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PROBLEM 6.99
Determine the force in members EH and GI of the
truss shown. (Hint: Use section aa.)
SOLUTION
Reactions:
0: 0
xx
FAΣ= =
0: 12 kips(45 ft) 12 kips(30 ft) 12 kips(15 ft) (90 ft) 0
Py
MAΣ= + + − =
12 kips
y=A
0: 12 kips 12 kips 12 kips 12 kips 0
y
FPΣ= −−−+=
24 kips=P
0: (12 kips)(30 ft) (16 ft) 0
G EH
MFΣ=− − =
22.5 kips
EH
F= −
22.5 kips
EH
FC=
0: 22.5 kips 0
x GI
FFΣ= − =
22.5 kips
GI
FT=
consent of McGraw-Hill Education.
PROBLEM 6.100
Determine the force in members HJ and IL of the
truss shown. (Hint: Use section bb.)
SOLUTION
Reactions:
0: 0
xx
FAΣ= =
0: 12 kips(45 ft) 12 kips(30 ft) 12 kips(15 ft) (90 ft) 0
Py
MAΣ= + + − =
12 kips
y=A
0: 12 kips 12 kips 12 kips 12 kips 0
y
FPΣ= −−−+=
24 kips=P
0: (16 ft) (12 kips)(15 ft) (24 kips)(30 ft) 0
L HJ
MFΣ= − + =
33.75 kips
HJ
F= −
33.8 kips
HJ
FC=
0: 33.75 kips 0
x IL
FFΣ= − =
33.75 kips
IL
F= +
33.8 kips
IL
FT=
consent of McGraw-Hill Education.
PROBLEM 6.101
The low-bed trailer shown is designed so that the
rear end of the bed can be lowered to ground level
in order to facilitate the loading of equipment or
wrecked vehicles. A 1400-kg vehicle has been
hauled to the position shown by a winch; the
trailer is then returned to a traveling position
where α = 0 and both AB and BE are horizontal.
Considering only the weight of the disabled
automobile, determine the force that must be
exerted by the hydraulic cylinder to maintain a
position with α = 0.
SOLUTION
Free Body: Trailer & Car
W = (1400 kg)(9.81 m/s2) = 13.734 kN
α =0
0: (6 m) (13.734 kN)(3.5 m) 0
A
ME
=−=
∑
E = +8.012 kN E =8.012 kN
consent of McGraw-Hill Education.
PROBLEM 6.102
The axis of the three-hinge arch ABC is a
parabola with the vertex at B. Knowing that
P = 112 kN and Q = 140 kN, determine (a) the
components of the reaction at A, (b) the
components of the force exerted at B on
segment AB.
SOLUTION
Free body: Segment AB:
(1)
0.75 (Eq. 1): (2)
Free body: Segment BC:
(3)
Add Eqs. (2) and (3):
(4)
From Eq. (1):
(5)
given that
(a) Reaction at A.
Considering again AB as a free body,
0: (3.2 m) (8 m) (5 m) 0
Ax y
MB B PΣ= − − =
(2.4 m) (6 m) (3.75 m) 0
xy
B BP−− =
0: (1.8 m) (6 m) (3 m) 0
Cx y
MB B QΣ= + − =
4.2 3.75 3 0
x
B PQ− −=
(3.75 3 )/4.2
x
B PQ= +
3.2
(3.75 3 ) 8 5 0
4.2
y
PQ B P+ − −=
( 9 9.6 )/33.6
y
B PQ=−+
112 kN and 140 kN.PQ= =
0: 0; 200 kN
x xx xx
F AB ABΣ= − = = =
200 kN
x
=A
0: 0
y yy
F A PBΣ = −− =
112 kN 10 kN 0
y
A− −=
122 kN
y
A= +
122.0 kN
y=A
consent of McGraw-Hill Education.
SOLUTION Continued
(b) Force exerted at B on AB.
From Eq. (4):
From Eq. (5):
(3.75 112 3 140)/4.2 200 kN
x
B= × +× =
200 kN
x
=B
( 9 112 9.6 140)/33.6 10 kN
y
B=−× + × =+
10.00 kN
y
=B
consent of McGraw-Hill Education.
PROBLEM 6.103
A 48-mm-diameter pipe is gripped by Stillson wrench shown.
Portions AB and DE of the wrench are rigidly attached to each other
and portion CF is connected by a pin at D. Assuming that no
slipping occurs between the pipe and the wrench, determine the
components of the forces exerted on the pipe at A and C.
SOLUTION
Free Body: Portion ABDE of Wrench
AD.
