Chapter 3
PROBLEM 3.12
C
PROBLEM 3.1
b
B
A
A 13.2-N force P is applied to the lever which controls the auger of a
snowblower. Determine the moment of P about A when D is equal to 30°.
D
E
d
It is known that a force with a moment of 900 N.m about D is required
to straighten the fence post CD. If a 200 mm, b 900 mm, and d 2.8
m, determine the tension that must be developed in the cable of winch
puller AB to create the required moment about point D.
a
SOLUTION
10.44
3
SOLUTION
0.9 m
10
First note
Px
Py
13.2 N sin 30q
P sin D
P cos D
13.2 N cos 30q
6.60 N
2.8 m
11.4315 N
Noting that the direction of the moment of each force component about A
is counterclockwise,
MA
xB/ A Py yB/ A Px
Slope of line EC =
0.9 m
3
=
2.8 m + 0.2 m 10
Then
TABx =
10
TAB
10.44
and
TABy =
3
TAB
10.44
0.086 m 11.4315 N 0.122 m 6.60 N 1.78831 N ˜ m
or M A
1.788 N ˜ m
W
0.2 m
Have
MD = TABx (y) + TABy (x)
\ 900 N.m =
10
3
TAB (0) +
TAB (2.8 m)
10.44
10.44
TAB = 1118.6 N
or TAB = 1119 N J
PROBLEM 3.21
Before the trunk of a large tree is felled, cables AB and BC are attached as
shown. Knowing that the tension in cables AB and BC are 777 N and
990 N, respectively, determine the moment about O of the resultant force
exerted on the tree by the cables at B.
y
8.4 m
7.2 m
PROBLEM 3.34
0.9 m
1.2 m
0.8 m
1m
Determine the value of a which minimizes the perpendicular distance
from point C to a section of pipeline that passes through points A and B.
5m
6m
2m
A
SOLUTION
x
z
y
a
C
3m
5.1 m
B
8m
x
Assuming a force F acts along AB,
MC
SOLUTION
rA/C u F
F d where
d
perpendicular distance from C to line AB
F
8 m i 7 m j 9 m k F
8 2 7 2 9 2 m
O AB F
F 0.57437 i 0.50257 j 0.64616 k
rA/C
? MC
MO
Have
where
8.4 m j
FB
TAB TBC
i
j
k
1
2.8
3a F
0.57437 0.50257 0.64616
ª¬ 0.30154 0.50257a i 2.3693 0.57437a j
rB/O u FB
rB/O
1 m i 2.8 m j a 3 m k
2.1108k @ F
Since
MC
rA/C u F 2
or
2
TAB
TBC
O BATAB
O BCTBC
0.9 m i 8.4 m j 7.2 m k
0.9 2 8.4 2 7.2 2
dF 2
rA/C u F 2
2
? 0.30154 0.50257a 2.3693 0.57437a 2.1108 777 N m
Setting
5.1 m i 8.4 m j 1.2 m k 990 N
5.12 8.4 2 1.2 2 m
d
d2
da
2
d2
0 to find a to minimize d
2 0.50257 0.30154 0.50257a 2 0.57437 2.3693 0.57437a Solving
a
0
2.0761 m
or a
2.08 m W
PROBLEM 3.39
PROBLEM 3.47
Steel framing members AB, BC, and CD are joined at B and C and are
braced using cables EF and EG. Knowing that E is at the midpoint of BC
and that the tension in cable EF is 330 N, determine (a) the angle
between EF and member BC, (b) the projection on BC of the force
exerted by cable EF at point E.
A fence consists of wooden posts and a steel cable fastened to each post
and anchored in the ground at A and D. Knowing that the sum of the
moments about the z axis of the forces exerted by the cable on the posts
at B and C is 66 N · m, determine the magnitude TCD when TBA 56 N.
SOLUTION
SOLUTION
k ˜ ª rB y u TBA º k ˜ ª rC y u TCD º
¬
¼
¬
¼
| Mz |
Based on
where
Mz
66 N ˜ m k
rB y
TBA
where
O BC
O EF
?
