Solucionario mecanica vectorial para ingenieros estatica edicion 9 ( pdf drive )

80 mm
PROBLEM 3.1
A foot valve for a pneumatic system is hinged at B. Knowing
that a - 28°, determine the moment of the 1 6-N force about
Point B by resolving the force into horizontal and vertical
components.
SOLUTION
Note that
and
= a- 20° = 28° -20° = 8°
Fx = (1 6 N)cos 8° = 1 5.8443 N
Fv =(16N)sin8° = 2.2268N
^£*. ©* — tL> Kl
* U^r^-^T^P,,
/|t^^^u^d
^r^^^^-,
Also x = (0. 1 7 m)cos 20° = 0. 1 59748 m
y = (0.17 m)sin 20° = 0.058143 m.
>n^y c
Noting that the direction of the moment of each force component about B is counterclockwise,
MB =xFy
+yFx
= (0.1 59748 m)(2.2268N)
+(0.058143 m)(l 5.8443 N)
= 1.277 N-m or MB =1.277N-m^)4
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153

PROBLEM 3.2
A foot valve for a pneumatic system is hinged at B. Knowing
that a = 28°, determine the moment of the 1 6-N force about
Point B by resolving the force into components along ABC and
in a direction perpendicular to ABC.
SOLUTION
First resolve the 4-lb force into components P and Q, where
g = (16 N)sin 28°
= 7.5115 N
Then MB = rmQ
°^7^
= (0.17m)(7.5115N)
= 1.277N-m or MB = 1.277 N-m^^
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154

200 in..,
25"
-100 mm- 200 mm
i
'">
mm
PROBLEM 3.3
A 300-N force is applied at A as shown. Determine (a) the moment
of the 300-N force about D, (b) the smallest force applied at B that
creates the same moment about D.
SOLUTION
(a)
O.2.
PC
0>T-**>
Fv =(300N)cos25°
= 27.1.89 N
Fy
=(300 N) sin 25°
= 126.785 N
F = (27 1 .89 N)i + (1 26.785 N)
j
r = ZM = -(0.1m)i-(0.2m)j
MD =rxF
MD = HO. 1 m)i - (0.2 m)j] x [(271 .89 N)i + (1.26.785 N)j]
= -(12.6785 N • m)k + (54.378 N •
m)k
= (41.700 N-m)k
M =41.7N-nO^
(b) The smallest force Q at B must be perpendicular to
DB at 45°^£L
MD =Q(DB)
41 .700 N m = (2(0.28284 m) Q = 147.4 N ^L 45° <
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155

200 mn.
25°
•-lOOnim-* -200 mm *.
125 il
•
C
H
PROBLEM3.4
A 300-N force is applied at A as shown. Determine (a) the
moment of the 300-N force about D, (b) the magnitude and
sense of the horizontal force applied at C that creates the
same moment about D, (c) the smallest force applied at C that
creates the same moment about D.
SOLUTION
(a) See Problem 3.3 for the figure and analysis leading to the determination of Md
M =41.7N-m^H
Cl^n
0>7.Siy
c = at.
(b) Since C is horizontal C = Ci
r = DC = (0.2 m)i - (0. 1 25 m)j
MD =rxCi = C(0.l25m)k
4l.7N-m = (0.l25m)(C)
C = 333.60 N
(c) The smallest force C must be perpendicular to DC; thus, it forms a with the vertical
C = 334N <
tan6^
0.125 m
0.2 m
a = 32.0°
MD = C(£>C); DC = V( - 2 m)
2
+ (°- 1 25 m)'
= 0.23585 m
41.70 Nm = C(0.23585m) C = 176.8 N^L 5HX)°<
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156

PROBLEM 3.5
An 8-1b force P is applied to a shift lever. Determine the moment of V about B
when a is equal to. 259.
SOLUTION
First note Px
= (8 lb) cos 25°
= 7.2505 lb
/^ =(8 lb) sin 25°
= 3.3809 lb
Noting that the direction of the moment of each force component about B is
clockwise, have
= -(8in.)(3.3809 1b)
- (22 in.)(7.2505 lb)
= -186.6 lb -in.
i*22. •**.
or Mj =186.6 lb -in.
J) ^
157

PROBLEM 3.6
For the shift lever shown, determine the magnitude and the direction of the smallest
force P that has a 21 0-lb • in. clockwise moment about B.
22 in.
SOLUTION
For P to be minimum it must be perpendicular to the line joining Points A. and B. Thus,
a = e
, 8 in ^T^s*. K
22 in. /
= 19.98° i
and MB =dP^n p ZZ i«.
Where d = rAIB fl
$ u
= ^m.y+(22m.y JB w
= 23.409 in.
Then
_210Ib-in.
'
min " 23.409 in.
-8.97 lb Pmin
=8.97 lb ^19.98° <
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158

PROBLEM 3.7
An 1 1 -lb force P is applied to a shift lever. The moment of P about B is clockwise
and has a magnitude of 250 lb •
in. Determine the value ofa.
22 in.
SOLUTION
By definition
where
and
also
Then
or
or
and
MB =rmPsm.6
= a + (9Q°-tf>)
_i 8 in.
<p - tan" 19.9831'
22 in.
rf/fl =V(8in.)
2
+(22in.)
2
= 23.409 in.
250lb-in = (23.409in.)(lllb)
xsin(tf + 90° -19.9831°)
sin (« + 70.01 69°) = 0.97088
a + 70.0 169° = 76. 1391°
a+ 70.0169° = 103.861° a = 6.12° 33.8° <
159

PROBLEM 3.8
It is known that a vertical force of 200 lb is required to remove the nail
at C from the board. As the nail first starts moving, determine (a) the
moment about B of the force exerted on the nail, (b) the magnitude of
the force P that creates the same moment about B if a ~ .1 0°, (c) the
smallest force P that creates the same moment about B.
V—A
r 4 in.
n
SOLUTION
(a) We have MB =raB FN
(4 in.)(200 lb)
800 lb -in.
or MB =H00b-m.)<
A- m.
(/;) By definition MB ~rA/B Psin
6» = 10° + (180°~70°)
= 120°
Then 800 lb • in. = (18 in.) x Psin 1
20°
or P = 51.3 lb <
(c) For P to be minimum, it must be perpendicular to the line joining
Points A and B. Thus, P must be directed as shown.
Thus
or
or
A*W™.
cl = f
A IB
800 lb • in. = (18 in.)Pm
^i„=44.4 1b Pmitl
=44.41b^l20 ^
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160

PROBLEM 3.9
A winch puller AB is used to straighten a fence post. Knowing that the tension in cable BC is 1040 N and
length d is 1 ,90 m, determine the moment about D of the force exerted by the cable at C by resolving that
force into horizontal and vertical components applied (a) at Point C, (b) at Point E.
0.2 m
0.875 m
SOLUTION
(a) Slope of line
Then
EC =
0.875 m
1.90 m + 0.2 m 12
*abx - ,~ (Tab)
12
13
960 N
(1040 N) 0,1«y
and *W=-0040N)
= 400 N
Then
(b) We have
MD
- TABx (0.875 m)~TABy (0.2 m)
= (960 N)(0.875 m) - (400 N)(0.2 m)
= 760 N •
m
MD ~rm(y) + TABx (x)
= (960 N)(0) + (400 N)(I .90 m)
= 760 N •
m
or M./J
=760N-m
>
)^
or M7>
= 760N-m
v
)<«
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161

PROBLEM 3.10
It is known that a force with a moment of 960 N •
m about D is required to straighten the fence post CD. If
d- 2,80 m, determine the tension that must be developed in the cable of winch puller AB to create the
required moment about Point D.
0.875 m
0.2 j.i»
SOLUTION
*.
»
£^
*AB
'My
o. a? 5^i
z-acw OiZ^
Slope of line
Then
and
We have
EC =
0.875 m 7
7'
2.80 m + 0.2 m 24
24
My
r/ffl)>
25
1
25
T/)B
rAH
24 7
960N-m= —7^(0) +—7^(2.80*)
7^= 1224 N or r^=1224N ^
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162

PROBLEM 3.11
It is known that a force with a moment of 960 N • m about D is required to straighten the fence post CD. If the
capacity of winch puller AB is 2400 N, determine the minimum value of distance d to create the specified
moment about Point D.
fe
0.2 in
0.875 in
SOLUTION
o.mtn
czom
The minimum value of d can be found based on the equation relating the moment of the force TAB about D:
MD ={TABmK ),(d)
where
Now
MD =960N-m
(^flmax )y = TAIHmx sin & = (2400 N)sfo
. . 0.875m
sin #
960 N •
m = 2400 N
^(t/ + 0.20)
2
+(0.875)
2
m
0.875
(d)
+ 0.20)
2
+(0.875)
2
or ^+ 0.20)
2
+ (0.875)
2
= 2. ! 875d
or (J + 0.20)
2
+ (0.875)
2
= 4.7852rf
2
or 3 .7852</
2
- 0.40c/ - .8056 =
Using the quadratic equation, the minimum values of d are 0.51719 m and -.41151 m.
Since only the positive value applies here, d — 0.5 1 7 1 9 m
or d ~ 5 1 7 mm ^
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163

5.3 in.
PROBLEM 3.12
12.0 in.
2.33 in.
mmM1 >iii. i
The tailgate of a car is supported by the hydraulic lift BC. If the
lift exerts a 125-lb force directed along its centerline on the ball
and socket at B, determine the moment of the force about A.
SOLUTION
First note
Then
and
Now
where
Then
dcB == 7(12.0 in.)
2
= 12.224 lin.
+ (2.33 in.)
2
cos 9 =
12.0 in.
12.2241 in.
sin 9-
2.33 in.
12.2241 in.
*cb = FCB cos 9 -FCB $m9l
1251b
12.2241 in.
[(12.0 in.)i- (2.33 in.) j]
M.A
= VB/A
X ^CB
XBIA
= (15.3in.)i-- (12.0 in. + 2.33 in.) j
M
= (15.3 in.) i--(14.33 in.) j
ru-nimn* 1251b
«5»,a> im.
Z.& >N.
2.33 ttvi.
—(12.01 — 2.331)
12.2241 in.
(1393.87 lb in.)k
(116.156 lb -ft)k or M„ = 116.2 lb- ft *)<
164