We have
= 4.3
20 mm 86 mm
y
xyx
A
AAA=
, 4.3
xxyy x
D AD A D= = =
(1)
Free Body: Portion CDF
( )
0: 38mm (20 mm) (400N)(398 mm) 0
xy
DD∑= − + =M
Substituting for Dy from (1):
( ) ( )
38 mm (4.3 ) 20 mm (400N)(398 mm) 0
(400)(398) 3316.7 N
48
4.3 14261.7 N
xx
x
yx
DD
D
DD
−+ =
= =
= =
0: 3316.7 N 400 N 0
xx
FC= + −=
∑
3717N 3.72 kN
xx
C= = ←C
0: 14261.7N 0
yy
FC= −=
∑
14260 N 14.26 kN
yy
C= = ↓C
From (1) and the free body diagram of ABDE:
3317N 3.32 kN
14260N 14.26 kN
xx x
yy y
AD
AD
= = = →
= = = ↑
A
A
consent of McGraw-Hill Education.
SOLUTION Continued
above:
3.32 kN , 14.26 kN
xy
= ←= ↓AA
3.72 kN , 14.26 kN
xy
= →= ↑CC
reactions R and M as shown.
consent of McGraw-Hill Education.
PROBLEM 6.104
The compound-lever pruning shears shown can be
adjusted by placing pin A at various ratchet positions on
blade ACE. Knowing that 300-lb vertical forces are
required to complete the pruning of a small branch,
determine the magnitude P of the forces that must be
applied to the handles when the shears are adjusted as
shown.
SOLUTION
We note that AB is a two-force member.
()
()
0.65 in. 0.55 in.
11
() ()
13
AB y
AB x
AB y AB x
F
F
FF
=
=
(1)
Free body: Blade ACE:
0: (300 lb)(1.6 in.) ( ) (0.5 in.) ( ) (1.4 in.) 0
C AB x AB y
M FFΣ= − − =
Use Eq. (1):
11
( ) (0.5 in.) ( ) (1.4 in.) 480 lb in.
13
AB x AB x
FF+=⋅
1.6846( ) 480
AB x
F=
( ) 284.9 lb
AB x
F=
11
( ) (284.9 lb)
13
AB y
F=
( ) 241.1lb
AB y
F=
Free body: Lower handle:
0: (241.1lb)(0.75 in.) (284.9 lb)(0.25 in.) (3.5 in.) 0
D
MPΣ= − − =
31.3 lbP=
consent of McGraw-Hill Education.
PROBLEM 6.105
A log weighing 800 lb is lifted by a pair of tongs as shown.
Determine the forces exerted at E and F on tong DEF.
SOLUTION
consent of McGraw-Hill Education.
PROBLEM 6.99
Determine the force in members EH and GI of the
truss shown. (Hint: Use section aa.)
SOLUTION
Reactions:
0: 0
xx
FAΣ= =
0: 12 kips(45 ft) 12 kips(30 ft) 12 kips(15 ft) (90 ft) 0
Py
MAΣ= + + − =
12 kips
y=A
0: 12 kips 12 kips 12 kips 12 kips 0
y
FPΣ= −−−+=
24 kips=P
0: (12 kips)(30 ft) (16 ft) 0
G EH
MFΣ=− − =
22.5 kips
EH
F= −
22.5 kips
EH
FC=
0: 22.5 kips 0
x GI
FFΣ= − =
22.5 kips
GI
FT=
consent of McGraw-Hill Education.
PROBLEM 6.100
Determine the force in members HJ and IL of the
truss shown. (Hint: Use section bb.)
SOLUTION
Reactions:
0: 0
xx
FAΣ= =
0: 12 kips(45 ft) 12 kips(30 ft) 12 kips(15 ft) (90 ft) 0
Py
MAΣ= + + − =
12 kips
y=A
0: 12 kips 12 kips 12 kips 12 kips 0
y
FPΣ= −−−+=
24 kips=P
0: (16 ft) (12 kips)(15 ft) (24 kips)(30 ft) 0
L HJ
MFΣ= − + =
33.75 kips
HJ
F= −
33.8 kips
HJ
FC=
0: 33.75 kips 0
x IL
FFΣ= − =
33.75 kips
IL
F= +
33.8 kips
IL
FT=
consent of McGraw-Hill Education.
PROBLEM 6.101
The low-bed trailer shown is designed so that the
rear end of the bed can be lowered to ground level
in order to facilitate the loading of equipment or
wrecked vehicles. A 1400-kg vehicle has been
hauled to the position shown by a winch; the
trailer is then returned to a traveling position
where α = 0 and both AB and BE are horizontal.
Considering only the weight of the disabled
automobile, determine the force that must be
exerted by the hydraulic cylinder to maintain a
position with α = 0.
SOLUTION
Free Body: Trailer & Car
W = (1400 kg)(9.81 m/s2) = 13.734 kN
α =0
0: (6 m) (13.734 kN)(3.5 m) 0
A
ME
=−=
∑
E = +8.012 kN E =8.012 kN
consent of McGraw-Hill Education.