16 m i 4.5 m j 12 m k
16 2 4.52 12 2 m
7 m i 6 m j 6 mk
7 2 6 2 6 2 m
20.5
and
T
TCD
3.0 m
TEF BC
CD
1
TCD 2i j 2k 3
cosT
? 66 N ˜ m ^
`
k ˜ 1 m j u ª¬ 24 N i 16 N j 48 N k º¼
­
ª1
º½
k ˜ ®1 m j u « TCD 2i j 2k » ¾
¬3
¼¿
¯
134.125q
or T
(b) By definition
O CDTCD
2 m i 1 m j 2 m k T
20.511.0 cosT
§ 157 ·
cos 1 ¨
¸
© 225.5 ¹
56 N 24 N i 16 N j 48 N k
1
7i 6 j 6k 11.0
11.0
16 7 4.5 6 12 6 O BATBA
3.5 m
1
16i 4.5j 12k 20.5
16i 4.5j 12k ˜ 7i 6 j 6k 1 m j
1.5 m i 1 m j 3 m k
11 cosT
O BC ˜ O EF
(a) By definition
rC y
134.1q W
TEF cosT
330 N cos134.125q
or
66
? TCD
229.26 N
24 2
TCD
3
3
66 24 N
2
or TCD
or TEF BC
230 N W
63.0 N W
PROBLEM 3.68
PROBLEM 3.53
y
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-N forces,
(b) the perpendicular distance between the 12-N forces if the resultant of
the two couples is zero, (c) the value of D if the resultant couple is
1.8 N ˜ m clockwise and d is 1.05 m.
The triangular plate ABC is supported by ball-and-socket joints at B and
D and is held in the position shown by cables AE and CF. If the force
exerted by cable AE at A is 1000 N, determine the moment of that force
about the line joining points D and B.
400
200
900
700
600
600
z
SOLUTION
350
900
300
450
600
M1
(a) Have
x
All dimensions in mm
where
SOLUTION
MDB = lDB ◊ (rA/D ¥ TAE)
Have
0.2 m
0.7 m
lDB =
where
b1.2 mg i - b0.35 mg j = 0.96 i – 0.28 j
0.6 m
TAE = lAETAE =
21 N
8.4 N ˜ m
M1 M 2
8.40 N ˜ m d 2 12 N or
. g - b0.2gb- 545.45g + b- 0.280g 0.2 b81818
. g-0
b01. gb18182
= 41.45 N.m
or MDB = 41.5 N.m J
M total
(c) Have
and
1.8 N ˜ m
W
0
? d2
or
0.960 - 0.280
0
0
0.2 N. m
- 0.1
818.18 - 545.45 18182
.
8.40 N ˜ m
0
11
. m
= (818.18 N) i – (545.45 N) j + (181.82 N) k
= (0.960)
F1
or M1
0.4 m
\ MDB =
0.4 m
0.4 m 21 N (b) Have
b0.9 mg i - b0.6 mg j + b0.2 mg k b1000 N g
0.9 m
0.3 m
d1
0.125 m
rA/D = – (0.1 m) j + (0.2 m) k
0.35 m
? M1
d1F1
0.700 m W
M1 M 2
8.40 N ˜ m 1.05 m sin D 12 N ? sin D
0.52381
D
31.588q
or D
31.6q W
PROBLEM 3.83
PROBLEM 3.85
A dirigible is tethered by a cable attached to its cabin at B. If the tension
in the cable is 1100 N, replace the force exerted by the cable at B with an
equivalent system formed by two parallel forces applied at A and C.
A force and a couple are applied to a beam. (a) Replace this system with
a single force F applied at point G, and determine the distance d.
(b) Solve part a assuming that the directions of the two 600-N forces are
reversed.
6 m 3.6 m
SOLUTION
(a)
SOLUTION
Require the equivalent forces acting at A and C be parallel and at an angle
of D with the vertical.
Then for equivalence,
1100 N
ÂFx: (1100 N) sin 30° = FA sin a + FB sin a
(1)
ÂFy : – (1100 N) cos 30° = – FA cos a – FB cos a
(2)
Dividing Equation (1) by Equation (2),
Have
b1100 Ng sin 30∞ = b F + F g sin a
- b F + F g cos a
- b1100 Ng cos 30∞
A
A
6m
6Fy : FC FD FE
Have
B
F
800 N 600 N 600 N
F
800 N
6M G : FC d 1.5 m FD 2 m 800 N d
B
1.5 m 600 N 2 m Simplifying yieldsa = 30°
d
1200 1200
800
Based on
d
3m
3.6 m
ÂMC:
b1100 Ng cos 30∞ b3.6 mg
F
or F
800 N W
or d
3.00 m W
or F
800 N W
0
0
(b)
= (FA cos 30°) (9.6 m)
\ FA = 412.5 N
or FA = 412.5 N
60° J
Based on
ÂMA:
b
Changing directions of the two 600 N forces only changes sign of the couple.
g
– 1100 N cos 30∞ (6 m) = (FC cos 30°) (9.6 m)
? F
and
\ FC = 687.5 N
or FC = 687.5 N
60° J
800 N
6M G : FC d 1.5 m FD 2 m 800 N d
d
0
1.5 m 600 N 2 m 1200 1200
800
0
or d
0 W
PROBLEM 3.93
y
900 N
1.2 m
To keep a door closed, a wooden stick is wedged between the floor and
the doorknob. The stick exerts at B a 200 N force directed along line AB.