20.5 in.
h
—
-
4.38 in
i
*Ti
7.62
1
t§|
1.7.2 hi.
PROBLEM 3.13
The tailgate of a car is supported by the hydraulic lift EC. If the lift
exerts a 125-lb force directed along its centerline on the ball and
socket at B, determine the moment of the force about A.
SOLUTION
First note
Then
dCB
:= 7(17.2 in.)
2
= 18.8123 in.
+ (7.62 in.)
2
cos -
17.2 in.
18.8123 in.
sin 6 -
7.62 in.
18.8123 in.
Z.O.'S «w.
and
Now
where
Then
fcd = (Ft:v*
cos 0) - (FCB sin 6>)j
=
S(,7' 2i"-)i + (7' 62i^
r^ = (20.5 in.)i- (4.38 in.)j
M, = [(20.5 in.)i - (4.38 in.)j] x t
.
1251b
(1 7.21 - 7.62j)
18.8 123 in.
n.z .«»».
4.?.e. >Ni.
XKvZ >KJ.
(1538.53 lb -in.)k
(128.2 lb -ft)k or M^ =128.2 lb- ft ^H
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165

120mm
65 mm
PROBLEM 3.14
A mechanic uses a piece of pipe AB as a lever when tightening an
alternator belt. When he pushes down at A, a force of 485 N is
exerted on the alternator at B. Determine the moment of that force
about bolt C if its line of action passes through O.
SOLUTION
We have Mc =rwc xFj,
Noting the direction of the moment of each force component about C is
clockwise.
Where
and
Mc =xFBy +yFBx
x - 1 20 mm - 65 mm = 55 mm
y - 72 mm + 90 mm - 1 62 mm
4 j,
F
65
lix
Fa
7(65)
2
+(72)
2
72
V(65)
2
+ (72)
3
-(485N) = 325N
.(485 N)- 360 N
iWc
= (55 mm)(360 N) + (1 62)(325 N)
= 72450 N- mm
= 72.450 N-m or Mc =72.5 N-m J) <
166

By definition:
Now
and
PROBLEM 3.15
Form the vector products B x C and B' * C, where B = B and use the results
obtained to prove the identity
sin a cos/? = -sin (a + P) + -sin (a - p).
SOLUTION
N(>te: B = £(cos/?i + sin/?j)
B' = 5(cos/?i-sin/?j)
C - C(cos ai + sin aj)
|BxC| = flCsin(a-jff)
|B'xC| = 5Csin(flf + y?)
B xC = Z?(cos /?i + sin y9j) x C(cos tfi + sin orj)
= BC(cos /?sin « - sin /?cos «)k
B'x C - /?(cos /?i - sin /?j)x C(cos ai -f sin #j)
- £C(cos yffsin ar + sin /?cos ar)k
Equating the magnitudes of BxC from Equations (I ) and (3) yields:
BCs'm(a -p)~ BC(cos ps'm a - sin pcos a)
Similarly, equating the magnitudes of B'xC from Equations (2) and (4) yields;
BCsm(a + p) = BC(cos ps'm a + sln pcos a)
Adding Equations (5) and (6) gives:
sin(a - p) + s'm(a + p) - 2cos /?sin a
0,)
(2)
(3)
(4)
(5)
(6)
or sin «cos /? - -sin(ar + /?) + -sin(flf - /?) ^
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167

PROBLEM 3.16
A line passes through the Points (20 m, 16 m) and (-1 m, ~4 m). Determine the perpendicular distance rffrom
the tine to the origin O of the system of coordinates.
SOLUTION
dAB = V[20m- (-1 m)]
2
+ [1 6 m - (-4 m)f
- 29 m
Assume that a force F, or magnitude F(N), acts at Point A and is
directed ixomA to B.
Then,,
Where
By definition
Where
Then
¥~FX m
'All
*B~ rA
d.41!
= -
—<21i + 20j)
29
V
MQ = r
A xF = dF
r,=-(lm)i-(4m)j
F
E> (Zom, K-rti^
M =[-(-1 m)i-(4 m)j]x—-[(21 m)i + (20 m)j]
29 m
a -(20)k + (84)k]
~F|k N-m
29
-*x
Finally
64
29
F = </(F)
. 64
rf- — m
29
</ = 2.21m ^
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168

PROBLEM 3.17
The vectors P and Q are two adjacent sides of a parallelogram. Determine the area of the parallelogram when
(tf)P--7i + 3j-3kandQ = 2i + 2j + 5k,(6)P = 6i-5j-2kandQ = -2i + 5j~k.
SOLUTION
(a) We have
where
Then
(b) We have
where
v4 = |PxQf
P = -7i + 3j-3k
Q = 2i + 2j + 5k
PxQ
k
-3
> J
-7 3
2 2 5
= [(15 + 6)i + (-6 + 35)j + (-14-6)k]
= (21)1 + (29)j(-20)k
^ = V(20)
2
+(29)
2
+(-20)
2
A = PxQ
P = 6i~5j-2k
Q = ~2i + 5in.j~lk
or 4 = 41.0 <4
Then PxQ
i fc
i
6 -5
-2 5
= [(5 + 1 0)i + (4 + 6)j + (30-1 0)k]
«(15)i + (10)j + (20)k
A=j(i5)-+(oy + (2oy or 4 = 26.9 ^
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169

PROBLEM 3.18
A plane contains the vectors A and B. Determine the unit vector normal to the plane when A and B are equal
to, respectively, (a) i + 2j - 5k and 4i - 7j - 5k, (b) 3i - 3j + 2k and -2i + 6j - 4k.
SOLUTION
{a) We have
where
Then
and
(b) We have
where
Then
and
AxB
A
B
AxB
|AxB|
:li + 2j-5k
= 4i - 7 j
- 5k
i
J *
1 +2 -5
4 -7 -5
(-1 ~ 35)i + (20 + 5)j + (-7 - 8)k
15(31 -1J -Ik)
|AxB|
X--
X-
A
B
AxB
15V(-3)
2
+H)2
+(-l)
2
-I5>/ri
15(-3f-lJ-lk)
or X
15VH
AxB
(-3i-j~k) 4
|AxB|
3i-3j + 2k
:-2i + 6j-4k
I
J k
3-3 2
-2 6 -4
(12-12)i + (-4 + l2)j + (18-6)k
(8j + 12k)
|AxB| 4^(2)
2
+(37 = 4^13
4(2j + 3k)
4^13
or X,
V^
(2j + 3k) <
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170

PROBLEM 3.19
Determine the moment about the origin O of the force F = 4i + 5j - 3 k that acts at a Point A. Assume that the
positioji vector ofA is {a) r - 2i - 3j + 4k, (£) r - 2i + 2.5j - 1 .5k, (c) r - 2i + 5j + 6k.
SOLUTION
(a) M,
(h) M,
' J k
2-3 4
4 5 -3
(9-20)i + (16 + 6)j + (I0 + 12)k
i j k
2 2.5 -1.5
4 5 -:s
Ma = -lli + 22j + 22k ^
(-7.5 + 7.5)i + (-6 + 6)j + (10-1 0)k M, «
(<0 M,
' J
2 5
4 5
(-1 5 - 30)i + (24 + 6)j + (10- 20)k M, -45i + 30j-10k A
Note: The answer to Part b could have been anticipated since the elements ofthe last, two rows of the
determinant are proportional.
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171

PROBLEM 3.20
Determine the moment about the origin of the force F == -2i + 3j + 5k that acts at a Point A. Assume that the
position vector ofA is (a) r == i + j + k,(6)r == 2i + 3j- 5k, (c)r == -4i + 6j + 10k.
SOLUTION
i j k
(a) M(}
= 1 1 1
-2 3 5
= (5-3)i + (-2-5)j + (3 + 2)k M = 2i-7j + 5k A
i j k
(h) M - 2 3 -f
-2 3 5
= (15 + 15)i + (10-10)j + (6 + 6)k Mo =30i + I2k A
1 J k
(c) Ma
- -4 6 10
-2 3 5
= (30 - 30)i + (-20 + 20)j + (-.12 + 1 2)k M = <
Note: The answer to Part c could have been anticipated since the elements of the last two rows of the
determinant are proportional.
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172

-VI
200 N
r
PROBLEM 3.21
A 200-N force is applied as shown to the bracket ABC. Determine
the moment of the force about A.
SOLUTION
We have
where
Then
M, =
%.i x *c
rCIA
= (0.06 m)i + (0.075 m)j
*c = ~(200 N)cos 30°j + (200 N)sin 30°k
i
j &
MA
-.
= 200 0.06 0.075
-cos 30° sin 30°
= 200[(0.075sin 30°)i - (0.06sin 30°)j ~ (0.06 cos 30°)k]
or M,, = (7.50 N m)i- (6.00 N-m)j-(l 0.39 N-m)k <4
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173

'/
PROBLEM 3.22
£
^-^
Before the trunk of a large tree is felled, cables AB and BC are
attached as shown. Knowing that the tensions in cables AB and BC
are 555 N and 660 N, respectively, determine the moment about O
. .'vjB 4.25 in
of the resultant force exerted on the tree by the cables at B.
/ ,.-"r
^J""
0.75 in/
-^1 "I
.T
SOLUTION
We have
where
and
M,
1
ff/O
r
l3/Q
X F#
(7m)j
Tdff + Tec
'yf/f ~ ,V
BA' AB
l
J!C
-(0.75m)i-(7m)J + (6m)k
(.75)
2
+(7)
2
+(6)
2
m
*-BC*BC
(555 N)
(4.25m)i-(7m)j + (lm)k
^(4.25)
2
+(7)
2
+(l)
2
m
-(660 N)
Fi?
= [-(45.00 N)i - (420.0 N)j -f- (360.0 N)k]
+[(340.0 N)i - (560.0 N)j + (80.00 M)k]
= (295.0 N)i - (980.0 N)j + (440.0 N)k
M,
i
J k
7
295 980 440
Nm
(3080 N •
m)i - (2070 N • m)k or M,}
= (3080 N-m)i- (2070 N-m)k <
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174

f
i in
PROBLEM 3.23
The 6-m boom ^5 has a fixed end /*. A steel cable is stretched from
the free end B of the boom to a Point C located on the vertical wall.
If the tension in the cable is 2.5 kN, determine the moment about A of
the force exerted by the cable at B.
SOLUTION
First note
Then
We have
where
Then
^c-VK>)
2
+(2.4)
2
+(-4)
:
= 7.6 m
V=-?4^(-« + 2.4j-4k)
7.6
^A ~ r/i//l
X TBC
l
B/A (6 m)i
M;4 ^(6m.)ix^4r^-(~6» + 2.4j-4k)
7.6
or M//== (7.89 kN-m)j + (4.74 kN-m)k <
175

3{ji)i.. C- 48itii
:'6-iftl
PROBLEM 3.24
A wooden board AB, which is used as a temporary prop to support
a small roof, exerts at Point A of the roof a 57-lb force directed
along BA. Determine the moment about C of that force.
SOLUTION
We have
where
and
rA/c
= (48 in.)i - (6 in.)j + (36 in.)k
<
BA
— xMpM
-(5in.)i + (90in.)j-(30in.)k
V(5)
2
+(90)
2
+ (30)
2
in.
= -(31b)i + (541b)j-(181b)k
(57 lb)
Mc lb -in.
i
J k
48 6 36
3 54 18
-(1 836 lb •
in.)i + (756 lb • in.)j + (2574 lb • in.)
or Mc = -(1 53 .0 lb • ft)i + (63.0 lb •
ft)j + (2 1 5 lb •
ft)k A
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176

