PROBLEM 3.78
Five separate force–couple systems act at the corners of a piece of sheet metal, which has been bent into
the shape shown. Determine which of these systems is equivalent to a force F = (10 lb)i and a couple of
moment M = (15 lb ⋅ ft)j + (15 lb ⋅ ft)k located at the origin.
SOLUTION
First note that the force–couple system at F cannot be equivalent because of the direction of the force [The
force of the other four systems is (10 lb)i]. Next, move each of the systems to the origin O; the forces
remain unchanged.
: (5 lb ft) (15 lb ft) (2 ft) (10 lb)
AO
A=Σ = ⋅+ ⋅ + ×MM j k k i
(25 lb ft) (15 lb ft)= ⋅+ ⋅jk
: (5 lb ft) (25 lb ft)
[(4.5 ft) (1 ft) (2 ft) ] 10 lb)
(15 lb ft) (15 lb ft)
DO
D=Σ =− ⋅+ ⋅
+ ++ ×
= ⋅+ ⋅
MM j k
ijk i
ik
: (15 lb ft) (15 lb ft)
: (15 lb ft) (5 lb ft)
[(4.5 ft) (1 ft) ] (10 lb)
GO
II
G
I
=Σ = ⋅+ ⋅
=Σ= ⋅− ⋅
+ +×
MM i j
MM j k
ij j
(15 lb ft) (15 lb ft)= ⋅− ⋅jk
The equivalent force–couple system is the system at corner D.
PROBLEM 3.79
Four forces act on a 700 × 375-mm plate as shown. (a) Find
the resultant of these forces. (b) Locate the two points where
the line of action of the resultant intersects the edge of the
plate.
SOLUTION
(a)
( 400 N 160 N 760 N)
(600 N 300 N 300 N)
(1000 N) (1200 N)
= Σ
=−+−
+ ++
=−+
RF
i
j
ij
22
(1000 N) (1200 N)
1562.09 N
1200 N
tan 1000 N
1.20000
50.194
R
θ
θ
= +
=
= −
= −
=−°
1562 N=R
50.2°
(b)
(0.5 m) (300 N 300 N)
(300 N m)
R
C=Σ×
= ×+
= ⋅
M rF
ij
k
(300 N m) (1200 N)
0.25000 m
250 mm
(300 N m) ( 1000 N)
0.30000 m
300 mm
x
x
x
y
y
y
⋅=×
=
=
⋅ = ×−
=
=
ki j
ji
consent of McGraw–Hill Education.
PROBLEM 3.80
A 32–lb motor is mounted on the floor. Find the resultant of
the weight and the forces exerted on the belt, and determine
where the line of action of the resultant intersects the floor.
SOLUTION
We have
: (60 lb) (32 lb) (140 lb)(cos30 sin30 )Σ − + °+ ° =F i j i jR
(181.244 lb) (38.0 lb)= +R ij
or
185.2 lb=R
11.84°
We have
:
O Oy
M M xRΣ Σ=
[(140 lb)cos30 ][(4 2cos30 )in.] [(140 lb)sin30 ][(2 in.)sin30 ]− °+ °− ° °
(60 lb)(2 in.) (38.0 lb)x−=
1( 694.97 70.0 120) in.
38.0
x= − −−
and
23.289 in.x= −
consent of McGraw–Hill Education.
PROBLEM 3.81
A couple of magnitude M = 54 lb ⋅ in. and the three forces shown are
applied to an angle bracket. (a) Find the resultant of this system of
forces. (b) Locate the points where the line of action of the resultant
intersects line AB and line BC.
SOLUTION
(a) We have
: ( 10 ) (30 cos 60 )
30 sin 60 ( 45 )
(30 lb) (15.9808 lb)
Σ =−+ °
+ °+−
=−+
FR j i
ji
ij
or
34.0 lb=R
28.0°
(b) First reduce the given forces and couple to an equivalent force-couple system
( , )
B
RM
at B.
below B.
consent of McGraw–Hill Education.
PROBLEM 3.82
A truss supports the loading shown. Determine the
equivalent force acting on the truss and the point of
intersection of its line of action with a line drawn through
Points A and G.