Replace that force with an equivalent force-couple system at C.
800 mm
60 mm
900 N
450 N
1350 N
1200 Nm
1560 Nm
900 N
270 N
1800 N
450 N
1800 Nm 1600 Nm 675 N
270 N.m
2700 Nm
600 Nm
2000 N
675 N
3050 Nm
450 N
200 Nm 1800 N 200 Nm
675 Nm
450 N
PROBLEM 3.98
A 1.2 m-long beam is subjected to a variety of loadings. (a) Replace each
loading with an equivalent force-couple system at end A of the beam.
(b) Which of the loadings are equivalent?
200 Nm
1800 mm
SOLUTION
 Fy:
(a) Have
(a)
– 900 N – 450 N = Ra
or R a = 1350 N
950 mm
A
O
and
600 mm
 M A:
J
1200 N.m – (450 N) (1.2 m) = Ma
or M a = 660 N.m J
x
110 mm
z
1.2 m
 Fy:
(b) Have
– 1350 N = Rb
or Rb = 1350 N J
SOLUTION
 MA: – 600 N.m = Mb
and
Have
0.8 m
or M b = 600 N.m J
ÂF: PAB = FC
0.06 m
 Fy:
(c) Have
or R c = 1325 N
PAB = lABPAB
1.8 m
=
b0.05 mg i + b0.95 mg j - b0.6 mg k
1125
.
m
and
or M c = 650 N.m
(200 N)
 Fy:
Have
or R d = 900 N J
and
0.6 m
 MA:
(1800 N) (1.2 m) – 1560 N.m = Md
ÂMC: rB/C ¥ PAB = MC
i
MC = 0.74
j
k
0 N. m
- 0.85
8.89 168.89 106.67
=
mb- 0.85gb- 106.67g i - b0.74gb- 106.67g j
+ [(0.74) (168.89) – (–0.85) (8.89)] k}N.m
or MC = (90.7 N.m) i + (78.9 N.m) j + (132.5 N.m) k J
J
– 900 N + 1800 N = Rd
or FC = (8.89 N) i + (168.89 N) j – (106.67 N) k J
0.95 m
J
 MA: – 3050 N.m – (2000 N) (1.2 m) = Mc
(d) Have
0.11 m
675 N – 2000 N = Rc
where
(e) Have
and
 Fy:
 MA :
or M d = 600 N.m
J
or R e = 1350 N
J
– 900 N – 450 N = Re
135 N.m + 270 N.m – (450 N) (1.2 m) = Me
or M e = 135 N.m
J
PROBLEM 3.98 CONTINUED
 Fy:
(f ) Have
2.6 kN
– 1800 N + 450 N = Rf
 M A:
z 30 mm 30 mm 4020mm
mm
20 mm
40 mm
200 N.m + 200 N.m + (450 N) (1.2 m) = Mf
or M f = 540 N.m J
 Fy:
(g) Have
or M g = 810 N.m J
 Fy:
 MA:
6F
(a) Have
R
ª¬ 2.6 j 5.25 j 10.5 cos 45qi sin 45q j 3.2i º¼ kN
?R
J
1600 N.m – 200 N.m – (675 N) (1.2 m) = Mh
or M h = 590 N.m
(b)
SOLUTION
R
675 N – 675 N = Rh
or R h = 1350 N
and
As four holes are punched simultaneously in a piece of aluminum sheet
metal, the punches exert on the piece the forces shown. Knowing that the
forces are perpendicular to the surfaces of the piece, determine (a) the
resultant of the applied forces when D 45q and the point of
intersection of the line of action of that resultant with a line drawn
through points A and B, (b) the value of D so that the line of action of the
resultant passes through fold EF.
Position the origin for the coordinate system along the centerline of the
sheet metal at the intersection with line EF.
