Cap 03 mecanica vectorial para ingenieros estatica 8ed.

COSMOS: Complete Online Solutions Manual Organization System
Chapter 3, Solution 1.
Resolve 90 N force into vector components P and Q
where Q = (90 N)sin 40°
= 57.851 N
Then MB = −rA/BQ
= −(0.225 m)(57.851 N)
= −13.0165 N⋅m
Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
Elliot R. Eisenberg, William E. Clausen, David Mazurek, Phillip J. Cornwell
© 2007 The McGraw-Hill Companies.
MB = 13.02 N⋅m W

Fx = (90 N)cos 25°
= 81.568 N
Fy = (90 N)sin 25°
= 38.036 N
x = (0.225 m)cos65°
= 0.095089 m
y = (0.225 m)sin 65°
= 0.20392 m
MB = xFy − yFx
= (0.095089 m)(38.036 N) − (0.20392 m)(81.568 N)
= −13.0165 N⋅m
MB = 13.02 N⋅m W

Px = (3 lb)sin 30°
= 1.5 lb
Py = (3 lb)cos30°
= 2.5981 lb
MA = xB/A Py + yB/A Px
= (3.4 in.)(2.5981 lb) + (4.8 in.)(1.5 lb)
= 16.0335 lb⋅in.
MA = 16.03 lb⋅in. W

For P to be a minimum, it must be perpendicular to the line joining points
A and B
with ( )2 ( )2 rAB = 3.4 in. + 4.8 in.
= 5.8822 in.
tan 1 y
α = θ = −  
x
 
−  
tan 1 4.8 in.
=  
3.4 in.
 
= 54.689°
Then MA = rAB Pmin
P M
or min
19.5 lb in.
5.8822 in.
A
AB
r
⋅
= =
= 3.3151 lb
∴Pmin = 3.32 lb 54.7°
or Pmin = 3.32 lb 35.3° W

By definition MA = rB/A Psinθ
where θ = φ + (90° −α )
φ −  
=  
and tan 1 4.8 in.
3.4 in.
 
= 54.689°
Also ( )2 ( )2
rB/A = 3.4 in. + 4.8 in.
= 5.8822 in.
Then (17 lb⋅in.) = (5.8822 in.)(2.9 lb)sin (54.689° + 90° −α )
or sin (144.689° −α ) = 0.99658
or 144.689° −α = 85.260°; 94.740°
∴α = 49.9°, 59.4° W

(a)
(b)
(a) MA = rB/A × TBF
MA xTBFy yTBFx = +
= (2 m)(200 N)sin 60° + (0.4 m)(200 N)cos60°
= 386.41 N⋅m
φ −  
=   = °
or MA = 386 N⋅m W
(b) For FC to be a minimum, it must be perpendicular to the line
joining A and C.
( )∴MA = d FC min
with ( )2 ( )2 d = 2 m + 1.35 m
= 2.4130 m
Then ( )( )min 386.41 N⋅m = 2.4130 m FC
( )min FC = 160.137 N
and tan 1 1.35 m 34.019
2 m
 
θ = 90 − φ = 90° − 34.019° = 55.981°
( )min ∴ FC = 160.1 N 56.0° W

(a)
(b)
(c)
MA = xTBF + yTBF
y x
= (2 m)(200 N)sin 60° + (0.4 m)(200 N)cos60°
= 386.41 N⋅m
M = xF
= = ⋅
= 193.205 N
⋅ = F
F =
θ − 
=   = °
AM = ⋅
or 386 N m
Have A C
or
386.41 N m
2 m
A
C
M
F
x
∴F = 193.2 N

C For B
F to be minimum, it must be perpendicular to the line joining A
and B
( )A B min
∴M = d F
with ( ) ( ) 2 2
d = 2 m + 0.40 m = 2.0396 m
Then 386.41 N m ( 2.0396 m
)( )C
min
( )min
189.454 N
C
and 1 2 m
tan 78.690
0.4 m
 
or ( )min
F = 189.5 N
78.7°
C

(a)
(b)
(c)
( ) /
M = r cos15
° W
B AB
= (14 in.)(cos15°)(5 lb)
= 67.615 lb⋅in.
M = r P °
M = r F
BM = ⋅
or 67.6 lb in.
/
sin85
B DB
67.615 lb⋅in. = (3.2 in.)Psin85°
or P = 21.2 lb
For ( )min,
F F must be perpendicular to BC.
Then, B C/B
67.615 lb⋅in. = (18 in.)F
or F = 3.76 lb 75.0°

(a) Slope of line
EC= =
T = T
ABx AB
= =
T = =
M = T − T
35 in. 5
+
76 in. 8 in. 12
Then ( ) 12
13
( ) 12
260 lb 240 lb
13
and ( ) 5
260 lb 100 lb
13 ABy
Then (35 in.) (8 in.)
D ABx ABy
= (240 lb)(35 in.) − (100 lb)(8 in.)
= 7600 lb⋅in.
DM = ⋅
or 7600 lb in.
(b) Have ( ) ( ) D ABx ABy M = T y + T x
= (240 lb)(0) + (100 lb)(76 in.)
= 7600 lb⋅in.
DM = ⋅
or 7600 lb in.