SOLUTION
We have
(240 lb)(cos70 sin 70 ) (160 lb)
(300 lb)( cos40 sin40 ) (180 lb)
= Σ
= °− ° −
+ − °− ° −
RF
R ij j
ij j
22
22
1
1
(147.728 lb) (758.36 lb)
(147.728) (758.36)
772.62 lb
tan
758.36
tan 147.728
78.977
xy
y
x
R RR
R
R
θ
−
−
=−−
= +
= +
=
=
−
=
−
= °
R ij
or
773 lb=R
79.0°
We have
Ay
M dRΣ=
where
[240 lbcos70 ](6 ft) [240 lbsin70 ](4 ft)
(160 lb)(12 ft) [300 lbcos40 ](6 ft)
[300 lbsin 40 ](20 ft) (180 lb)(8 ft)
7232.5 lb ft
A
MΣ=− ° − °
− +°
− °−
=−⋅
7232.5 lb ft
758.36 lb
9.5370 ft
d−⋅
=−
=
or
9.54 ft to the right of dA=
consent of McGraw–Hill Education.
PROBLEM 3.83
A machine component is subjected to the forces and
couples shown. The component is to be held in place by a
single rivet that can resist a force but not a couple. For
P = 0, determine the location of the rivet hole if it is to be
located (a) on line FG, (b) on line GH.
SOLUTION
We have
First replace the applied forces and couples with an equivalent force–couple system at G.
Thus,
: 200cos15 120cos 70
xx
F PR
Σ °− °+ =
or
(152.142 ) N
x
RP
= +
: 200sin 15 120sin 70 80
yy
FRΣ − °− °− =
or
244.53 N
y
R= −
: (0.47 m)(200 N)cos15 (0.05 m)(200 N)sin15
(0.47 m)(120 N)cos70 (0.19 m)(120 N)sin70
(0.13 m)( N) (0.59 m)(80 N) 42 N m
40 N m
G
G
M
P
M
Σ − °+ °
+ °− °
− − +⋅
+ ⋅=
or
(55.544 0.13 ) N m
G
MP=−+ ⋅
(1)
Setting
0P=
in Eq. (1):
Now with R at I,
: 55.544 N m (244.53 N)
G
MaΣ − ⋅=−
or
0.227 ma=
and with R at J,
: 55.544 N m (152.142 N)
G
MbΣ − ⋅=−
or
0.365 mb=
consent of McGraw–Hill Education.
PROBLEM 3.84
Solve Problem 3.83, assuming that P = 60 N.
PROBLEM 3.83 A machine component is subjected to
the forces and couples shown. The component is to be held
in place by a single rivet that can resist a force but not a
couple. For P = 0, determine the location of the rivet hole
if it is to be located (a) on line FG, (b) on line GH.
SOLUTION
See the solution to Problem 3.83 leading to the development of Equation (1):
and
(55.544 0.13 ) N m
(152.142 ) N
G
x
MP
RP
=−+ ⋅
= +
For
60 NP=
we have
(152.142 60)
212.14 N
[55.544 0.13(60)]
63.344 N m
x
G
R
M
= +
=
=−+
=−⋅
Then with R at I,
: 63.344 N m (244.53 N)
G
MaΣ − ⋅=−
or
0.259 m
a=
and with R at J,
: 63.344 N m (212.14 N)
G
MbΣ − ⋅=−
or
0.299 mb=
(a) The rivet hole is 0.299 m above G.
(b) The rivet hole is 0.259 m to the right of G.
consent of McGraw–Hill Education.
PROBLEM 3.85
As an adjustable brace BC is used to bring a wall into plumb,
the force–couple system shown is exerted on the wall. Replace
this force–couple system with an equivalent force–couple system
at A if
21.2 lbR=
and
13.25 lb · ft.M=
SOLUTION
We have
:
A BC
λ
Σ==
F RR R
where
(42 in.) (96 in.) (16 in.)
106 in.