/.)
«'
0.6 id
0.6 m "X
PROBLEM 3.25
The ramp ABCD is supported by cables at corners C and D.
The tension in each of the cables is 810 N. Determine the
moment about A of the force exerted by (a) the cable at D,
(b) the cable at C.
SOLUTION
(a) We have M^riy
,xTM
where **/•/,(= (2-3 m)j
Tw: = *"DE^f.)E
-
(0.6m)l + (3.3in)J-(3m)k
V(0.6)
2
+(3.3)
2
+(3)
2
m
= (1 08 N)i + (594 N)j - (540 N)k
i
J k
M,,= 2.3
108 594 -540
Nm
= -(1 242 N -
m)i - (248.4 N •
m)k
or M/i
=-(1242N-m)i-(248N-m)k <
(b) We have ^A=rG//l
xTCG
where r(
,,=(2.7m)i + (2.3m)j
T — 1 T
*CO ~~
*"CG'CG
= -(.6m)i + (3.3m)j-(3m)k
(810N)
V(.6)
2
+(3.3)
2
+(3)
2
m
= -(108 N)i + (594 N)j - (540 N)k
i
J k
MA
= 2.7 2.3
-108 594 -540
N-m
= -(1 242 N •
m)i + (1458 N •
m)j + (1 852 N •
m)k
or MA = -(1242 N • m)i + (1 458 N • m)j + (1 852 N •
m)k A
177

PROBLEM 3.26
A smaJI boat hangs from two davits, one of which is shown in the
figure. The tension in line ABAD is 82 lb. Determine the moment
about C of the resultant force R,, exerted, on the davit at A.
SOLUTION
We have
where
and
Thus
Also
Using Eq. (3.21):
R.=2R(/)
. + F,D
,„~-(82 1b)j
AD
AD I
AD —= (82 lb)
AD
6i~7.75j-3k
10.25
F,/3
=(48lb)i-(62 1b)j-(24 1b)k
RA = 2¥AB + FAD = (48 lb)i - (226 lb)j - (24 lb)k
l
A/C
Mc
(7.75ft)j + (3ft)k
1
J
7.75
48 -226
k
3
-24
(492 lb • ft)i + (1 44 lb •
ft)j - (372 lb • ft)k
Mc = (492 lb •
ft)i + (1 44.0 lb ft)j - (372 lb • ft)k <
178

PROBLEM 3.27
In Problem 3.22, determine the perpendicular distance from
Point O to cable AB.
PROBLEM 3.22 Before the trunk of a large tree is felled,
cables AB and BC are attached as shown. Knowing that the
tensions in cables AB and BC are 555 N and 660 N,
respectively, determine the moment about O of the resultant
force exerted on the tree by the cables at B.
SOLUTION
We have
where
Now
and
IM TBA d
d = perpendicular distance from O to line AB.
M, r
B/0 X *BA
rBIO ={lm)
BA' AB
(0.75m)i-(7m)j + (6m)k
*BA ~~
'"BA
1
AB
M,
and
or
(555 N)
'(0.75)
2
+(7)
2
+(6)
2
m
-(45.0 N)i - (420 N)j + (360 N)k
i j k
7 N-m
-45 -420 360
(2520.0 N •
m)i + (315.00 N •
m)k
|M 1
= V(252o-°)
2
+ (31 5.00)
;
= 2539.6 N-m
2539.6 N-m = (555 N)rf
d = 4.5759 m or d = 4.58 m <
179

PROBLEM 3,28
Point O to cable BC.
PROBLEM 3.22 Before the trunk of a large tree is felled,
cables AB and BC are attached as shown. Knowing that the
tensions in cables AB and BC are 555 N and 660 N,
respectively, determine the moment about O of the resultant
force exerted on the tree by the cables at.#.
SOLUTION
We have
where
[M, Tscd
d — perpendicular distance from O to line BC.
M,
l
B/0
rB/0 X ^liC
7mj
P — 1 T
'BC ™ *"BC* HC
(4.25m)i-(7m)j + (lm)k
(660 N)
M,
and
V(4.25)
2
+(7)
2
+(l)
2
m
(340 N)i - (560 N)j + (80 N)k
I
J k
7
340 -560 80
(560 N • m)i - (2380 N • m)k
|M !
= 7(560)
2
+(2380)
= 2445.0 N-m
2445.0 N-m = (660 N)rf
d = 3.7045 m or d~ 3.70 m
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, inc. All rights reserved. No part of this Manual may be displayed,
180

A
."(i in.
•IS in.
Cj-iii.
.90 in,:
()(> in.
~~*S B ~~^~£
PROBLEM 3.29
In Problem 3.24, determine the perpendicular distance from Point D
to a line drawn through Points A and B.
PROBLEM 3.24 A wooden board AB, which is used as a temporary
prop to support a small roof, exerts at Point A of the roof a 57-lb
force directed along BA. Determine the moment about C of that
force.
SOLUTION
We have
where
and
IMJ^rf
d — perpendicular distance from D to line AB.
M.l}
^rAlii
x¥]iA
r,/D =-(6in.)j + (36in.)k
^liA
~ ^BA^BA
(-(5in.)i + (90in.)j-(30 in.)k)
V(5)
2
+(90)
2
+ (30)
2
in.
-(31b)i + (541b)j-(181b)k
(57 lb)
M,;)
I
J k
-6 36 lb -in.
-3 54 -18
-(1 836.00 lb • in.)i - (1 08.000 lb •
in.)j - (1 8.0000 lb •
in.)k
IM 836.00)
2
+(108.000)
2
+ (18.0000)'
= 1839.26 lb -in.
1839.26 lb -in =(57 lb)e/
</ = 32.268 in. or </ = 32.3in. <
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181

(if) in.
B^-<
PROBLEM 3.30
In Problem 3.24, determine the perpendicular distance from Point C to a
line drawn through Points A and B.
PROBLEM 3.24 A wooden board AB, which is used as a temporary
prop to support a small roof, exerts at Point A of the roof a 57-3 b force
directed along BA. Determine the moment about C of that force.
SOLUTION 4
We have
where
Mc ~FBA d
d - perpendicular distance from C to line AB.
3fc.. .
M.c ^rA/c x¥BA ^%f M / >l
D ^-.
rAIC
- (48 in.)i - (6 in,)j + (36 in.)k ^^ //S^
fylA
~ ^BA^IIA
f*
_ (~(5 in.)i + (90 in.)j - (30 in.)k) fe*
7(5)
2
+(90)
2
+(30)
2
in.
= ™(3 lb)i + (54 lb)j - (1 8 lb)k
i
J k
Mc = 48 -6 36
-3 54 -18
lb -in.
and
= -(1 836.001b -i
|M
n.)i - (756.00 lb • in.)j + (2574.0 lb •
in.)k
2
c |
= V(l 836.00)
2
+ (756.00)
2
+ (2574.0)
-3250.8 lb -in.
3250.8 lb in. = 57 lb
d- 57.032 in. or d = 57.0 in. ^
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182

0.6 m
PROBLEM 3.31
Point A to portion DE of cable DEF.
PROBLEM 3.25 The ramp ABCD is supported by cables at
comers C and D, The tension in each of the cables is 810 N.
Determine the momentabout A of the force exerted by (a) the
cable at D, (b) the cable at C. .
SOLUTION
We have
where
IM TDEd
and
d
—
perpendicular distance from A to line DE,
M
r„
HIA (2.3 m)j
'DE ~ A"DE*-DE
(0.6m)i + (3.3m)j-(3m)k
(8I0N)
M
V(0.6)
2
+ (3.3)
2
+(3)
2
m
(108 N)i + (594 N)j - (540 N)k
i
J b
2.3 N-m
108 594 540
~(1 242.00 N • m)i - (248.00 N •
m)k
|MJ =-
N
/(1242.00)
2
+(248.00)
:
= 1266.52 N-m
1266.52 Nm= (8 10 N)d
d = 1 .56360 m
O'fcn,
or d = 1.564 m
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183

PROBLEM 3.32
Point A to a line drawn through Points C and G.
PROBLEM 3.25 The ramp ABCD is supported by cables at
corners C and D. The tension in each of the cables is 810 N.
Determine the moment about A of the force exerted by (a) the
cable at D, (h) the cable at C.
0.6 m
SOLUTION
We have
where d - perpendicular distance from A to line CG.
rG/A
X Ice
M
IV
u/A
l
CG
M
(810 N)
(2.7m)i + (2.3m)j
-(0.6.m)i + (3.3m)j-(3 m)k
A/(0.6)
2
+(3.3)
2
+ (3)
2
m
-(108N)i + (594N)j-(540N)k
i J k
2.7 2.3 N-m
-108 594 -540
-(1242.00 N • m)i + (1458.00 N •
m)j +(1 852.00 N m)k
and
= 2664.3 N-m
2664.3 N-m = (8 ION)d
d = 3.2893 m or d = 3.29 m 4
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184

PROBLEM 3.33
.In Problem 3.26, determine the perpendicular distance from
Point C to portion AD of the line ABAD.
PROBLEM 3.26 A small boat hangs from two davits, one of
which is shown in the figure. The tension in line ABAD is
82 lb. Determine the moment about C of the resultant force
R4 exerted on the davit at A.
SOLUTION
First compute the moment about C of the force FDA exerted by the line on D:
From Problem 3.26:
= -(48 lb)i + (62 lb)j + (24 lb)k
Mc =rm; xF^
= +(6 ft)i x[-(48 lb)i + (62 lb)j + (24 lb)k]
= -(1441b-ft)j + (372 1b-ft)k
Mc =V044>
2
+(372)
2
= 398.90 lb •
ft
Then Mc
= VMd
Since F,w = 82 1b
398.90 lb -ft
82 1b
d = 4.86 ft <
yon are using it without permission.
185

*-* PROBLEM 3.34
Determine the value of a that minimizes the
i
16 ft perpendicular distance from Point C to a section
of pipeline that passes through Points A and B.
SOLUTION
Assuming a force F acts along AB,
M.c MrA/c xF^F(d)
Where d - perpendicular distance from C to line AB
%ABF
(24ft)i + (24ft)j-(28)k
F = W
7(24)
2
+(24)
2
+(18)
2
ft
•F
(6)i + (6)j-(7)k
AK;
= (3 ft)i - (1 ft)j - {a - 1 ft)k
i
J k
3 -10 10a
6 6-7
Mc
.[(10 + 6fl)i + (81-6.ai)j + 78k]
11
Since
12!
^/c xr or r^xF2
|
= W
(1 + 6a)
2
+(81- 6a)
1
+ (78)
2
= d*
d / j2
Setting -j-{d )- to find a to minimize c/
1
[2(6)(l + 6a) + 2(-6)(8 1 - 6a)] =
Solving
121
a = 5.92 ft or o - 5.92 ft ^
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. Afo par/ o//Aw Minna/ »m>' fc displayed
186

