PROBLEM 3.49
Two parallel 60–N forces are applied to a lever as shown.
Determine the moment of the couple formed by the two forces
(a) by resolving each force into horizontal and vertical components
and adding the moments of the two resulting couples, (b) by using
the perpendicular distance between the two forces, (c) by summing
the moments of the two forces about Point A.
SOLUTION
(a) We have
12
:
B xy
dC dCΣ −+ =MM
where
1
2
(0.360 m)sin55°
0.29489 m
(0.360 m)sin55
0.20649 m
d
d
=
=
= °
=
(60 N)cos20
56.382 N
(60 N)sin 20
20.521 N
x
y
C
C
= °
=
= °
=
(0.29489 m)(56.382 N) (0.20649 m)(20.521 N)
(12.3893 N m)
=−+
=−⋅
M kk
k
or
12.39 N m
= ⋅M
(b) We have
()
60 N[(0.360 m)sin(55 20 )]( )
Fd= −
= °− ° −
Mk
k
(12.3893 N m)=−⋅k
or
12.39 N m= ⋅M
(c) We have
//
:( )
A A BA B CA C
Σ Σ×= ×+ × =
M rFr Fr F M
or
12.39 N m= ⋅M
consent of McGraw–Hill Education.
PROBLEM 3.50
A plate in the shape of a parallelogram is acted upon by two
couples. Determine (a) the moment of the couple formed by the
two 21–lb forces, (b) the perpendicular distance between the 12–lb
forces if the resultant of the two couples is zero, (c) the value of
α
if the resultant couple is
72 lb in.⋅
clockwise and d is 42 in.
SOLUTION
PROBLEM 3.51
Two 80–N forces are applied as shown to the corners B and D of
a rectangular plate. (a) Determine the moment of the couple
formed by the two forces by resolving each force into horizontal
and vertical components and adding the moments of the two
resulting couples. (b) Use the result obtained to determine the
perpendicular distance between lines BE and DF.
SOLUTION
(a) Resolving forces into components:
(80 N)sin50 61.284 NP= =
(80 N)cos50 51.423 NQ= =
(51.423 N)(0.5 m) (61.284 N)(0.3 m)
7.3263 N m
M= −
= ⋅
7.33 N m= ⋅
M
(b) Distance between lines BE and DF
Fd=
M
or
7.3263 N m=(80 N)
0.091579 m
d
d
⋅
=
91.6 mmd=
PROBLEM 3.52
A piece of plywood in which several holes are being drilled
successively has been secured to a workbench by means of two
nails. Knowing that the drill exerts a 12–N∙m couple on the
piece of plywood, determine the magnitude of the resulting
forces applied to the nails if they are located (a) at A and B,
(b) at B and C, (c) at A and C.
SOLUTION
(a)
12 N m (0.45 m)
M Fd
F
=
⋅=
26.7 NF=
(b)
12 N m (0.24 m)
M Fd
F
=
⋅=
50.0 NF=
(c)
22
(0.45 m) (0.24 m)
0.510 m
M Fd d
= = +
=
12 N m (0.510 m)F⋅=
23.5 NF=
consent of McGraw–Hill Education.
PROBLEM 3.53
Four 1
1
2
–in.–diameter pegs are attached to a board as shown.
Two strings are passed around the pegs and pulled with the
forces indicated. (a) Determine the resultant couple acting on
the board. (b) If only one string is used, around which pegs
should it pass and in what directions should it be pulled to
create the same couple with the minimum tension in the string?
(c) What is the value of that minimum tension?
SOLUTION
(a)
(60 lb)(10.5 in.) (40 lb)(13.5 in.)
630 lb in. 540 lb in.
M= +
= ⋅+ ⋅
1170 lb in.M= ⋅
have
9
tan 36.9 90 53.1
12
θθ θ
= = ° °− = °
Direction of forces:
With pegs A and D:
53.1
θ
= °
With pegs B and C:
53.1
θ
= °
(c) The distance between the centers of the two pegs is
22
12 9 15 in.+=
Therefore, the perpendicular distance d between the forces is
3
15 in. 2 in.
4
16.5 in.
d
= + 