PROBLEM 6.102
The axis of the three-hinge arch ABC is a
parabola with the vertex at B. Knowing that
P = 112 kN and Q = 140 kN, determine (a) the
components of the reaction at A, (b) the
components of the force exerted at B on
segment AB.
SOLUTION
Free body: Segment AB:
(1)
0.75 (Eq. 1): (2)
Free body: Segment BC:
(3)
Add Eqs. (2) and (3):
(4)
From Eq. (1):
(5)
given that
(a) Reaction at A.
Considering again AB as a free body,
0: (3.2 m) (8 m) (5 m) 0
Ax y
MB B PΣ= − − =
(2.4 m) (6 m) (3.75 m) 0
xy
B BP−− =
0: (1.8 m) (6 m) (3 m) 0
Cx y
MB B QΣ= + − =
4.2 3.75 3 0
x
B PQ− −=
(3.75 3 )/4.2
x
B PQ= +
3.2
(3.75 3 ) 8 5 0
4.2
y
PQ B P+ − −=
( 9 9.6 )/33.6
y
B PQ=−+
112 kN and 140 kN.PQ= =
0: 0; 200 kN
x xx xx
F AB ABΣ= − = = =
200 kN
x
=A
0: 0
y yy
F A PBΣ = −− =
112 kN 10 kN 0
y
A− −=
122 kN
y
A= +
122.0 kN
y=A
consent of McGraw-Hill Education.
SOLUTION Continued
(b) Force exerted at B on AB.
From Eq. (4):
From Eq. (5):
(3.75 112 3 140)/4.2 200 kN
x
B= × +× =
200 kN
x
=B
( 9 112 9.6 140)/33.6 10 kN
y
B=−× + × =+
10.00 kN
y
=B
consent of McGraw-Hill Education.
PROBLEM 6.103
A 48-mm-diameter pipe is gripped by Stillson wrench shown.
Portions AB and DE of the wrench are rigidly attached to each other
and portion CF is connected by a pin at D. Assuming that no
slipping occurs between the pipe and the wrench, determine the
components of the forces exerted on the pipe at A and C.
SOLUTION
Free Body: Portion ABDE of Wrench
AD.
We have
= 4.3
20 mm 86 mm
y
xyx
A
AAA=
, 4.3
xxyy x
D AD A D= = =
(1)
Free Body: Portion CDF
( )
0: 38mm (20 mm) (400N)(398 mm) 0
xy
DD∑= − + =M
Substituting for Dy from (1):
( ) ( )
38 mm (4.3 ) 20 mm (400N)(398 mm) 0
(400)(398) 3316.7 N
48
4.3 14261.7 N
xx
x
yx
DD
D
DD
−+ =
= =
= =
0: 3316.7 N 400 N 0
xx
FC= + −=
∑
3717N 3.72 kN
xx
C= = ←C
0: 14261.7N 0
yy
FC= −=
∑
14260 N 14.26 kN
yy
C= = ↓C
From (1) and the free body diagram of ABDE:
3317N 3.32 kN
14260N 14.26 kN
xx x
yy y
AD
AD
= = = →
= = = ↑
A
A
consent of McGraw-Hill Education.
SOLUTION Continued
above:
3.32 kN , 14.26 kN
xy
= ←= ↓AA
3.72 kN , 14.26 kN
xy
= →= ↑CC
reactions R and M as shown.
consent of McGraw-Hill Education.
PROBLEM 6.104
The compound-lever pruning shears shown can be
adjusted by placing pin A at various ratchet positions on
blade ACE. Knowing that 300-lb vertical forces are
required to complete the pruning of a small branch,
determine the magnitude P of the forces that must be
applied to the handles when the shears are adjusted as
shown.
SOLUTION
We note that AB is a two-force member.
()
()
0.65 in. 0.55 in.
11
() ()
13
AB y
AB x
AB y AB x
F
F
FF
=
=
(1)
Free body: Blade ACE:
0: (300 lb)(1.6 in.) ( ) (0.5 in.) ( ) (1.4 in.) 0
C AB x AB y
M FFΣ= − − =
Use Eq. (1):
11
( ) (0.5 in.) ( ) (1.4 in.) 480 lb in.
13
AB x AB x
FF+=⋅
1.6846( ) 480
AB x
F=
( ) 284.9 lb
AB x
F=
11
( ) (284.9 lb)
13
AB y
F=
( ) 241.1lb
AB y
F=
Free body: Lower handle:
0: (241.1lb)(0.75 in.) (284.9 lb)(0.25 in.) (3.5 in.) 0
D
MPΣ= − − =
31.3 lbP=
consent of McGraw-Hill Education.
PROBLEM 6.105
A log weighing 800 lb is lifted by a pair of tongs as shown.
Determine the forces exerted at E and F on tong DEF.
SOLUTION
consent of McGraw-Hill Education.
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