PROBLEM 3.35
Given the vectors P =3i -
P S, and Q • S.
-J + 2k,Q == 4i + 5j-3k, and S = -2i + 3j - k, compute the scalar products P •
Q,
SOLUTION
P-Q = (31-1j + 2k)-(4i-5J-3k)
= (3)(4) + (-l)(~5) + (2)(-3)
= 1 or P Q = I A
P • S = (3i - lj + 2k) •
(-2i + 3j - Ik)
= (3)(-2) + H)(3) + (2)H)
= -U. or P-S = -U <
Q-S-(4i-5j~3k)-(-2i + 3j-l.k)
= (4K-2) + (5)(3) + (-3X-l)
= 10 or Q-S = 10 <
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187

PROBLEM 3.36
Form the scalar products B • C and B' •
C, where B = B', and use the
results obtained to prove the identity
cos a cos P ~ —cos (a + /?) +—cos (a - ft).
SOLUTION
y By definition
where
(1)
B-C = £Ccos(a~/?)
B = /?[(cos/?)i + (sin/?)j]
C = C[(cos a) + (sin a)j]
(B cos /?)(Ccos a) + (B sin /?)(Csin a) - BCcos(a~ (3)
or cos /?cos a+ sin /?sin a = cos(a - j3)
By definition B' • C - BCcos (or + /?)
where B' = [(cos fi)i ~ (sin /?)j]
(B cos p) (C cos or) + (~B sin /?)(C sin or) = BCcos (a + /?)
or cos /?cos « - sin ft sin a = cos {a + /?)
Adding Equations (1) and (2),
2 cos /?cos a = cos (a-fi) + cos (a + /?)
or cosacosfi = ~-cos(a + fi) + ~cos(a~ ft) A
(2)
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188

PROBLEM 3,37
Section AB of a pipeline lies in the yz plane and forms an angle
of 37° with the z axis. Branch lines CD and EF join AB as
shown. Determine the angle formed by pipes AB and CD.
SOLUTION
First note AB = AB{$m 37°j - cos 37°k)
CD = CD(-cos 40° cos 55°j + sin 40°j- cos 40° sin 55°k)
fc.b
Now
or
or
AB- CD = (AB)(CD) cos
AB(sm 37°j - cos 37°k) • CD(-cos 40° cos 55°i + sin 40°j - cos 40°sin 55°k)
~(AB)(CD) cos
cose? = (sin 37°)(sin 40°) + (~cos 37°)(--cos 4()°sin 55°)
= 0.88799
or (9 = 27.4° A
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189

PROBLEM 3.38
Section AB of a pipeline lies in the yz plane and forms an angle
of 37° with the z axis. Branch lines CD and EF join AB as
shown. Determine the angle formed by pipes AB and EF.
SOLUTION
First note AB = /f£?(sm37 j-cos37 k)
__.
= £F(cos 32° cos 45°i + sin. 32°j - cos 32° sin 45°k)
1
£
F
l/^(€t *W A
Now AB-EF--= (AB)(EF)cosO
or AB(sm
:
17°j - cos 37°k) - £F(cos 32° cos 45°j + sin 32°j - cos 32° sin 45°k)
= (AB)(EF)cos0
or cos = (sin 37°)(s.in 32°) + (-cos 37°)(-cos 32° sin 45°)
= 0.79782
or (9 = 37.1° ^
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1 90

PROBLEM 3.39
Consider the volleyball net shown.
Determine the angle formed by guy
wires AB and AC,
SOLUTION
First note
and
By definition
or
or
AB - V(-6.5)
2
+ (-8)
2
+ (2)
2
= 1 0.5 ft
AC^yj(0f+(~S)2
+ (6)
2
=]0 ft
AB = -(6.5 ft)i - (8 ft)j + (2 ft)k
7c = -(8ft)j + (6ft)k
AB-AC = (ABXAC)cos0
(-6,51 - 8j + 2k) • (-8j + 6k) = (1 0.5)0 0) cos -
(-6.5)(0) + (-8)(-8) + (2)(6) - 1 05 cos 6
cos = 0.72381 or = 43.6° <
191

sit
PROBLEM 3.40
Consider the volleyball net shown.
Determine the angle formed by guy
wires AC and AD.
SOLUTION
First note
and
By definition
or
or
^C = V(0)
2
+(~8)
2
+(6)
2
-10 ft
AD = J(4)
2
+(-&)
2
+ (l?
= 9 ft
IC = ~(8ft)j + (6ft)k
75 = (4ft)j-(8ft)j + (lft)k
AC AD = (AC)(AD)cos0
(-8j + 6k) •
(4i - 8j + k) = (1 0)(9) cos t
(0)(4) + (-8)(-8) + (6)(1) = 90cos 6
cos0 = 0.77778 or # = 38.9° <
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192

1.2 in
2.-1
PROBLEM 3.41
Knowing that the tension in cable AC is 1260 N, determine
(a) the angle between cable AC and the boom AB, (b) the
projection on AB ofthe force exerted by cable AC at Points.
-f
2.6 in
2.4 m
SOLUTION
(a) First note
and
By definition
or
or
or
(b) We have
AC~-=^j(~2A)
2
+ (0.8)
2
+(1.2)
2
= 2.8m
AB = yj(-2 A)
2
+(-].&? +(Q)
2
= 3.0m
^C = -(2.4 m)i + (0.8 m)j + (1 .2 m)k
I/? = -(2.4m)i-(K8m)j
AC -AB = (AC){AB) cos
(-2.41 + 0.8j + 1 .2k) • (-2.4i - 1 .8j) = (2.8)(30) x cos
(-24X-2.4) + (0.8X-1 .8) + (1 .2X0) - 8.4cos
cos# = 0.51429
= 7^c cos<9
= (1260N)(0.51429)
or = 59.O
C
or C^cU=648N
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193

2.4 m
PROBLEM 3.42
Knowing that the tension in cable AD is 405 N, determine (a) the
angle between cable AD and the boom AB, (b) the projection on
AB of the force exerted by cable AD at Point A.
SOLUTION
{a) First note AD = >/(-2.4)
2
+ (i ,2)
2
+ (-2.4)
2
= 3.6m
AB = V(-2.4)
2
+ (-1 .8)
2
+ (0)
2
-3.0m
and AD = -(2.4 m)i + (1 .2 m)j - (2.4 m)k
AB = -(2.4m)i~(1.8m)j
By definition, ADAB = {AD)(AB)cos0
(-2.4i + 1 .2j - 2.4k) (-2.4i - .1 .8j) = (3.6)(3.0)cos#
(-2.4)(-2.4) 4- (1 .2)(-I.8) + (-2.4X0) = 10.8 cos
cos $ — —
3
^^70.5° <
(b) (* AD'AB ~ *AD '
^Ali
= TAD cos&
= (405N)^ Pad)ab = 135.0 N <
PROPRIETARY MATERIAL. © 20 ]0 The McGraw-Hill Companies, Inc. AH rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or. by any means, without the prior written permission of the publisher, or used beyond the limited
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194

PROBLEM 3.43
Slider P can move along rod OA. An elastic cord PC is
attached to the slider and to the vertical member BC. Knowing
that the distance from O to P is 6 in. and that the tension in the
cord is 3 lb, determine (a) the angle between the elastic cord
and the rod OA, (b) the projection on OA of the force exerted
by cord PC at Point P.
SOLUTION
First note
Then
^= >/(12)
2
+(12)
2
+(-6)
2
=18in.
» OA 1
,
= i(2i + 2j~k)
Now OP = 6 in, => OP = -(04)
The coordinates ofPoint P are (4 in., 4 in., -2 in.)
PC = (5 in.)i + (1 .1 in..)j + (14 in.)k
PC = 7(5)
2
+ (1 1)
2
+ (14)
2
= V342 in.
~PCXOA ^(PC)cosO
so that
and
(a) We have
or
or
(5i + llj + 14k)--(2i + 2(i-k) = 7342cos^
cos
]
3V342
0.32444
[(5)(2) + (ll)(2) + (14)(~l)j
(b) We have
= {T
pc'K]>c )-XOA
PC
ft
pc OA
— Tpr COS &
= (3 1b)(0.32444)
or = 71.1° ^
or (Tpc)oa =0-973 lb ^
=^^S~S5£=SS£=f£S
195

PROBLEM 3.44
Slider P can move along rod OA. An elastic cord PC is
attached to the slider and to the vertical member BC.
Determine the distance from O to P for which cord PC and
rod OA are perpendicular.
SOLUTION
First note
Then
(M^/(12)
2
+(12)
2
+(-6)- =18 in.
OA 1
XCH
~ —(12i + 12j-6k)
OA
OA 18
(2i + 2j-k)
Let the coordinates of Point P be (x in., j> in., z in.). Then
PC = [(9 - x)in.]i + (15- y)in.]j + [(12- z)in.]k
Also,
and
OP = rfo^o, = -^(21 + 2J-k)
OP ~ (x in.)i + (>' in.)j + (z in.)k
2 2
dOP
The requirement that CM and PC" be perpendicular implies that
^PC-0
or -(2j + 2j-k)-[(9-.v)i + (15-y)j + (12-2)k] =
or (2)|9--^>| + (2) 'l5-|rfw |
+ H) 12 -rf0P
or Jnp = 12.00 in. ^
W
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196

PROBLEM 3.45
Determine the volume of the parallelepiped of Fig. 3.25 when
(a) P = 4i - 3j + 2k, Q = -2i - 5j + k, and S = 7i + j - k,
(/;) P = 5i - j + 6k, Q = 2i + 3j + k5 and S = -3i - 2j + 4k.
SOLUTION
Volume of a parallelepiped is found using the mixed triple product.
O) VoI = P-(QxS)
4-3 2
-2 -5 1 in.
3
7 1 -1
(20-21.-4 + 70 + 6-4)
67
or Volume =67.0 4
(b) Vol=P-(QxS)
5 -1 6
2 3 I in.
3
-3 -2 4
(60 + 3-24 + 54 + 8+10)
111
or Volunie = 111.0 A
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197

PROBLEM 3.46
Given the vectors P - 4i - 2j + 3k,Q = 2i + 4j - 5k, and S = SJ - j + 2k, determine the value of Sx for which
the three vectors are coplanar.
SOLUTION
If P, Q, and S are coplanar, then P must be perpendicular to (Qx S).
P-(QxS) = ()
(or, the volume of a parallelepiped defined by P, Q, and S is zero).
4-2 3
Then 2 4 -5-0
Sx
-1 2
or 32 + lO.S: -6-20 + 8-125;. =0 S =7 <
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198

