PROBLEM 6.73
A 100–lb force directed vertically downward is applied to the
toggle vise at C. Knowing that link BD is 6 in. long and that
a = 4 in., determine the horizontal force exerted on block E.
PROBLEM 6.74
A 100–lb force directed vertically downward is applied to the
toggle vise at C. Knowing that link BD is 6 in. long and that
a = 8 in., determine the horizontal force exerted on block E.
SOLUTION
Free body: Entire Frame
We have
Horizontal force exerted on block.
()
8 in. sin15 2.0706 in.
2.0706 in.
sin = 20.188
6 in.
BF
ββ
= =
=
()
( )
8 in. cos15 7.7274 in.
6 in. cos20.188 5.6314 in.
13.359 in.
AF
FD
AD AF FD
= =
= =
=+=
293.33 lb
BD
F=
( )
( ) cos (293.33 lb)cos 20.188
BD x BD
FF
β
= =
( ) 275 lb
BD x
=F
Copyright © McGraw–Hill Education. All rights reserved. No reproduction or distribution without the prior written
consent of McGraw–Hill Education.
PROBLEM 6.75
The shear shown is used to cut and trim electronic circuit board
laminates. For the position shown, determine (a) the vertical
component of the force exerted on the shearing blade at D, (b) the
reaction at C.
SOLUTION
We note that BD is a two–force member.
Free body: Member ABC: We have the components:
(400 N)sin30 200 N
x
= °=
P
(400 N)cos30 346.41 N
y
= °=P
25
()65
BD x BD
F=F
60
()65
BD y BD
F=F
0: ( ) (45) ( ) (30) (45 300sin30 )
(30 300cos30 ) 0
C BD x BD y x
y
MF F P
P
Σ= + − + °
− + °=
3
25 60
(45) (30) (200)(195) (346.41)(289.81)
65 65
45 139.39 10
3097.6 N
BD BD
BD
BD
FF
F
F

+=+


= ×
=
60 60
( ) (3097.6 N) 2859.3 N
65 65
BD y BD
FF= = =
( ) 2860 N
BD y =F
Copyright © McGraw–Hill Education. All rights reserved. No reproduction or distribution without the prior written
consent of McGraw–Hill Education.
SOLUTION Continued
(b) Returning to FB diagram of member ABC,
68.5°
0: ( ) 0
x BD x x x
F F PCΣ= −− =
25
() 65
25 (3097.6) 200
65
991.39
x BD x x BD x
x
CF P FP
C
= −= −
= −
= +
991.39 N
x
=C
0: ( ) 0
y BD y y y
F F PCΣ= −− =
60 60
( ) (3097.6) 346.41
65 65
y BD y y BD y
CF P FP
= −= −= −
2512.9 N
y
C= +
2512.9
y
=C
2295 NC=
2700 N=C
Copyright © McGraw–Hill Education. All rights reserved. No reproduction or distribution without the prior written
consent of McGraw–Hill Education.
PROBLEM 6.76
Water pressure in the supply system exerts a downward force of 135 N
on the vertical plug at A. Determine the tension in the fusible link DE
and the force exerted on member BCE at B.
SOLUTION
Free body: Entire linkage:
0: 135 0
y
FC+Σ = − =
135 NC= +
Free body: Member BCE:
0: 0
xx
FBΣ= =
0: (135 N)(6 mm) (10 mm) 0
B DE
MTΣ= − =
81.0 N
DE
T=
0: 135 81 0
yy
FBΣ= +− =
216 N
y
B= +
216 N=B
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PROBLEM 6.77
A 39–ft length of railroad rail of weight 44 lb/ft is lifted by the tongs
shown. Determine the forces exerted at D and F on tong BDF.
SOLUTION
PROBLEM 6.78
The tongs shown are used to apply a total upward force of 45 kN on a pipe cap.
Determine the forces exerted at D and F on tong ADF.
SOLUTION
FBD whole:
FBD ADF:
By symmetry,
so
22.5 kN= =AB
0: (75 mm) (100 mm)(22.5 kN) 0
F
MDΣ= − =
30.0 kN=D
0: 0
xx
F FDΣ= −=
30 kN
x
FD= =
0: 22.5 kN 0
yy
FFΣ= − =
22.5 kN
y
F=
37.5 kN=F
36.9°
consent of McGraw–Hill Education.