=
We must have
1170 lb in. (16.5 in.)M Fd F= ⋅=
70.9 lbF=
consent of McGraw–Hill Education.
PROBLEM 3.54
Four pegs of the same diameter are attached to a board as
shown. Two strings are passed around the pegs and pulled with
the forces indicated. Determine the diameter of the pegs
knowing that the resultant couple applied to the board is
1132.5 lb·in. counterclockwise.
SOLUTION
1132.5 lb in. [(9 ) in.](60 lb) [(12 ) in.](40 lb)
AD AD BC BC
MdF dF
dd
= +
⋅= + + +
1.125 in.d=
consent of McGraw–Hill Education.
PROBLEM 3.55
In a manufacturing operation, three holes are drilled simultaneously
in a workpiece. If the holes are perpendicular to the surfaces of the
workpiece, replace the couples applied to the drills with a single
equivalent couple, specifying its magnitude and the direction of
its axis.
SOLUTION
123
222
(1.5 N m)( cos20 sin20 ) (1.5 N m)
(1.75 N m)( cos25 sin25 )
(4.4956 N m) (0.22655 N m)
(0) ( 4.4956) (0.22655)
4.5013 N m
M
=++
= ⋅ − °+ ° − ⋅
+ ⋅ − °+ °
=− ⋅+ ⋅
= +− +
= ⋅
MM M M
jk j
jk
jk
4.50 N mM= ⋅
axis
(0.99873 0.050330 )
cos 0
cos 0.99873
cos 0.050330
x
y
z
M
θ
θ
θ
==−+
=
= −
=
Mjk
λ
90.0 , 177.1 , 87.1
xy z
θθ θ
=°=°=°
PROBLEM 3.56
The two shafts of a speed–reducer unit are subjected to
couples of magnitude M1 = 15 lb∙ft and M2 = 3 lb∙ft,
respectively. Replace the two couples with a single
equivalent couple, specifying its magnitude and the
direction of its axis.
SOLUTION
1(15 lb ft)M= ⋅ k
2
(3 lb ft)M= ⋅ i
22
12
22
(15) (3)
15.30 lb ft
M MM= +
= +
= ⋅
15
tan 5
3
x
θ
= =
78.7
x
θ
= °
90
y
θ
= °
90 78.7
11.30
z
θ
= °− °
= °
15.30 lb ft; 78.7 , 90.0 , 11.30
xyz
M
θθθ
= ⋅ =°=°= °
consent of McGraw–Hill Education.
PROBLEM 3.57
Replace the two couples shown with a single equivalent
couple, specifying its magnitude and the direction of its axis.
SOLUTION
Replace the couple in the ABCD plane with two couples P and Q shown:
160 mm
(50 N) (50 N) 40 N
200 mm
CD
PCG

= = =


120 mm
(50 N) (50 N) 30 N
200 mm
CF
QCG

= = =


Couple vector M1 perpendicular to plane ABCD:
1
(40 N)(0.24 m) (30 N)(0.16 m) 4.80 N mM
=−=⋅
Couple vector M2 in the xy plane:
2(12.5 N)(0.192 m) 2.40 N mM=− =−⋅
144 mm
tan 36.870
192 mm
θθ
= = °
1
(4.80 cos36.870 ) (4.80 sin36.870 )
3.84 2.88
= °+ °
= +
M jk
jk
2
2.40= −Mj
12
1.44 2.88=+= +MM M j k
3.22 N m; 90.0 , 53.1 , 36.9
xyz
θθθ
= ⋅ =°=°=°M
consent of McGraw–Hill Education.
PROBLEM 3.58
Solve Prob. 3.57, assuming that two 10–N vertical forces
have been added, one acting upward at C and the other
downward at B.
PROBLEM 3.57 Replace the two couples shown with a
single equivalent couple, specifying its magnitude and the
direction of its axis.
SOLUTION
160 mm
(50 N) (50 N) 40 N
200 mm
CD
PCG