0.1 1 in...
PROBLEM 3.47
The 0.61x].00-m lid ABCD of a storage 'bin is hinged
along side AB and is held open by looping cord DEC over a
frictionless hook at E. If the tension in the cord is 66 N,
determine the moment about each of the coordinate axes of
the force exerted by the cord at D.
SOLUTION
First note
Then
and
Now
where
Then
z = Vi°-61)
2
-(0.Jl)
2
0.60 m
0.11m
dm = V(0.3)
2
+(0.6)
2
+(-0.6r
= 0.9 m
1
l)E
66 N
0.9
(0.3i + 0.6j-0.6k)
= 22[(lN)i + (2N)j-(2N)k]
MA = rD/A xTDf:
r/)//(
= (0.ll m)j + (0.60m)k
i
J k
M/(
= 220 0.11 0.60
1 2 -2
= 22[(-0.22 ~ 1 .20)i + 0.60j - 0. 1 1
k
]
= - (3 1 .24 N • m)i + (1 3.20 N •
m)j - (2.42 N •
m)k
-3 1 .2 N • m, Mv
= 1 3.20 N •
m, A/2
= -2.42 N -
m A
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199

PROBLEM 3.48
The 0.61xl.00-m lid ABCD of a storage bin is hinged
along side AB and is held open by looping cord DEC over a
frictionless hook at E. If the tension in the cord is 66 N,
determine the moment about each of the coordinate axes of
the force exerted by the cord at C.
SOLUTION
First note
Then
and
Now
where
Then
x6y-(o.uy
0.60 m
= l.lm
l
CE
66 N
1.1
(-0.7i + 0.6j-0.6k)
= 6[-(7N)i + (6N)j-(6N)k]
MA ^vm xE
r£M = (0.3m)I + (0.71m)J
i
J
k
MA -6 0.3 0.71
-7 6 -6
= 6[-4.26i + 1 .8j + (1 .8 + 4.97)k]
= - (25.56 N • m)i + (1 0.80 N • m)j + (40.62 N • m)k
Mx
- -25.6 N •
m, My
= 1 0.80 N • m, Afz
= 40.6 N • m -4
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200

PROBLEM 3,49
To lift a heavy crate, a man uses a block and tackle attached to the
bottom of an I-beam at hook B. Knowing that the moments about the y
and the z axes of the force exerted at B by portion AB of the rope are,
respectively, 120 N • m and -460 N •
m, determine the distance a.
SOLUTION
First note
Now
where
Then
Thus
^= (2.2m)i-(3.2 m)j-(am)k
™*D ~ VAID X '«/j
'AID (2.2m}i + (1.6m)j
T
T,
BA
I! A
dISA
M,
T,
BA
dBA
(2.2i-3.2j-ak)(N)
i
J k
2.2 1.6
2.2 -3.2 -a
T,
BA
dt .
{- 1 .6a + 22a + [(2.2)(~3 .2) - (1 .6)(2.2)]k}
Mv =2.2-^-a
BA
Then forming the ratio
M,
it
M.
-I0.56-
7*4
d
(N •
m)
(N •
m)
HA
120 N-m 2-2 7Z"(N-m)
dB,t
-460 N-m -10.56-^- (N-m)
or a = 1 .252 m ^j
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201

PROBLEM 3.50
To lift a heavy crate, a man uses a block and tackle attached to the
bottom of an I-beam at hook B. Knowing that the man applies a 195-N
force to end A of the rope and that the moment of that force about they
axis is 132 N • m, determine the distance a.
SOLUTION
First note
and
Now
where
Then
Substituting for My and dt
HA
dBA
- J(22f+(-3.2)
2
+ (-af
= Vl5.08 + «
2
m
195 N
Ts
,=^^-!
-(2.2i-3.2j-ok)
My
~ -(rA/D
xTw)
Vf/0
M,
(2.2m)i + (1.6m)j
195
195
d
1
2.2 1.6
2.2 -3.2 -a
(2.2a) (N • m)
HA
132 N-m
195
JJm+a*
(2.2a)
or 0.30769^1 5.08 + a* - a
Squaring both sides of the equation
0.094675(15.08 + a
2
) = a
2
or a = 1.256 m A
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202

PROBLEM 3.51
A small boat hangs from two davits, one of which is shown in
the figure. It is known that the moment about the z axis of the
resultant force R t
exerted on the davit at A must not exceed
279 lb • ft in absolute value. Determine the largest allowable
tension in line ABAD when x ~ 6 ft.
SOLUTION
First note R 21^ +TAD
Also note that only TAD will contribute to the moment about the z axis.
Now
Then,
Now
where
Then for 71.
= 10.25 ft
'AD
M..
y
AIC
279
T
AD
AD
r
(6i-7.75j-3k)
10.25
(7.75ft)j + (3ft)k
7
10.25
1
7,75 3
6 -7.75 -3
~^H-(1)(7.75)(6)|
10.25'
A ;I
or 7" =61.5 lb <
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203

PROBLEM 3.52
For the davit of Problem 3.51, determine the largest allowable
distance x when the tension in line ABAD is 60 lb.
SOLUTION
From the solution of Problem 3.51, TAD is now
l
AI)
AD
AD
60 lb
yjX
2
+(-lJS)
2
+(-lf
(xi-7.75j-3k)
Then M = k • (rA/c xT,(0 ) becomes
279
279.
60
V*
2
+(-7.75)
2
+(-3)'
60
1
7.75 3
jc -7.75 -3
six
2
+69.0625
(D(7.75)(x)
279>/x
2
+ 69^0625 = 465.x
0.6Vjc
2
+ 69.0625 =x
Squaring both sides: 0.36x
2
+ 24.8625 = r
x
1
=38.848 x = 6.23 ft 4
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204

PROBLEM 3,53
To loosen a frozen valve, a force F of magnitude 70 lb is
applied to the handle of the valve. Knowing that 6 = 25°,
Mx - -6.1 lb • ft, and Mz
~ -43 lb • ft, determine (j) and d.
SOLUTION
We have
where
Pol-
and
From Equation (3)
From Equation (1)
rivi
(
VAIO
'
F
F
ML:
M.
My
M.
^ = cos
-(4in.)i + (llin.)j-(rf)k
F(cos #eos fi - sin $ + cos #sin <J)k)
70 1b, 9 = 25°
-
(70 lb)[(0.9063 1 cos 0)i - 0.42262j + (0.9063 1 sin 0)k]
i
j k
(701b) -4 11 -J in.
-0.90631 cos -0.42262 0.90631 sin
(70 lb)[(9.9694sin - 0.42262^)1 + (-0.9063 Wcos + 3.6252sin <p) j
+ (1.69048 -9.9694 cos 0)k] in.
(70 lb)(9.9694sin - 0.42262d)m. = -(61 lb •
ft)(I2 in./ft) (1)
(70 lb)(-0.9063 Irfcos (j) + 3.6252 sin <f>)
in. (2)
(70 lb)(l .69048 - 9.9694cos ^ in. - -43 lb •
ft(l 2 in./ft) (3)
U 634.33
697.86
24.636'
d
1022.90
29.583
=
34.577 in.
or (p = 24.6° <
or c/ = 34.6in. ^
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205

i I in.
PROBLEM 3.54
When a force F is applied to the handle of the valve shown, its
moments about the x and z axes are, respectively, Mx ~ -11 lb • ft
and M.z
- -81 lb •
ft. For d~21 in., determine the moment My of
F about the y axis.
SOLUTION
We have
Where
and
ZM : rw xF = M(
r^=-(4in.)i + (lim.)j-(27in.)k
F = F(cos 0cos <p - sin. $ + cos sin flk)
i J k
-4 1 .1 -27
cos 0cos (p -sin 6 cos #sin
F[(11 cos sin 0-27 sin 0)i
+ (-27 cos Bcos <p+ 4 cos #sin 0)j
+ (4sin - 1 1 cos 6* cos ^)k](lb • in.)
Ma =F lb in.
Mx
= F(l. 1 cos (9 sin 0-27 sin 0)(lb • in.)
MJ?
= F(-21 cos 0cos0 + 4cos 0sin 0) (lb • in.)
Mz
= F(4 sin <9 - 1 .1 cos cos (j)) (lb •
in.)
Now, Equation (1) cos #sin <f>-
11
M
F
^ + 27sin#
cos 0cos 0- — 4sin —
—
-
n{ F
and Equation (3)
Substituting Equations (4) and (5) into Equation (2),
0)
(2)
(3)
(4)
(5)
M, =/N-27 4sin0-
F
+ 4
1 ( M
111 F
*- + 27sin0
or M.
11
(27MZ
+4.MV )
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206

PROBLEM 3.54 (Continued)
Noting that the ratios yp and ± are the ratios of lengths, have
Mv
=—(-81 lb -ft) +—(-77 lb -ft)
-
11 11
= 226.82 lb •
ft or Mv
-= -227 lb -
ft A
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207

0.35 in
0.75 m
0.75 in
*
PROBLEM 3.55
The frame ACD is hinged at A and D and is supported by a
cable that passes through a ring at B and is attached to hooks
at G and H. Knowing that the tension in the cable is 450 N,
determine the moment about the diagonal AD of the force
exerted on the frame by portion BH ofthe cable.
SOLUTION
MAD ~ "'AD '
VBIA
X *Bll)
Where X „,D ---(4i-3k)
'a/a (0.5 m)i
and
Then
dm - vv0.375)
2
+(0.75)
2
+(-0.75)
2
= 1.125 m
T.SH
450 N
—(0.375i + 0.751 - 0,75k)
1.1.25
J
(1 50 N)i + (300 N)j - (300 N)k
Finally Hu> =
4 -3 |
0.5 !
150 300 -300;
[(-3X0.5X300)]
or MAD 90.0 N-m
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208

0.35 m
0.75 m
PROBLEM 3.56
In Problem 3.55, determine the moment about the diagonal AD
of the force exerted on the frame by portion BG of the cable.
SOLUTION
Where
and
Then
Finally
mad=^ad-(*biaXTbg)
M/) =-(4i-3k)
r»
MA (0.5 m)j
BG - V(-°-5) + (0.925)
2
+ (-0.4)
2
= 1.125 ra
450 N
TliC!
=- -(-0.5i + 0.925j-0.4k)
1.125
J '
= -(200 N)i + (370 N)j - (1 60 N)k
-3
u
»'i
4
0.5
-200 370 -160
[(-3X0.5X370)] MAD -Ul.ON-m
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209