PROBLEM 6.79
If the toggle shown is added to the tongs of Problem 6.141 and a single
vertical force is applied at G, determine the forces exerted at D and F on
tong ADF.
SOLUTION
Free body: Toggle:
By symmetry,
AG is a two–force member.
Free body: Tong ADF:
1(45 kN) 22.5 kN
2
y
A= =
22.5 kN
22 mm 55 mm
56.25 kN
x
x
A
A
=
=
0: 22.5 kN 0
yy
FFΣ= − =
22.5 kN
y
F= +
150 kND= +
150.0 kN=D
0: 56.25 kN 150 kN 0
xx
FFΣ= − + =
93.75 kN
x
F=
96.4 kN=F
13.50°
consent of McGraw–Hill Education.
0: (75 mm) (22.5 kN)(100 mm) (56.25 kN)(160 mm) 0
F
MDΣ= − − =
PROBLEM 6.80
A small barrel weighing 60 lb is lifted by a pair of tongs as shown. Knowing
that
5a=
in., determine the forces exerted at B and D on tong ABD.
SOLUTION
We note that BC is a two–force member.
Free body: Tong ABD:
3
15 5
y
xxy
B
BBB= =
0: (3 in.) 3 (5 in.) (60 lb)(9 in.) 0
Dy y
MB BΣ= + − =
30 lb
y
B=
3 : 90 lb
x yx
B BB= =
0: 90 lb 0
xx
FD
Σ= − + =
90 lb
x=D
0: 60 lb 30 lb 0
yy
FDΣ= − − =
30 lb
y=D
94.9 lb=B
18.43°
94.9 lb=D
18.43°
consent of McGraw–Hill Education.
PROBLEM 6.81
A force P of magnitude 2.4 kN is applied to the piston of the engine system shown. For each of the two
positions shown, determine the couple M required to hold the system in equilibrium.
See solution on the next page…
consent of McGraw–Hill Education.
PROBLEM 6.74
A 100–lb force directed vertically downward is applied to the
toggle vise at C. Knowing that link BD is 6 in. long and that
a = 8 in., determine the horizontal force exerted on block E.
SOLUTION
Free body: Entire Frame
We have
Horizontal force exerted on block.
()
8 in. sin15 2.0706 in.
2.0706 in.
sin = 20.188
6 in.
BF
ββ
= =
=
()
( )
8 in. cos15 7.7274 in.
6 in. cos20.188 5.6314 in.
13.359 in.
AF
FD
AD AF FD
= =
= =
=+=
293.33 lb
BD
F=
( )
( ) cos (293.33 lb)cos 20.188
BD x BD
FF
β
= =
( ) 275 lb
BD x
=F
Copyright © McGraw–Hill Education. All rights reserved. No reproduction or distribution without the prior written
consent of McGraw–Hill Education.
PROBLEM 6.75
The shear shown is used to cut and trim electronic circuit board
laminates. For the position shown, determine (a) the vertical
component of the force exerted on the shearing blade at D, (b) the
reaction at C.
SOLUTION
We note that BD is a two–force member.
Free body: Member ABC: We have the components:
(400 N)sin30 200 N
x
= °=
P
(400 N)cos30 346.41 N
y
= °=P
25
()65
BD x BD
F=F
60
()65
BD y BD
F=F
0: ( ) (45) ( ) (30) (45 300sin30 )
(30 300cos30 ) 0
C BD x BD y x
y
MF F P
P
Σ= + − + °
− + °=
3
25 60
(45) (30) (200)(195) (346.41)(289.81)
65 65
45 139.39 10
3097.6 N
BD BD
BD
BD
FF
F
F

+=+


= ×
=
60 60
( ) (3097.6 N) 2859.3 N
65 65
BD y BD
FF= = =
( ) 2860 N
BD y =F
Copyright © McGraw–Hill Education. All rights reserved. No reproduction or distribution without the prior written
consent of McGraw–Hill Education.