BC
−−
=i jk
λ
21.2 lb (42 96 16 )
106
A
= −−R i jk
or
(8.40 lb) (19.20 lb) (3.20 lb)
A
=−−R i jk
We have
/
:
A CA A
Σ ×+ =M r RMM
where
/
1
(42 in.) (48 in.) (42 48 )ft
12
(3.5 ft) (4.0 ft)
CA
=+=+
= +
r i k ik
ik
(8.40 lb) (19.50 lb) (3.20 lb)
42 96 16 (13.25 lb ft)
106
(5.25 lb ft) (12 lb ft) (2 lb ft)
BC M
λ
=−−
= −
−+ +
= ⋅
=− ⋅+ ⋅+ ⋅
Ri jk
M
i jk
i jk
Then
3.5 0 4.0 lb ft ( 5.25 12 2 ) lb ft
8.40 19.20 3.20
A
⋅ +− + + ⋅ =
−−
ijk
i jk M
(71.55 lb ft) (56.80 lb ft) (65.20 lb ft)
A
=⋅+⋅−⋅M ijk
or
(71.6 lb ft) (56.8 lb ft) (65.2 lb ft)
A
= ⋅+ ⋅− ⋅M ijk
consent of McGraw–Hill Education.
PROBLEM 3.86
As plastic bushings are inserted into a 60–mm–diameter cylindrical sheet
metal enclosure, the insertion tools exert the forces shown on the enclosure.
Each of the forces is parallel to one of the coordinate axes. Replace these
forces with an equivalent force–couple system at C.
SOLUTION
For equivalence,
:ΣF
(17 N) (12 N) (16 N) (21 N)
(21 N) (29 N) (16 N)
ABCD
=+++
=−− − −
=−− −
RFFFF
jjki
i jk
///
:
C ACABCBDCD
MΣ = ×+ ×+ ×
Mr Fr Fr F
[(0.11 m) (0.03 m) ] [ (17 N)]
[(0.02 m) (0.11 m) (0.03 m) ] [ (12 N)]
[(0.03 m) (0.03 m) (0.03 m) ] [ (21 N)]
(0.51 N m) [ (0.24 N m) (0.36 N m) ]
[(0.63 N m) (0.63 N m) ]
M= − ×−
+ + − ×−
+ + − ×−
=− ⋅ +− ⋅ − ⋅
+ ⋅+ ⋅
jk j
ijk j
ijk i
i ki
kj
∴ The equivalent force–couple system at C is
(21.0 N) (29.0 N) (16.00 N)=−− −R ij k
(0.870 N m) (0.630 N m) (0.390 N m)=−⋅+⋅+⋅M ijk
consent of McGraw–Hill Education.
PROBLEM 3.78
Five separate force–couple systems act at the corners of a piece of sheet metal, which has been bent into
the shape shown. Determine which of these systems is equivalent to a force F = (10 lb)i and a couple of
moment M = (15 lb ⋅ ft)j + (15 lb ⋅ ft)k located at the origin.
SOLUTION
First note that the force–couple system at F cannot be equivalent because of the direction of the force [The
force of the other four systems is (10 lb)i]. Next, move each of the systems to the origin O; the forces
remain unchanged.
: (5 lb ft) (15 lb ft) (2 ft) (10 lb)
AO
A=Σ = ⋅+ ⋅ + ×MM j k k i
(25 lb ft) (15 lb ft)= ⋅+ ⋅jk
: (5 lb ft) (25 lb ft)
[(4.5 ft) (1 ft) (2 ft) ] 10 lb)
(15 lb ft) (15 lb ft)
DO
D=Σ =− ⋅+ ⋅
+ ++ ×
= ⋅+ ⋅
MM j k
ijk i
ik
: (15 lb ft) (15 lb ft)
: (15 lb ft) (5 lb ft)
[(4.5 ft) (1 ft) ] (10 lb)
GO
II
G
I
=Σ = ⋅+ ⋅
=Σ= ⋅− ⋅
+ +×
MM i j
MM j k
ij j
(15 lb ft) (15 lb ft)= ⋅− ⋅jk
The equivalent force–couple system is the system at corner D.
PROBLEM 3.79
Four forces act on a 700 × 375-mm plate as shown. (a) Find
the resultant of these forces. (b) Locate the two points where
the line of action of the resultant intersects the edge of the
plate.
SOLUTION
(a)
( 400 N 160 N 760 N)
(600 N 300 N 300 N)
(1000 N) (1200 N)
= Σ
=−+−
+ ++
=−+
RF
i
j
ij
22
(1000 N) (1200 N)
1562.09 N
1200 N
tan 1000 N
1.20000
50.194
R
θ
θ
= +
=
= −
= −
=−°
1562 N=R
50.2°
(b)
(0.5 m) (300 N 300 N)
(300 N m)
R
C=Σ×
= ×+
= ⋅
M rF
ij
k
(300 N m) (1200 N)
0.25000 m
250 mm
(300 N m) ( 1000 N)
0.30000 m
300 mm
x
x
x
y
y
y
⋅=×
=
=
⋅ = ×−
=
=
ki j
ji
consent of McGraw–Hill Education.