Slope of line
35 in. 7
EC= =
+
112 in. 8 in. 24
Then
24
T = T
ABx AB
25
and
7
T = T
ABy AB
25
Have ( ) ( ) D ABx ABy M = T y + T x
( ) ( ) 24 7
∴ ⋅ = T + T
7840 lb in. 0 112 in.
AB AB
25 25
250 lb
T =
AB
T =
or 250 lb
AB

The minimum value of d can be found based on the equation relating the moment of the force TAB about D:
( ) ( ) max D ABy
M = T d
where MD = 1152 N⋅m
( ) ( ) ABmax y ABmax sin 2880 N sin T = T θ = θ
Now
sin 1.05 m
( )2 ( )2
d 0.24 1.05 m
θ =
+ +
 
1152 N m 2880 N 1.05
∴ ⋅ =    
( ) 2 2
( 0.24 ) ( 1.05
)
d
d
 + + 
or ( )2 ( )2 d + 0.24 + 1.05 = 2.625d
or (d + 0.24)2 + (1.05)2 = 6.8906d 2
or 5.8906d 2 − 0.48d − 1.1601 = 0
Using the quadratic equation, the minimum values of d are 0.48639 m and −0.40490 m.
Since only the positive value applies here, d = 0.48639 m
or d = 486 mm W

with ( )2 ( )2 dAB = 42 mm + 144 mm
= 150 mm
sin 42 mm
150 mm
θ =
cos 144 mm
150 mm
θ =
and FAB = − FAB sinθ i − FAB cosθ j
2.5 kN ( 42 mm) (144 mm)
150 mm
=  − i − j
= −(700 N)i − (2400 N) j
Also ( ) ( ) rB/C = − 0.042 m i + 0.056 m j
Now MC = rB/C × FAB
= (−0.042i + 0.056 j) × (−700i − 2400 j)N⋅m
= (140.0 N⋅m)k
or MC = 140.0 N⋅m W

with ( )2 ( )2 dAB = 42 mm + 144 mm
= 150 mm
sin 42 mm
150 mm
θ =
cos 144 mm
150 mm
θ =
FAB = − FAB sinθ i − FAB cosθ j
2.5 kN ( 42 mm) (144 mm)
150 mm
=  − i − j
= −(700 N)i − (2400 N) j
Also ( ) ( ) rB/C = − 0.042 m i − 0.056 m j
Now MC = rB/C × FAB
= (−0.042i − 0.056j) × (−700i − 2400j)N⋅m
= (61.6 N⋅m)k
or MC = 61.6 N⋅m W

: (0.090 m) 88 80 N (0.280 m) 105 80 N
D D 137 137 ΣM M =  ×  −  × 
   
= −12.5431 N⋅m
or MD = 12.54 N⋅m W

Note: B = B(cosβ i + sinβ j)
B′ = B(cosβ i − sinβ j)
C = C(cosαi + sinα j)
By definition: B × C = BCsin (α − β ) (1)
B′ × C = BCsin (α + β ) (2)
Now ... B × C = B(cosβ i + sinβ j) × C(cosα i + sinα j)
= BC(cosβ sinα − sinβ cosα )k (3)
and B′ × C = B(cosβ i − sinβ j) × C(cosαi + sinα j)
= BC(cosβ sinα + sinβ cosα )k (4)
Equating the magnitudes of B × C from equations (1) and (3) yields:
BCsin (α − β ) = BC(cosβ sinα − sinβ cosα ) (5)
Similarly, equating the magnitudes of B′ × C from equations (2) and (4) yields:
BCsin (α + β ) = BC(cosβ sinα + sinβ cosα ) (6)
Adding equations (5) and (6) gives:
sin (α − β ) + sin (α + β ) = 2cosβ sinα
or sin cos 1 sin ( ) 1 sin ( )
α β = α + β + α − β W
2 2

Have d = λAB × rO/A
r
B A
where /
rB A
/
AB
=
λ
and rB/A = (−210 mm − 630 mm)i
+ (270 mm − (−225 mm))j
= −(840 mm)i + (495 mm) j
( )2 ( )2
rB/A = −840 mm + 495 mm
= 975 mm
−
(840 mm) + (495 mm)
Then =
i j
AB 975 mm λ
1 ( 56 33 )
65
= − i + j
Also rO/A = (0 − 630)i + (0 − (−225)) j
= −(630 mm)i + (225 mm) j
1 ( 56 33 ) (630 mm) (225 mm)
65
∴ d = − i + j × − i + j
= 126.0 mm
d = 126.0 mm W