SOLUTION Continued
(b) Returning to FB diagram of member ABC,
68.5°
0: ( ) 0
x BD x x x
F F PCΣ= −− =
25
() 65
25 (3097.6) 200
65
991.39
x BD x x BD x
x
CF P FP
C
= −= −
= −
= +
991.39 N
x
=C
0: ( ) 0
y BD y y y
F F PCΣ= −− =
60 60
( ) (3097.6) 346.41
65 65
y BD y y BD y
CF P FP
= −= −= −
2512.9 N
y
C= +
2512.9
y
=C
2295 NC=
2700 N=C
Copyright © McGraw–Hill Education. All rights reserved. No reproduction or distribution without the prior written
consent of McGraw–Hill Education.
PROBLEM 6.76
Water pressure in the supply system exerts a downward force of 135 N
on the vertical plug at A. Determine the tension in the fusible link DE
and the force exerted on member BCE at B.
SOLUTION
Free body: Entire linkage:
0: 135 0
y
FC+Σ = − =
135 NC= +
Free body: Member BCE:
0: 0
xx
FBΣ= =
0: (135 N)(6 mm) (10 mm) 0
B DE
MTΣ= − =
81.0 N
DE
T=
0: 135 81 0
yy
FBΣ= +− =
216 N
y
B= +
216 N=B
consent of McGraw–Hill Education.
PROBLEM 6.77
A 39–ft length of railroad rail of weight 44 lb/ft is lifted by the tongs
shown. Determine the forces exerted at D and F on tong BDF.
SOLUTION
PROBLEM 6.78
The tongs shown are used to apply a total upward force of 45 kN on a pipe cap.
Determine the forces exerted at D and F on tong ADF.
SOLUTION
FBD whole:
FBD ADF:
By symmetry,
so
22.5 kN= =AB
0: (75 mm) (100 mm)(22.5 kN) 0
F
MDΣ= − =
30.0 kN=D
0: 0
xx
F FDΣ= −=
30 kN
x
FD= =
0: 22.5 kN 0
yy
FFΣ= − =
22.5 kN
y
F=
37.5 kN=F
36.9°
consent of McGraw–Hill Education.
PROBLEM 6.79
If the toggle shown is added to the tongs of Problem 6.141 and a single
vertical force is applied at G, determine the forces exerted at D and F on
tong ADF.
SOLUTION
Free body: Toggle:
By symmetry,
AG is a two–force member.
Free body: Tong ADF:
1(45 kN) 22.5 kN
2
y
A= =
22.5 kN
22 mm 55 mm
56.25 kN
x
x
A
A
=
=
0: 22.5 kN 0
yy
FFΣ= − =
22.5 kN
y
F= +
150 kND= +
150.0 kN=D
0: 56.25 kN 150 kN 0
xx
FFΣ= − + =
93.75 kN
x
F=
96.4 kN=F
13.50°
consent of McGraw–Hill Education.
0: (75 mm) (22.5 kN)(100 mm) (56.25 kN)(160 mm) 0
F
MDΣ= − − =
PROBLEM 6.80
A small barrel weighing 60 lb is lifted by a pair of tongs as shown. Knowing
that
5a=
in., determine the forces exerted at B and D on tong ABD.
SOLUTION
We note that BC is a two–force member.
Free body: Tong ABD:
3
15 5
y
xxy
B
BBB= =
0: (3 in.) 3 (5 in.) (60 lb)(9 in.) 0
Dy y
MB BΣ= + − =
30 lb
y
B=
3 : 90 lb
x yx
B BB= =
0: 90 lb 0
xx
FD
Σ= − + =
90 lb
x=D
0: 60 lb 30 lb 0
yy
FDΣ= − − =
30 lb
y=D
94.9 lb=B
18.43°
94.9 lb=D
18.43°
consent of McGraw–Hill Education.
PROBLEM 6.81
A force P of magnitude 2.4 kN is applied to the piston of the engine system shown. For each of the two
positions shown, determine the couple M required to hold the system in equilibrium.
See solution on the next page…
consent of McGraw–Hill Education.