PROBLEM 3.80
A 32–lb motor is mounted on the floor. Find the resultant of
the weight and the forces exerted on the belt, and determine
where the line of action of the resultant intersects the floor.
SOLUTION
We have
: (60 lb) (32 lb) (140 lb)(cos30 sin30 )Σ − + °+ ° =F i j i jR
(181.244 lb) (38.0 lb)= +R ij
or
185.2 lb=R
11.84°
We have
:
O Oy
M M xRΣ Σ=
[(140 lb)cos30 ][(4 2cos30 )in.] [(140 lb)sin30 ][(2 in.)sin30 ]− °+ °− ° °
(60 lb)(2 in.) (38.0 lb)x−=
1( 694.97 70.0 120) in.
38.0
x= − −−
and
23.289 in.x= −
consent of McGraw–Hill Education.
PROBLEM 3.81
A couple of magnitude M = 54 lb ⋅ in. and the three forces shown are
applied to an angle bracket. (a) Find the resultant of this system of
forces. (b) Locate the points where the line of action of the resultant
intersects line AB and line BC.
SOLUTION
(a) We have
: ( 10 ) (30 cos 60 )
30 sin 60 ( 45 )
(30 lb) (15.9808 lb)
Σ =−+ °
+ °+−
=−+
FR j i
ji
ij
or
34.0 lb=R
28.0°
(b) First reduce the given forces and couple to an equivalent force-couple system
( , )
B
RM
at B.
below B.
consent of McGraw–Hill Education.
PROBLEM 3.82
A truss supports the loading shown. Determine the
equivalent force acting on the truss and the point of
intersection of its line of action with a line drawn through
Points A and G.
SOLUTION
We have
(240 lb)(cos70 sin 70 ) (160 lb)
(300 lb)( cos40 sin40 ) (180 lb)
= Σ
= °− ° −
+ − °− ° −
RF
R ij j
ij j
22
22
1
1
(147.728 lb) (758.36 lb)
(147.728) (758.36)
772.62 lb
tan
758.36
tan 147.728
78.977
xy
y
x
R RR
R
R
θ
−
−
=−−
= +
= +
=
=
−
=
−
= °
R ij
or
773 lb=R
79.0°
We have
Ay
M dRΣ=
where
[240 lbcos70 ](6 ft) [240 lbsin70 ](4 ft)
(160 lb)(12 ft) [300 lbcos40 ](6 ft)
[300 lbsin 40 ](20 ft) (180 lb)(8 ft)
7232.5 lb ft
A
MΣ=− ° − °
− +°
− °−
=−⋅
7232.5 lb ft
758.36 lb
9.5370 ft
d−⋅
=−
=
or
9.54 ft to the right of dA=
consent of McGraw–Hill Education.
PROBLEM 3.83
A machine component is subjected to the forces and
couples shown. The component is to be held in place by a
single rivet that can resist a force but not a couple. For
P = 0, determine the location of the rivet hole if it is to be
located (a) on line FG, (b) on line GH.
SOLUTION
We have
First replace the applied forces and couples with an equivalent force–couple system at G.
Thus,
: 200cos15 120cos 70
xx
F PR
Σ °− °+ =
or
(152.142 ) N
x
RP
= +
: 200sin 15 120sin 70 80
yy
FRΣ − °− °− =
or
244.53 N
y
R= −
: (0.47 m)(200 N)cos15 (0.05 m)(200 N)sin15
(0.47 m)(120 N)cos70 (0.19 m)(120 N)sin70
(0.13 m)( N) (0.59 m)(80 N) 42 N m
40 N m
G
G
M
P
M
Σ − °+ °
+ °− °
− − +⋅
+ ⋅=
or
(55.544 0.13 ) N m
G
MP=−+ ⋅
(1)
Setting
0P=
in Eq. (1):
Now with R at I,
: 55.544 N m (244.53 N)
G
MaΣ − ⋅=−
or
0.227 ma=
and with R at J,
: 55.544 N m (152.142 N)
G
MbΣ − ⋅=−
or
0.365 mb=
consent of McGraw–Hill Education.