 M A: 270 N.m + 2700 N.m – (1800 N) (1.2 m) = Mg
(h) Have
20 mm
E
– 450 N – 1800 N = Rg
or R g = –2250 N J
and
5.25 kN
60 mm
or R f = 1350 N J
and
PROBLEM 3.108
y
R
10.6246 kN i 15.2746 kN j
10.6246 2 15.2746 2
Rx2 Ry2
J
18.6064 kN
Therefore, none of the loadings are equivalent. J
T
§ Ry ·
tan 1 ¨
¸
© Rx ¹
§ 15.2746 ·
tan 1 ¨
¸
© 10.6246 ¹
or R
M EF
Have
55.179q
18.61 kN
55.2q W
6M EF
where
MEF
2.6 kN 90 mm 5.25 kN 40 mm 10.5 kN 20 mm 3.2 kN ª¬ 40 mm sin 45q 40 mm º¼
? MEF
15.4903 N ˜ m
To obtain distance d left of EF,
M EF
Have
?d
dRy
15.4903 N ˜ m
15.2746 u 103 N
d 15.2746 kN 1.01412 u 103 m
or d
1.014 mm left of EF W
PROBLEM 3.119
PROBLEM 3.108 CONTINUED
M EF
(b) Have
M EF
0
A portion of the flue for a furnace is attached to the ceiling at A. While
supporting the free end of the flue at F, a worker pushes in at E and pulls
out at F to align end E with the furnace. Knowing that the 45 N force at
F lies in a plane parallel to the yz plane, determine (a) the angle D the
force at F should form with the horizontal if duct AB is not to tend to
rotate about the vertical, (b) the force-couple system at B equivalent to
the given force system when this condition is satisfied.
250 mm
0
850 mm
2.6 kN 90 mm 5.25 kN 40 mm 10.5 kN 20 mm z
3.2 kN ª¬ 40 mm sin D 40 mm º¼
325 mm
45 N
a
900 mm
450 mm
? 128 N ˜ m sin D
sin D
D
C
F
D
700 mm
225 mm
E
22.5 N
106 N ˜ m
0.828125
SOLUTION
55.907q
(a) Duct AB will not have a tendency to rotate about the vertical or y-axis if:
or D
55.9q W
R
M By
j ˜ 6M RB
j ˜ rF /B u FF rE/B u FE
0
where
rF/B = (1.125 m) i – (0.575 m) j + (0.7 m) k
rE/B = (1.35 m) i – (0.85 m) j + (0.7 m) k
b
g b
g
FF = 45 N sin a j + cos a k
FE = – (22.5 N) k
i
j
k
i
j
k
\ ÂM RB = (45 N) 1125
. m 0.575 m 0.7 m + (22.5 N) 1.35 m 0.85 m 0.7 m
0
sin a
cos a
0
0
-1
=
Thus,
. cos a - 315
. sin a + 19.125g i - b- 50.625 cos a - 30.375g j + b50.625 sin a g k N. m
b- 25875
R
M By
= – 50.625 cos a + 30.375 = 0
cos a = 0.60
a = 53.130°
or a = 53.1° J
PROBLEM 3.119 CONTINUED
PROBLEM 3.126
(b) R = FE + FF
The forces shown are the resultant downward loads on sections of
the flat roof of a building because of accumulated snow. If the snow
represented by the 580-kN force is shoveled so that the this load
acts at E, determine a and b knowing that the point of application
of the resultant of the four loads is then at B.
where
FE = – (22.5 N) k
FF = (45 N) (sin 53.130° j + cos 53.130° k) = (36 N) j + (27 N) k
\ R = (36 N) j + ( 27 N) k J
b g
b g
b g
SOLUTION
b g
.
N 0.6 + 315
. N 0.8 - 19.125 N i - 50.625 N 0.6 - 30.375 j + 50.625 0.8 k
and M = ÂM BR = - 25875
= – (21.6 N.m) i – (0) j + (40.5 N.m) k
or M = – (21.6 N.m) i + (40.5 N.m) k J
Have
6F : FB FC FD FE
R
2350 kN j 330 kN j 140 kN j 580 kN j
? R
Have
R
3400 kN j 6M x : FB z B FC zC FD z D FE z E R zB 2350 kN 16 m 330 kN 6 m 140 kN 33.5 m 580 kN b 3400 kN 16 m ? b
Have
or b
17.4655 m
6M z : FB xB FC xC FD xD FE xE 17.47 m W
R xB 2350 kN 32 m 330 kN 54 m 140 kN 32 m 580 kN a 3400 kN 32 m ? a
19.4828 m
or a
19.48 m W