–kg block may be supported by a small frame in each of the four ways shown. The diameter of the
pulley is 250mm. For each case, determine (a) the force components and the couple representing the
reaction at A, (b) the force exerted at D on the vertical member.
SOLUTION
Load:
2
(400kg)(9.81 m / ) 3924NWs= =
(a) Free Body: Frame
0:
x
F=
∑
0 = 0
xx
A=A
0: 7848 N 0 7.85 kN
yy y
FA=−== ↑
∑A
0: (7848N)(2m) 0
AA
MM=−=
∑
15696 N m =15.70 kN m
AA
M=⋅⋅M
Free Body: Frame: Member BEC
0: ( cos45 )(1 m) (7848N)(2m) 0
B DE
MF= −=
∑
DE
F =22197 N
(b) Force on vertical member:
22.2 kN=D
45
consent of McGraw–Hill Education.
SOLUTION Continued
(a) Free Body: Frame
0: 0 =0
xx x
FA= =
∑
A
0: 3924 N 0 3.92 kN
yy y
FA=−= = ↑
∑A
0: (3924N)(2.125 m) 0
AA
MM=−=
∑
8338.5 N m =8.34 kN m
AA
M=⋅⋅M
Free Body: Pulley
0: 3924 N
0: 3924 N
xx
yy
FC
FC
= =
= =
∑
∑
Free Body: Member BEC
0:( cos4.5 )(1 m) (3924 N)(2 m) 0
B DE
MF= −=
∑
11099 N
DE
F=
(b) Force on Vertical member:
11.10 kN
=
D
45
(a) Free Body: Frame
0: 0 =0
xx x
FA= =
∑
A
0: 3924N 0 3.92 kN
yy y
FA=−= = ↑
∑
A
0: (3924N)(2.125 m) 0
AA
MM=−=
∑
A
8339 N m =8.34 kN m
A
M=⋅⋅M
Free Body: Pulley
0: (3924 N)cos45 0
xx
FC=−=
∑
2774.7N
x
C=
consent of McGraw–Hill Education.
SOLUTION Continued
0: 3924 N (3924 N)sin 45 0
yy
FC=−− =
∑
6699N
y
C=
3) Free Body: Member BEC
0: ( cos45 )(1 m) (6699 N)(2m) 0
B DE
MF= −=
∑
18948 N
DE
F=
(b) Force on vertical member:
18.95 kN
=D
45
(a) Free Body: Frame
0: 3924 N 0 3.92 kN
yx x
FA=−= = →
∑
A
0: 3924 N 0 3.92 kN
yy y
FA=−== ↑
∑A
0: (3924N)(2.725m) (3924N)(2.125m) 0
AA
MM=+−=
∑
2354.4 N m =2.35 kN m
AA
M=−⋅ ⋅M
Free Body: Pulley
0: 3924 N
0: 3924 N
xx
yy
FC
FC
= =
= =
∑
∑
Free Body: Member BEC
0:( cos45 )(1 m) (3924 N)(2 m) 0
B DE
MF= −=
∑
11099N
DE
F=
(b) Force on Vertical member:
11.10 kN=D
45
consent of McGraw–Hill Education.
SOLUTION Continued
Free body: Member CDE:
From above, we have
(b)
4.00 kips
x=C
0: (10 kips)(7 ft) (4 kips)(6 ft) (8 ft) 0
Ey
MCΣ= − − =
5.75 kips,
y
C= +
5.75 kips
y=
C
(a)
3
0: 4 kips (10 kips) 0
5
xx
FEΣ= − + + =
2 kips,
x
E= −
2.00 kips
x=E
4
0: 5.75 kips (10 kips) 0,
5
yy
FEΣ= − + =
2.25 kips
y=E
consent of McGraw–Hill Education.
PROBLEM 6.107
For the bevel–gear system shown, determine the required value of α
if the ration of MB to MA is to be 3.
Gear A:
A
r
Gear B:
tan
AB
rr
a
=
0: 0
A AA
M M Jr=−==
∑
(1)
AA
M Jr=
0: 0
B BB
M M Jr= −=
∑
(2)
BB
M Jr=
Ratio:
3 : 3
B BB BA
A AA
M Jr r rr
M Jr r
= = = =
3 tan
BB
rr
a
=
1
tan =18.4
3
aa
=
consent of McGraw–Hill Education.
PROBLEM 6.108
A 400–kg block may be supported by a small frame in each of the four ways shown. The diameter of the
pulley is 250mm. For each case, determine (a) the force components and the couple representing the
reaction at A, (b) the force exerted at D on the vertical member.
SOLUTION
Load:
2
(400kg)(9.81 m / ) 3924NWs= =
(a) Free Body: Frame
0:
x
F=
∑
0 = 0
xx
A=A
0: 7848 N 0 7.85 kN
yy y
FA=−== ↑
∑A
0: (7848N)(2m) 0
AA
MM=−=
∑
15696 N m =15.70 kN m
AA
M=⋅⋅M
Free Body: Frame: Member BEC
0: ( cos45 )(1 m) (7848N)(2m) 0
B DE
MF= −=
∑
DE
F =22197 N
(b) Force on vertical member:
22.2 kN=D
45
consent of McGraw–Hill Education.
SOLUTION Continued
(a) Free Body: Frame
0: 0 =0
xx x
FA= =
∑
A
0: 3924 N 0 3.92 kN
yy y
FA=−= = ↑
∑A
0: (3924N)(2.125 m) 0
AA
MM=−=
∑
8338.5 N m =8.34 kN m
AA
M=⋅⋅M
Free Body: Pulley
0: 3924 N
0: 3924 N
xx
yy
FC
FC
= =
= =
∑
∑
Free Body: Member BEC
0:( cos4.5 )(1 m) (3924 N)(2 m) 0
B DE
MF= −=
∑
11099 N
DE
F=
(b) Force on Vertical member:
11.10 kN
=
D
45
(a) Free Body: Frame
0: 0 =0
xx x
FA= =
∑
A
0: 3924N 0 3.92 kN
yy y
FA=−= = ↑
∑
A
0: (3924N)(2.125 m) 0
AA
MM=−=
∑
A
8339 N m =8.34 kN m
A
M=⋅⋅M
Free Body: Pulley
0: (3924 N)cos45 0
xx
FC=−=
∑
2774.7N
x
C=
consent of McGraw–Hill Education.
SOLUTION Continued
0: 3924 N (3924 N)sin 45 0
yy
FC=−− =
∑
6699N
y
C=
3) Free Body: Member BEC
0: ( cos45 )(1 m) (6699 N)(2m) 0
B DE
MF= −=
∑
18948 N
DE
F=
(b) Force on vertical member:
18.95 kN
=D
45
(a) Free Body: Frame
0: 3924 N 0 3.92 kN
yx x
FA=−= = →
∑
A
0: 3924 N 0 3.92 kN
yy y
FA=−== ↑
∑A
0: (3924N)(2.725m) (3924N)(2.125m) 0
AA
MM=+−=
∑
2354.4 N m =2.35 kN m
AA
M=−⋅ ⋅M
Free Body: Pulley
0: 3924 N
0: 3924 N
xx
yy
FC
FC
= =
= =
∑
∑
Free Body: Member BEC
0:( cos45 )(1 m) (3924 N)(2 m) 0
B DE
MF= −=
∑
11099N
DE
F=
(b) Force on Vertical member:
11.10 kN=D
45
consent of McGraw–Hill Education.