0.7 m
.O.G.Ti]
PROBLEM 3.57
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 55 N, determine the
moment of that force about the line joining Points D and B.
SOLUTION
First note
Then
Also
Then
Now
where
Then
4E = V(°-9) + (~0-6) + (0-2)
2
=1.1 m
«=-yp(0.9i-0.6J + 0.2k)
= 5[(9N)i-(6N)j + (2N)k]
DB - y[( 2f + (-0.35)
2
+ (0)
2
X
1.25m
DB
OB
DB
1.25
1
(1.2I-0.35J)
25
(241 - 7j)
MDB ~ '"DB
"*DB 'VAID X T Ui)
TO/}
=-(0.1m)j + (0.2m)k
!
M»-^(5)
24 -7
-0.1 0.2
9 -6 2
H1.8- 12.6 + 28.8)
or MDfl
=2.28N-m ^
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210

PROBLEM 3.58
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 CF at C is 33 N, determine the
moment of that force about the line joining Points D and B.
SOLUTION
First note dCF
--
0.2)
2
= 1.1 m
=
V(0.6)
2
+(-0.9)
2
+ (-
Then :f
'-
33 N
-_ (0.6i-0.9j + 0.2k)
Also DB :
= 3[(6N)i~(9N)j-(2N)k]
=
V0-2)
2
+(-0.35)
2
+(0)
2
~ 1 .25 m
Then ^DB
~~
_ DB
'
DB
=—(I.2i- 0.35j)
1.25
J
= _L(24i-7j)
25
V J;
Now Mm -A'OB '(rC/D X *cp)
where yan :-(0.2m)j-(0.4m)k
24 -7
Then Mm = -(3)
25
0.2 -0.4
6 -9 -2
= —(-9.6 + 16.8-86.4)
25
or Mm ^~9.50N-m <
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211

l>
PROBLEM 3.59
A regular tetrahedron has six edges of length a. A force P is
directed as shown along edge BC. Determine the moment of P
about edge OA.
>-
SOLUTION
We have
where
From triangle OBC
Since
or
Then
and
MoA=*OA<*aoX*)
{OA)x
2
f i
(OA)z
=(OA)x tan W n/3, 2>/3
vv-1
/
(OA)
2
=(OA)l +(OA)l +{OAz f
2 [
a
2
+ {OA)l +
( Y
a
a , 12 . a
4,0
2 h 2>/3
J. 2. 1 ,
x- =
l
,+
Vl
J+
ivr
k
p = = (a sin30°)i-(a coS 30°)k
(/)) =
P
(|
_ ^
« 2
rc/0 =oi
M(A!
2 V3
I
1
2V3
Kf)
]
-73"
2
'
T
v - 3
y
0X->/3) =
aP
Moa =
aP_
72"
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212

PROBLEM 3.60
A regular tetrahedron has six edges of length a. (a) Show that two
opposite edges, such as OA and BC, are perpendicular to each
other, (b) Use this property and the result obtained in Problem 3.59
to determine the perpendicular distance between edges OA and BC.
SOLUTION
(a) For edge OA to be perpendicular to edge BC
OA-BC = Q
where
From triangle OBC (OA)x
(OA)z
=(OA)x tm30° = -
04 = I-|I + (CM)J +
( i
>

V3 J 2^3
k
a
2Sj
and BC = (asm 30°) i - (a cos 30°) k
Then
or
so that
a .
2
2
i+(.oa) v i+
2
c
a
^+ (O^)v
(0)-~ =
4 }
4
OA-BC^0
(i-V3k)~ =
<9/i is perpendicular to #C.
(6) Have A^fW = Pt/, with P acting along BC and d the perpendicular distance from OA to 5C.
From the results of Problem 3.57
ft?
M0A
Pa
4i
4i
~Pd or d = —T^ A
72
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213

y
PROBLEM 3.61
y^ji 45 in. A sign erected on uneven ground is
4 V guyed by cables EF and EG. If the
11 _^#g& force exerted by cable £7** at E is 46 lb,
si /,: .^-^.,-f
?;"" .--..
determine the moment of that force
96 ii r. 8»v jm^M^^^W% about the line join ing Poi nts A and D.
;|*:
=
i|
r
.
1 fPI
Ml
'/
L#j >' M^:
^îÎbr>>V -
47 in .
l:-;iy';!t'^ " ':>-^ :
':
:
:
'"
'
'
<*><
. -
.--•'- •
. •• -'• -/••- ,"- ~-J
"
"^v "^
.--^ &%8 iftô^
i 1
^Xl7in. >
• v-ii:—;T*V ->''' T. •
-'*'"- I'-i i"»?
»*"
^' fT /^ "''
i
"x*'' '
SOLUTION
First note that EC ~
-V(48)
2
+ (36)
2
- 60 in. and that
J§ = -g = |. The coordinates of Point E are
then(fx48,96, J> 36) or (36 in., 96 in,, 27 in,). Then
^,=>/H5)
2
+ H10)
2
+(30)
2
= 115 in.
Then T^==^(-l5I-110J + 30k)
Also
= 2[-(3 lb)i - (22 lb)j + (6 lb)k]
AD = 7C48)
2
+ (-12)
2
+ (36)
2
= 3.2726 in.
Then
_
= —!—(48i-12j + 36k)
12^26
=ir
(4M+3k)
Now MAD =-Kiiy(-rEI,4
XTEF)
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214

where
Then M
rm = (36 in.)i + (96 in.)j + (27 in.)k
1
2
(2)
4 -I 3
36 96 27
-3 -22 6
(2304 + 81-2376 + 864 + 216 + 2376)
or M^, =1359 lb -in. <«
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215

>J
PROBLEM 3.62
% in.
-1 > in. A sign erected on uneven ground is
guyed by cables EF and EG. If the
^0^ force exerted by cable EG at E is 54 lb,
,, ,,^'y^^ determine the moment of that force
Is. ^M^M^M:-}-^:-^ about the line joining Points A and D.
"&ftK?'?^^T?^i*K;
: V^lS ]ff ;1 - :
i'
p":
>^' *
*- "' " ^^^x in.
XJ7in-
?
|V; 36iri7 ^
|12
1 ®^,
SOLUTION
First note that BC = - 60 in. and that ~~ ~— ~. The coordinates of Point E are
BC 60 4
=V(48>
2
+ (36)
2
=
then (Jx 48, 96, f <36) or (36 in , 96 in., 27 in.). Then
4*;=Vai)
2
+(-88)
2
+(-44)
2
= 99 in.
Then T*;= ^(lli-88j -44k)
Also
= 6[(llb)i-(8 1b)j-(4 1.b)k]
^/> - s/(48)
2
+ (~1 2)
2
+ (36)
2
= 12>/26 in.
Then AD
AD
= —!=(48i~I2j + 36k)
12^26
J
= ~=-(4i-j + 3k)
V26
Now MAD ^kAD <rm xTEG )
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21fi

where
Then
rBA = (36 in.)i + (96 in.)j + (27 in.)k
(6)
(-1536 -27 »864 -.288 -144 + 864)
MAD
26
_6_
V26
4 -I 3
36 96 27
I -8 -4
or M,n =~23501b-in. <
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217

PROBLEM 3.63
Two forces F| and F, in space have the same magnitude F, Prove that the moment of F, about the line of
action of F2 is equal to the moment of F2
about the line of action of F,
,
SOLUTION
First note that ¥x
~fAy and F2
~F2
a
2
Let Mj = moment of F2
about the line of action of M, and M2
= moment of F, about the line of
action of M,
Now, by definition
Since
Using Equation (3.39)
so that
itfi=4-(rBM xF2 )
= A-(i
:wa x A2
)f2
Fi=F2 =F and xm
Mi=^r(»*^xi2
)F
M2 =X2
-{-*BIA xAx
)F
h i?BIA X X2 ) = ^2 HfiM X ^1
)
M2
= A}
-(rm xZ2
)F
-r»
BM
M[2
=M2i
A
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218

0.35 m
^m
0.925 m
0.875 m
//
0.75 in
L _y_
0.75 m
0.5 n>*K-^^ 1 C,<
0.5 m"A,
PROBLEM 3.64
In Problem 3.55, determine the perpendicular distance between
portion BH of the cable and the diagonal AD.
PROBLEM 3.55 The frame ACD is hinged at A and D and is
supported by a cable that passes through a ring at B and is
attached to hooks at G and H. Knowing that the tension in the
cable is 450 N, determine the moment about the diagonal AD
ofthe force exerted on the frame by portion BH of the cable.
SOLUTION
From the solution to Problem 3.55: 7)W =450N
TBH = (1 50 N)i + (300 N)j - (300 N)k
|A/^! = 90,0N-m
X ,D =-(4i.-3k)
Based on the discussion of Section 3.1 1, it follows that only the perpendicular component of TBH will
contribute to the moment of Tbh about line AD.
Now 'lill )para!lei
~~
*IIH '"'AD
= (1 501 + 300j - 300k) • I(4i - 3k)
= ^l'050)(4) + (-300)(-3)]
= 300 N
Also
so that
T
*liH
v BH 'perpendicular
= V */>•// /parallel
+ ( '«// ^perpendicular
= V(450)
2
- (300)
2
= 335.41 N
Since %AD and (TM )pcrpcndicuiQr are perpendicular, it follows that
MAD ~ "('BH Jpcipendiculftr
or 90.0 N-m
d
= rf(335.41N)
= 0.26833 m d = 0.268 m A
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219

0.35 in
(;:v.l 0.875 m
PROBLEM 3.65
portion BG of the cable and the diagonal AD,
PROBLEM 3.56 In Problem 3.55, determine the moment
about the diagonal AD of the force exerted on the frame by
portion BG of the cable.
SOLUTION
From the solution to Problem 3.56: T}iG
- 450 N
Tbg =-(200M)i + (370N)j-(160N)k
MAD
= WH-m
^D--(4i-3k)
Based on the discussion of Section 3.11, it follows that only the perpendicular component of TSo will
contribute to the moment of Tbg about line AD.
Now (Tbg ) parallel
~ *
BO ' ^ 41)
Also
so that
- (-2001 + 370j - 1 60k) -
~(4i - 3k)
= |[(-200X4) + (-I60X-3)]
= -64 N
*BG ~ '
*/J6" /parallel
"*~
V *BG ) perpendicular
(TBG )pcrpenc.icuiar
= V(450)
2
- (~64)
2
= 445.43 N
Since XAD and (Tso )pC1
.
p,ndicular
are perpendicular, it follows that
™AD ~ "('ag)perpendicular
llJN-m = J(445.43N)
</ = 0.24920 m
or
d = 0.249 m A
220