= = =


120 mm
(50 N) (50 N) 30 N
200 mm
CF
QCG

= = =


Couple vector M1 perpendicular to plane ABCD.
1
(40 N)(0.24 m) (30 N)(0.16 m) 4.80 N mM
=−=⋅
144 mm
tan 36.870
192 mm
θθ
= = °
1
(4.80cos36.870 ) (4.80sin36.870 )
3.84 2.88
= °+ °
= +
M jk
jk
2
(12.5 N)(0.192 m) 2.40 N m
2.40
M=− =−⋅
= − j
3 / 3/
; (0.16 m) (0.144 m) (0.192 m)
(0.16 m) (0.144 m) (0.192 m) ( 10 N)
1.92 1.6
BC BC
M=×= + −
= + − ×−
=−−
Mr r i j k
i j kj
ik
consent of McGraw–Hill Education.
PROBLEM 3.50
A plate in the shape of a parallelogram is acted upon by two
couples. Determine (a) the moment of the couple formed by the
two 21–lb forces, (b) the perpendicular distance between the 12–lb
forces if the resultant of the two couples is zero, (c) the value of
α
if the resultant couple is
72 lb in.⋅
clockwise and d is 42 in.
SOLUTION
PROBLEM 3.51
Two 80–N forces are applied as shown to the corners B and D of
a rectangular plate. (a) Determine the moment of the couple
formed by the two forces by resolving each force into horizontal
and vertical components and adding the moments of the two
resulting couples. (b) Use the result obtained to determine the
perpendicular distance between lines BE and DF.
SOLUTION
(a) Resolving forces into components:
(80 N)sin50 61.284 NP= =
(80 N)cos50 51.423 NQ= =
(51.423 N)(0.5 m) (61.284 N)(0.3 m)
7.3263 N m
M= −
= ⋅
7.33 N m= ⋅
M
(b) Distance between lines BE and DF
Fd=
M
or
7.3263 N m=(80 N)
0.091579 m
d
d
⋅
=
91.6 mmd=
PROBLEM 3.52
A piece of plywood in which several holes are being drilled
successively has been secured to a workbench by means of two
nails. Knowing that the drill exerts a 12–N∙m couple on the
piece of plywood, determine the magnitude of the resulting
forces applied to the nails if they are located (a) at A and B,
(b) at B and C, (c) at A and C.
SOLUTION
(a)
12 N m (0.45 m)
M Fd
F
=
⋅=
26.7 NF=
(b)
12 N m (0.24 m)
M Fd
F
=
⋅=
50.0 NF=
(c)
22
(0.45 m) (0.24 m)
0.510 m
M Fd d
= = +
=
12 N m (0.510 m)F⋅=
23.5 NF=
consent of McGraw–Hill Education.
PROBLEM 3.53
Four 1
1
2
–in.–diameter pegs are attached to a board as shown.
Two strings are passed around the pegs and pulled with the
forces indicated. (a) Determine the resultant couple acting on
the board. (b) If only one string is used, around which pegs
should it pass and in what directions should it be pulled to
create the same couple with the minimum tension in the string?
(c) What is the value of that minimum tension?
SOLUTION
(a)
(60 lb)(10.5 in.) (40 lb)(13.5 in.)
630 lb in. 540 lb in.
M= +
= ⋅+ ⋅
1170 lb in.M= ⋅
have
9
tan 36.9 90 53.1
12
θθ θ
= = ° °− = °
Direction of forces:
With pegs A and D:
53.1
θ
= °
With pegs B and C:
53.1
θ
= °
(c) The distance between the centers of the two pegs is
22
12 9 15 in.+=
Therefore, the perpendicular distance d between the forces is
3
15 in. 2 in.
4
16.5 in.
d
= + 