PROBLEM 3.84
Solve Problem 3.83, assuming that P = 60 N.
PROBLEM 3.83 A machine component is subjected to
the forces and couples shown. The component is to be held
in place by a single rivet that can resist a force but not a
couple. For P = 0, determine the location of the rivet hole
if it is to be located (a) on line FG, (b) on line GH.
SOLUTION
See the solution to Problem 3.83 leading to the development of Equation (1):
and
(55.544 0.13 ) N m
(152.142 ) N
G
x
MP
RP
=−+ ⋅
= +
For
60 NP=
we have
(152.142 60)
212.14 N
[55.544 0.13(60)]
63.344 N m
x
G
R
M
= +
=
=−+
=−⋅
Then with R at I,
: 63.344 N m (244.53 N)
G
MaΣ − ⋅=−
or
0.259 m
a=
and with R at J,
: 63.344 N m (212.14 N)
G
MbΣ − ⋅=−
or
0.299 mb=
(a) The rivet hole is 0.299 m above G.
(b) The rivet hole is 0.259 m to the right of G.
consent of McGraw–Hill Education.
PROBLEM 3.85
As an adjustable brace BC is used to bring a wall into plumb,
the force–couple system shown is exerted on the wall. Replace
this force–couple system with an equivalent force–couple system
at A if
21.2 lbR=
and
13.25 lb · ft.M=
SOLUTION
We have
:
A BC
λ
Σ==
F RR R
where
(42 in.) (96 in.) (16 in.)
106 in.
BC
−−
=i jk
λ
21.2 lb (42 96 16 )
106
A
= −−R i jk
or
(8.40 lb) (19.20 lb) (3.20 lb)
A
=−−R i jk
We have
/
:
A CA A
Σ ×+ =M r RMM
where
/
1
(42 in.) (48 in.) (42 48 )ft
12
(3.5 ft) (4.0 ft)
CA
=+=+
= +
r i k ik
ik
(8.40 lb) (19.50 lb) (3.20 lb)
42 96 16 (13.25 lb ft)
106
(5.25 lb ft) (12 lb ft) (2 lb ft)
BC M
λ
=−−
= −
−+ +
= ⋅
=− ⋅+ ⋅+ ⋅
Ri jk
M
i jk
i jk
Then
3.5 0 4.0 lb ft ( 5.25 12 2 ) lb ft
8.40 19.20 3.20
A
⋅ +− + + ⋅ =
−−
ijk
i jk M
(71.55 lb ft) (56.80 lb ft) (65.20 lb ft)
A
=⋅+⋅−⋅M ijk
or
(71.6 lb ft) (56.8 lb ft) (65.2 lb ft)
A
= ⋅+ ⋅− ⋅M ijk
consent of McGraw–Hill Education.
PROBLEM 3.86
As plastic bushings are inserted into a 60–mm–diameter cylindrical sheet
metal enclosure, the insertion tools exert the forces shown on the enclosure.
Each of the forces is parallel to one of the coordinate axes. Replace these
forces with an equivalent force–couple system at C.
SOLUTION
For equivalence,
:ΣF
(17 N) (12 N) (16 N) (21 N)
(21 N) (29 N) (16 N)
ABCD
=+++
=−− − −
=−− −
RFFFF
jjki
i jk
///
:
C ACABCBDCD
MΣ = ×+ ×+ ×
Mr Fr Fr F
[(0.11 m) (0.03 m) ] [ (17 N)]
[(0.02 m) (0.11 m) (0.03 m) ] [ (12 N)]
[(0.03 m) (0.03 m) (0.03 m) ] [ (21 N)]
(0.51 N m) [ (0.24 N m) (0.36 N m) ]
[(0.63 N m) (0.63 N m) ]
M= − ×−
+ + − ×−
+ + − ×−
=− ⋅ +− ⋅ − ⋅
+ ⋅+ ⋅
jk j
ijk j
ijk i
i ki
kj
∴ The equivalent force–couple system at C is
(21.0 N) (29.0 N) (16.00 N)=−− −R ij k
(0.870 N m) (0.630 N m) (0.390 N m)=−⋅+⋅+⋅M ijk
consent of McGraw–Hill Education.