PROBLEM 3.66
cable AE and the line joining Points D and B.
PROBLEM 3,57 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 55 N, determine the moment of that force about the line
joining Points D and B,
SOLUTION
From the solution to Problem 3.57 T„? =55N
T^=5[(9N)i-(6N)j + (2N)k]
|MD/i |
= 2.28N-m
X/«-'—(24i-7j)
Based on the discussion of Section 3.11, it follows that only the perpendicular component of TAE will
contribute to the moment of TAe about line DB.
Now (*AE )parallel ~ ^AE * ^ Dli
5(9i-6j + 2k)-—(24i~7j)
[(9)(24) + (-6)(-7)]
Also
so that
5
= 51. 6N
*AE ~ ( *Ae)parallel
+ ( *AEJperpendicular
(''./: ),,, neodfcul* - J(5S)
2
+(51.6)
2
- 1 9.0379 N
' perpendicuiar
Since A,
ra and (T (/i )pcrpendicular
MDB ^"(T^Operpendieiilar
or 2.28 N-m = </(!. 9.0379 N)
d~ 0.1 19761 J = 0.1198 m ^
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Ill

PROBLEM 3.67
In Problem 3.58, determine the perpendicular distance
between cable CF and the line joining Points D and B.
PROBLEM 3.58 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 CF at
C is 33 N, determine the moment of that force about the line
joining Points D and B.
:0.3iJi
SOLUTION
From the solution to Problem 3.58 Tcr = 33N
Tc/
,=3[(6N)i-(9N)j-(2N)k]
jA*D«| = 9.50N-m
Xm =
25
(241 ~ 7,)
Based on the discussion of Section 3.11, it follows that only the perpendicular component of TCp will
contribute to the moment of Tcr about line DB,
Now ( i-CF )parallcl
~" *CF " * o«
= 3(61-9j-2k)~ (241 - 7j)
= ^[(6)(24) + (-9)(-7)]
- 24.84 N
AlSO Tc y;- — (TcF )jwral le!
+ ( *-C/' )perpendicular
s° that (Ta.Wendicuiar = V(33)
2
- (24.84)
2
= 2.1 .725 N
Since XDB and (TCF )pcrpendicillar
MDB I "V'CY'Vperpcndic
perpendicular
or 9,50N-m = </x21.725N
or d~ 0.437 m A
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Ill

% ill
PROBLEM 3.68
In Problem 3.61, determine the perpendicular
distance between cable EF and the line joining
Points A and D.
PROBLEM 3.61 A sign erected on uneven
ground is guyed by cables EF and EG. If the
force exerted by cable EF at E is 46 lb,
determine the moment of that force about
the line joining Points A. and D.
SOLUTION
From the solution to Problem 3.61 7V„=46 1b
El-
M
X
T^-2H3 1b)i-(22 1b)j + (6 1b)k]
lb in.
(4i-j + 3k)
1Di -1359 1b-in.
1
AD
Based on the discussion of Section 3.11, it follows that only the perpendicular component of TEF will
contribute to the moment of TEF about line AD.
Now (TeF /parallel ~ ^EF ' ^,
AD
2(-3i-22j + 6k)
2
(4i-j + 3k)
Also
so that
[(~3)(4) + (-22)H) + (6)(3)j
'26
= 10.9825 lb
i EF - ( lEF )para||c]
+ (lEF )perpemiicular
(^perpendicular = V(46)
2
- (10.9825)
2
= 44.670 lb
Since AD and OW ^^are perpendicular, it follows that
or
MAD - d{TEl, )pcrpeiidicular
1359 lb • in. = rfx 44.670 lb or t/ = 30.4in. <
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223

PROBLEM 3.69
In Problem 3.62, determine the perpendicular
distance between cable EG and the line joining
Points A and D.
PROBLEM 3.62 A sign erected on uneven
ground is guyed by cables EF and EG. If the
force exerted by cable EG at E is 54 lb,
determine the moment of that force about, the
line joining Points A and D.
SOLUTION
From the solution to Problem 3.62 TB0 = 54 lb
TfiC
=6[(llb)i-(81b)i-(41b)k]
|M/JD |
= 23501b-in.
^=4^(41 -j + 3k)
V26
Based on the discussion of Section 3.11, it follows that only the perpendicular component of TgG will
contribute to the moment of Teg about line AD.
N»w (TEG ) para„e,
= TEG XAD
- 6(i - 8j - 4k) •
-fL(4i - j + 3k)
V26
= ~™L[(I)(4) + (-8X-D + (-4X3)] -
Thus, (TBG )perpendiciliar
= T£G = 54 lb
Since %A0 and (T£6 pe[pcildicutar
1 ™AD
~ "*EG /perpendicular
or 2350 lb • in. = dx 54 lb
or rf = 43.5 in. 4
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224

PROBLEM 3.70
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 Points.
SOLUTION
(a) We have
where
(b) We have
(c) We have
SM8 : -t/,Cv +rf2
C,,=M
</, = (0.360 m)sin 55°
= 0.29489 m
d2
=(0.360 m)sin 55°
= 0.20649 m
C, =(60 N)cos 20°
- 56.382 N
C,, =(60 N) sin 20°
= 20.521 N
M = -(0.29489 m)(56.382 N)k + (0.20649 m)(20.52 1 N)k
--=-(12.3893 N-m)k
M = Fd(-k)
= 60 N[(0.360 m)sin(55° - 20°)](-k)
= -(12.3893 N-m)k
£MA : £(r t
x F) = rm x FB + rCIA x Fc = M
i
J k
M = (0.520 m)(60 N) cos 55° sin 55°
-cos 20° -sin 20°
I
J k
cos 55° sin 55°
cos 20° sin 20°
(1 7.8956 N •
m - 30.285 N •
m)k
-(12.3892 N-m)k
or M = 12.39 N-mJ^
or M = 12.39 N-m
+(0.800 m)(60N)
or M = 12.39 N-m
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225

1211./
A"
1 ft
21. II >
1
1
'
lb 16 in.
PROBLEM 3.71
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, (/?) the perpendicular distance between the 12-lb
forces if the resultant of the two couples is zero, (c) the value of a
if the resultant couple is 72 lb- in. clockwise and dfc 42 in.
SOLUTION
Z»V?
(a) We have Mx
=d^
c where dx
~ 1 6 in.
Fj=211b
M, =(16in.)(21 lb)
-336 lb -in.
(/?) We have M,+M.2
=0
1 2. Ho
or M, -336 lb -in. H
</, = 28.0 in. <
or 336 lb • in. -^(1 2 lb) =
(c) We have Mlolll
=M1
+M2
or -72 lb -in. = 336 lb -in. - (42 in.)(sin a){ 2 lb)
sin a~ 0.80952
and a = 54.049° or a = 54.0° ^
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226

100 imn KiOi
140 mm
160 mm
2-10 mil
V^
SOLUTION
P D
1
n
P A
(a) We have
or
(A)
We have
(c)
,32- .tv,
o
r%^ We have
PROBLEM 3,72
A couple M of magnitude 18 N-m is applied to the
handle of a screwdriver to tighten a screw into a
block of wood. Determine the magnitudes of the
two smallest horizontal forces that are equivalent to
M if they are applied (a) at corners A. and D, (b) -at
corners B and C, (c) anywhere on the block.
M = Pd
18N-m = P(.24m)
P = 75.0N
dBC ^yj(BE)
2
+{ECf
= V(-24m)
2
+(.08m)
2
= 0.25298 m
M = Pd
18N-m = P(0.25298m)
P-71J52N
dAC =yl(ADf+(DCf
or Pmin =75.0N <
or P = 71.2N <
).24m)
2
+ (0.32 m)2
0.4 m
M = PdAC
18N-m = />(0.4m)
P = 45.0N or P = 45.0N A
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227

.',:.) I!>
6 in.
25 lii
:
25 ih
L Sin.-
PROBLEM 3.73
Four 1 -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.
(/;) 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 ofthat minimum, tension?
SOLUTION
35"&
sty*
(a) +) M = (35 lb)(7 in.) + (25 lb)(9 in.)
-245 lb -in. + 225 lb -in.
M = 470 lb- in. ^)<
(/;) With only one string, pegs A and D, or B and C should be used. We have
6
tan
8
= 36.9< 90°- = 53.1°
Direction of forces:
With pegs /I andD:
With pegs/? and C:
(c) The distance between, the centers ofthe two pegs is
53.1° -4
53.1° <
F
A s
<J&+6
2
=10 in.
Therefore, the perpendicular distance d between the forces is
We must have
c/ = 10in. + 2|-in.
= 1 1 in.
M = Fd 4701b-in. = F(llin.) F = 42.7 lb <
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228

25 lb
PROBLEM 3.74
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 485 lb in.
counterclockwise.
SOLUTION
M ^ dADpAD+dBCFBC
485 lb • in. = [(6 + rf)in.](35 lb) + [(8 + rf)in.j(25 lb) rf = 1.250 in. <
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reproduced or distributed in any jorm or by any means, without the prior written permission of the publisher, or used beyond the limited
distribution to teachers andeducatorspermitted by McGraw-Hillfor their individual coursepreparation. Ifwit are a student using this Manual
you are using if withoutpermission.
229

SllH'i
PROBLEM 3.75
The shafts of an angle drive are acted upon by the two couples shown.
Replace the two couples with a single equivalent couple, specifying its
magnitude and the direction of its axis.
filb.fl
SOLUTION
Based on
where
I2M
M == M,+M2
M,== -(81b-ft)j
M2
== -(61b-ft)k
M == -(8tb-ft)j-(61b-ft)k
|M| ==
>/(8)
2
+(6)
2
=10 lb -ft
3l =
M
"|m:|
-(8 lb •
ft)j - (6 lb • fl)k
or M = 10.00 lb- ft <
or M
10ib-ft
-0.8j-0.6k
jMjl = (101b-ft)(~0.8j-0.6k)
cos0v
=O #v
=90°
cos<9v
=-0.8 6>
(
.
=143.130°
cos0z
--O,6 Z
=126.870°
or 9X
- 90.0° 6 = 143. 1
° Oz
= 26.9° <
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230

170 mm
160 mm
18 N
150 mm
.1.50 mm
!8N
3-3 N
PROBLEM 3.76
If P = 0, replace the two remaining couples with a single
equivalent couple, specifying its magnitude and the direction
of its axis.
SOLUTION
We have
where
Also,
M = M, + M,
M,
l
GIC
rt?/c
x *i
-(0.3 m)i
(18N)k
M,=-(0.3m)ix(18N)k
= (5.4N-m)J
M2 =rD/r xF2
'D/F -(.15m)i + (.08m)j
(.15m)i + (.08m)j + (.17m)k
M
(.15)
2
+(.08)
2
+(.1.7)
2
m
= 141.421 N-m(.15i + .08j + .l 7k)
i
J k
= 141,421 N-m -.15 .08
-.15 .08 .17
- 141 .42 1(.01 361 + 0.0255j)N •
m
(34 N)
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231

and M = [(5.4 N -m)j] + [14 1.42.1 (.01 36i + . 0255j) N • m]
= (1 .92333 N • m)i + (9.0062 N • m)j
|m:|-^(m,)
2
+(m;v )
2
= V(l-92333)
2
+(9.0062)
2
= 9.2093 N •
m or M = 9,2 1 N • m <
M = (1 .92333 N •
m)i + (9.0062 N • m)j
~|Mj~ 9.2093 N-m
= 0.20885 + 0.97795
cos 6>
v
=0.20885
0,= 77.945° or 9^11.9° <
cos
y
=0.97795
0, =12.054° or 0, =12.05°*
cos Z
=0.0
=90° or 6' =90.0° <
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232