=
We must have
1170 lb in. (16.5 in.)M Fd F= ⋅=
70.9 lbF=
consent of McGraw–Hill Education.
PROBLEM 3.54
Four pegs of the same diameter are attached to a board as
shown. Two strings are passed around the pegs and pulled with
the forces indicated. Determine the diameter of the pegs
knowing that the resultant couple applied to the board is
1132.5 lb·in. counterclockwise.
SOLUTION
1132.5 lb in. [(9 ) in.](60 lb) [(12 ) in.](40 lb)
AD AD BC BC
MdF dF
dd
= +
⋅= + + +
1.125 in.d=
consent of McGraw–Hill Education.
PROBLEM 3.55
In a manufacturing operation, three holes are drilled simultaneously
in a workpiece. If the holes are perpendicular to the surfaces of the
workpiece, replace the couples applied to the drills with a single
equivalent couple, specifying its magnitude and the direction of
its axis.
SOLUTION
123
222
(1.5 N m)( cos20 sin20 ) (1.5 N m)
(1.75 N m)( cos25 sin25 )
(4.4956 N m) (0.22655 N m)
(0) ( 4.4956) (0.22655)
4.5013 N m
M
=++
= ⋅ − °+ ° − ⋅
+ ⋅ − °+ °
=− ⋅+ ⋅
= +− +
= ⋅
MM M M
jk j
jk
jk
4.50 N mM= ⋅
axis
(0.99873 0.050330 )
cos 0
cos 0.99873
cos 0.050330
x
y
z
M
θ
θ
θ
==−+
=
= −
=
Mjk
λ
90.0 , 177.1 , 87.1
xy z
θθ θ
=°=°=°
PROBLEM 3.56
The two shafts of a speed–reducer unit are subjected to
couples of magnitude M1 = 15 lb∙ft and M2 = 3 lb∙ft,
respectively. Replace the two couples with a single
equivalent couple, specifying its magnitude and the
direction of its axis.
SOLUTION
1(15 lb ft)M= ⋅ k
2
(3 lb ft)M= ⋅ i
22
12
22
(15) (3)
15.30 lb ft
M MM= +
= +
= ⋅
15
tan 5
3
x
θ
= =
78.7
x
θ
= °
90
y
θ
= °
90 78.7
11.30
z
θ
= °− °
= °
15.30 lb ft; 78.7 , 90.0 , 11.30
xyz
M
θθθ
= ⋅ =°=°= °
consent of McGraw–Hill Education.
PROBLEM 3.57
Replace the two couples shown with a single equivalent
couple, specifying its magnitude and the direction of its axis.
SOLUTION
Replace the couple in the ABCD plane with two couples P and Q shown:
160 mm
(50 N) (50 N) 40 N
200 mm
CD
PCG

= = =


120 mm
(50 N) (50 N) 30 N
200 mm
CF
QCG

= = =


Couple vector M1 perpendicular to plane ABCD:
1
(40 N)(0.24 m) (30 N)(0.16 m) 4.80 N mM
=−=⋅
Couple vector M2 in the xy plane:
2(12.5 N)(0.192 m) 2.40 N mM=− =−⋅
144 mm
tan 36.870
192 mm
θθ
= = °
1
(4.80 cos36.870 ) (4.80 sin36.870 )
3.84 2.88
= °+ °
= +
M jk
jk
2
2.40= −Mj
12
1.44 2.88=+= +MM M j k
3.22 N m; 90.0 , 53.1 , 36.9
xyz
θθθ
= ⋅ =°=°=°M
consent of McGraw–Hill Education.
PROBLEM 3.58
Solve Prob. 3.57, assuming that two 10–N vertical forces
have been added, one acting upward at C and the other
downward at B.
PROBLEM 3.57 Replace the two couples shown with a
single equivalent couple, specifying its magnitude and the
direction of its axis.
SOLUTION
160 mm
(50 N) (50 N) 40 N
200 mm
CD
PCG

= = =


120 mm
(50 N) (50 N) 30 N
200 mm
CF
QCG

= = =


Couple vector M1 perpendicular to plane ABCD.
1
(40 N)(0.24 m) (30 N)(0.16 m) 4.80 N mM
=−=⋅
144 mm
tan 36.870
192 mm
θθ
= = °
1
(4.80cos36.870 ) (4.80sin36.870 )
3.84 2.88
= °+ °
= +
M jk
jk
2
(12.5 N)(0.192 m) 2.40 N m
2.40
M=− =−⋅
= − j
3 / 3/
; (0.16 m) (0.144 m) (0.192 m)
(0.16 m) (0.144 m) (0.192 m) ( 10 N)
1.92 1.6
BC BC
M=×= + −
= + − ×−
=−−
Mr r i j k
i j kj
ik
consent of McGraw–Hill Education.