PROBLEM 3.77
If P~0, replace the two remaining couples with a single
equivalent couple, specifying its magnitude and the direction
of its axis.
SOLUTION
dOE
E,
M = M, + M2 ;
/'J
= 16 lb, F2
= 40 lb
M, = rc x F, - (30 in.)* x H)6 lb)jj = -(480 lb •
in.)k
M2 = rm x F2 ; rm = (1 5 in.)i - (5 in.)j
>/(0)
2
+ (5)
2
+(10)
2
= 5^ in.
= 8>/5[(llb)j-(2ib)k]
i j k
M2 -8n/5 15 ~5
1 -2
- 8>/5[(10 lb • in.)i + (30 lb • in.)j + (1 5 lb •
in.)k]
-(480 lb in.)k + 8V5[(1 lb •
in.)i + (30 lb •
in.)j + (1 5 lb • in.)k]
(178.885 lb •
in.)i + (536.66 lb •
in.)j - (21 1 .67 lb •
in.)k
^/(178.885)
2
+(536.66)
2
+(-21 1.67)
2
M
M
K
603.99 lb in
M
A/ = 604 lb -in. <
cos0v
COS
COS 0..
M
0.29617
0.88852
-0.35045
0.29617! + 0.88852j - 0.35045k
72.8° 0=27.3° 0,,=110.5° A
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233

PROBLEM 3.78
If P = 20 lb, replace the three couples with a single
equivalent couple, specifying its magnitude and the
direction of its axis.
SOLUTION
From the solution to Problem. 3.77
M, = -(480 lb • in.)k
M2
= 8>/5 [(10 lb • in.)i + (30 lb • in.)j + (1 5 lb • in.)k]
16 lb force:
40 lb force:
F - 20 lb M3
= rc x P
= (30in.)ix(20lb)k
= (600 lb - in.)j
M-M, +M2 +M3
= ~(480)k + 8V5 (1 Oi + 30j + 1 5k) + 600j
-(178.885 lb -in)i + (1136.66 lb -in.)j -(211.67 lb-in.)k
M = ./{178.885)
2
+ (1
1
3.66)
2
+(211 .67)
2
= 1169.96 lb in M = 1170 lb- in. <
X„
M
M
0. 1 52898i + 0.97 1 54j - 0. 1 8092 1 k
cos ex
= 0.152898
cos ^ = 0.97 154
cos 0.= -0.1 8092.1 9. =81.2° 0=13.70° 6'=100.4° <
234

160 nun
,-
18 N
PROBLEM 3.79
If P = 20 N, replace the three couples with a single
equivalent couple, specifying its magnitude and the
SOLUTION
We have
where
M-M, + M2 +M3
M^r^xF,
i
J k
0.3
18
N-m = (5.4N-m)j
M3
=r /fXF2
i
J k
,15 .08 141.421 N-m
.15 .08 .17
141.421(.0136i + .0255j)N-m
(See Solution to Problem 3.76.)
M3= r(X4
xF3= 0.3 0.17 N-m
20
= -(3.4N-m)i + (6N-m)k
M=[(1.92333-3.4)i + (5.4 + 3.6062)j + (6)k]N.m
= -(1 .47667 N • m)i + (9.0062 N • m)j + (6 N in)
llVff— 1**2 , nj2 , xj2
M; + ML
V
+ m;
V(l .47667) + (9.0062) + (6)
2
10.9221 N-m or M = 10.92N-m A
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235

X
M -1.47667 + 9.0062 + 6
jM| 10.9221
= -0. 1 35200i + 0.82459J + 0.54934k
cos(?T
= -0.135200 6X
=97.770 or <9
V
-97.8° <
cos V
= 0.82459 #,,=34.453 or
y
= 34.5° ^
cos#. = 0.54934 Z
=56.678 or <9
2
= 56.7° ^
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236

z A } 1600 N-m
] 200 N-iii
1 1.20 N«m
PROBLEM 3.80
Shafts A and B connect the gear box to the wheel assemblies
of a tractor, and shaft C connects it to the engine. Shafts A and
B lie in the vertical yz plane, while shaft C is directed along
the x axis. Replace the couples applied to the shafts with a
single equivalent couple, specifying its magnitude and the
SOLUTION
Represent the given couples by the following couple vectors:
M^ = -1 600sin 20°j + 1 600cos20°k
M:
Mc
-(547.232 N m)j + (1 503.5 1 N •
m)k
1200sin 20°j + 1200cos 20°k
(41 0.424 N • m)j + (11 27.63 N •
m)k
-(1120N-m)i
The single equivalent couple is
M^M,, +M5 +MC
= -(1 120 N • m)i - (1 36.808 N • m)j + (263 1.1 N- m)k
M = yj( 1 20)
2
+ (1 36.808)
2
+ (263 1 .
I)
2
= 2862.8 N-m
-1120
cos 6,
COS tfy ^
cos 0„
2862.8
-136.808
2862.8
2631.1
2862.8
M = 2860N-m 0=113.0° B'
=92.7* 23.2° <
237

Aim W
20°
A*'
PROBLEM 3.81
The tension in the cabJe attached to the end C of an
adjustable boom ABC is 560 lb. Replace the force
exerted by the cable at C with an equivalent force-
couple system (a) at A, (b) at B.
SOLUTION
{a) Based on ZF: F,,=r = 560lb
mK^*^t^^ or ¥A = 560 lb ^ 20° ^
X^^S TscnSo ZMA : M^^sinSO )^,)
= (560lb)sin50°(18ft)
^A
(b)
or
Based on
= 7721.7 lb -ft
M/(
=7720.1b-ftjH
XF: 7^ = 7* = 560 lb
Ffi
= 560 lb ^C 20° <
ZMB : MB =(Tsm50°)(dB )
I & or
«eA = (560 lb) sin 50° (10 ft)
3*6
^MB c
or
= 4289.8 lb ft
Mg =42901b-ftjH
£&<!2t
h c*** F
*
PROPRIETARY MATSRIAL, © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed,
you are using if without permission.
238

PROBLEM 3.82
A 1 60-lb force P is applied at Point A of a structural member.
Replace P with (a) an equivalent force-couple system at C,
(b) an equivalent system consisting of a vertical force at B
and a second force at D.
SOLUTION
(a) Based on
where
(b) Based on
2.F: Pc ^P = 60b
XMC : Mc —Pxdcy + P
y
dCx
Px =(160 lb) cos 60°
= 80 lb
/;=(1601b)sin60°
= 138.564 lb
rf
tt =4ft
JCl,
= 2.75 ft
Mc = (80 lb)(2.75 ft) + (1 38.564 lb)(4 ft)
= 220 lb -ft + 554.26 lb- ft
= 334.26 lb -ft
EFV
: PDx =Pcos 60°
or P, =160 lb ^T.60°^
or Mc =334 lb •
ft )<
= (160lb)cos60c
= 80 lb
TM, (Pcos60°)(clDA ) = PB (dDB )
[(160 lb)cos60°](1.5 ft) = PB (fi ft)
PB = 20.0 lb or P„=20.01bH
PROPRIETARY MATERIAL. €> 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of (his Manual may be displayed,
reproduced or distributed in any form or by any means, without the prior written permission of the. publisher, or used beyond the limited
distribution to teachers andeducators permitted by McGraw-Hillfor their individual coursepreparation, ifyou are a student using this Manual,
239

ZFV
: Psin60o
= PB + PI)},
(160 lb) sin 60° = 20.0 lb + PDy
PDy =118.564 lb
2
^=>/('k)
2
+('y
= 7(8°)
2
+0 18- 564)
2
= 143.029 lb
# = tan
_1
(p
p
J118.564^
- tan
]
— —
I 80 )
= 55.991° or PD ,= 143.0 lb ^56.0°^
distribution to teachers and educatorspermitted by McGraw-Hillfor their individual course preparation. Ifyou are a student using this Manual,
240

-*-r-HS A PROBLEM 3.83
50 nun F,- i
«i
100 mm
The 80-N horizontal force. P acts on a bell crank, as shown.
(a) Replace P with an equivalent force-couple system at B.
(b) Find the two vertical forces at C and D that are equivalent
to the couple found in Part a.
SOLUTION
JvJ:
A
(a) Based on IF: F/?
=F = 80N
XM: Mn =Fdr
80 N (.05 m)
4.0000 N •
m
or FB
= 80.0 N --•••-
^
or M«-4,00N-m }^
5
«*
^
1
(b) If the two vertical forces are to be equivalent to MB, they must be
a couple. Further, the sense of the moment of this couple must be
counterclockwise.
Then, with Fc and FD acting as shown,
XM: MD =Fcd
4.0000 N-m = Fc (. 04m)
Fc =100.000 N or Fc =100.0nH
2Fy
: = FD -Fc
Fft =l 00.000 N or F„ =100.0 nH
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may he displayed,
241

PROBLEM 3.84
ts
iSsif^S& &^-fift^i isKSS ......' A dirigible is tethered by a cable attached to its cabin at B. If
the tension in the cable is 1 040 N, replace the force exerted by
the cable at B with an equivalent system formed by two
A B C parallel forces applied at A and C,
6.7 m 4 m
60°/
......,lA.^ ,
SOLUTION
Require the equivalent forces acting at A and C be parallel and at an
angle of awith the vertical.
Then for equivalence,
XFX : (1 040 N) sin 30° = FA si n a + FB sin a
ZFy
: -(1 040 N) cos 30° = -FA cos a - FB cos a
Dividing liquation (1) by Equation (2),
(1040 N) sin 30° __
{FA +FB )sna
-(.1040 N) cos 30° ~
-{FA + FB )cosa
Simplifying yields a ~ 30°
Based on
XMC : [(1040 N)cos30°](4 m) = {FA cos30°)(l 0.7 m)
FA = 388.79 N
(1)
(2)
or
Based on
or
F,, = 389 N "^ 60° M
XM.A :
- [(1 040 N) cos 30°](6.7 m) = (Fc cos 30°)(l 0.7 m)
Fc =651.21 M
FC =651N ^60°^
PROPRIETARY MATERIAL. © 2010 The McGraw-Hill Companies, Inc. Ali rights reserved. No part of this Manual may be displayed,
reproduced or distributed in any form or by any means, without (he prior written permission of the publisher, or used beyond the limited
distribution to teachers and educators permitted by McGraw-Hillfor their individual coursepreparation. Ifyou are a student using this Manual,
242

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