(a)
= ×
A B
×
A B
λ
where A = 12i − 6j + 9k
B = −3i + 9j − 7.5k
Then
i j k
× = −
12 6 9
− −
3 9 7.5
A B
= (45 − 81)i + (−27 + 90) j + (108 − 18)k
= 9(−4i + 7j + 10k)
And 2 2 2 A × B = 9 (−4) +(7) +(10) = 9 165
9( − 4 + 7 +
10 )
∴ = i j k λ
9 165
or ( ) 1
A B
i j k
× = − −
−
− + −
∴ = i j k λ
λ = − i + j − k
λ = − 4 i + 7 j + 10
k
165
(b)
= ×
×
A B
λ
where A = −14i − 2j + 8k
B = 3i + 1.5j − k
Then
14 2 8
3 1.5 1
A B
= (2 − 12)i + (24 − 14) j + (−21 + 6)k
= 5(−2i + 2j − 3k)
and 2 2 2 A × B = 5 (−2) + (2) + (−3) = 5 17
5( 2 2 3 )
5 17
1
or ( 2 2 3
) 17

(a) Have A = P × Q
i j k
3 7 2 in.2
5 1 3
P Q
× = −
−
= [(21 + 2)i + (10 − 9)j + (3 + 35)k]in.2
= (23 in.2 )i + (1 in.2 )j + (38 in.2 )k
∴ A = (23)2 + (1)2 + (38)2 = 44.430 in.2
i j k
or A = 44.4 in.2 W
(b) A = P × Q
2 4 3 in.2
6 1 5
× = −
−
P Q
= [(−20 − 3)i + (−18 − 10)j + (−2 + 24)k] in.2
= −(23 in.2 )i − (28 in.2 )j + (22 in.2 )k
∴ A = (−23)2 + (−28)2 + (22)2 = 42.391 in.2
or A = 42.4 in.2 W

(a) Have MO = r × F
i j k
6 3 1.5 N m
7.5 3 4.5
= − ⋅
−
= [(−13.5 − 4.5)i + (11.25 − 27)j + (−18 − 22.5)k] N⋅m
= (−18.00i − 15.75j − 40.5k) N⋅m
or MO = −(18.00 N⋅m)i − (15.75 Ν⋅m)j − (40.5N⋅m)k W
(b) Have MO = r × F
i j k
2 0.75 1 Nm
7.5 3 4.5
= − − ⋅
−
= [(3.375 + 3)i + (−7.5 + 9)j + (6 + 5.625)k] N⋅m
= (6.375i + 1.500j + 11.625k) N⋅m
or MO = (6.38 N⋅m)i + (1.500 Ν⋅m)j + (11.63 Ν⋅m)k W
(c) Have MO = r × F
i j k
= − − ⋅
2.5 1 1.5 N m
7.5 3 4.5
= [(4.5 − 4.5)i + (11.25 − 11.25)j + (−7.5 + 7.5)k] N⋅m
or MO = 0 W
This answer is expected since r and F are proportional (F = −3r). Therefore, vector F has a line of action
passing through the origin at O.

(a) Have MO = r × F
i j k
= − − ⋅
7.5 3 6 lb ft
3 −
6 4
= [(12 − 36)i + (−18 + 30)j + (45 − 9)k] lb⋅ft
or MO = −(24.0 lb⋅ft)i + (12.00 lb⋅ft) j + (36.0 lb⋅ft)k W
(b) Have MO = r × F
i j k
= − − ⋅
7.5 1.5 1 lb ft
3 −
6 4
= [(6 − 6)i + (−3 + 3)j + (4.5 − 4.5)k] lb⋅ft
or MO = 0 W
(c) Have MO = r × F
i j k
= − − ⋅
8 2 14 lb ft
3 −
6 4
= [(8 − 84)i + (−42 + 32)j + (48 − 6)k] lb⋅ft
or MO = −(76.0 lb⋅ft)i − (10.00 lb⋅ft) j + (42.0 lb⋅ft)k W

With TAB = −(369 N) j
( ) ( ) ( ) ( )
i − j −
k
2.4 m 3.1 m 1.2 m
( )2 ( )2 ( )2
369 N
+ − + −
2.4 m 3.1 m 1.2 m
JJJG
T AD
= =
T
AB AD
AD
TAD = (216 N)i − (279 N) j − (108 N)k
Then RA = 2 TAB + TAD
= (216 N)i − (1017 N) j − (108 N)k
Also ( ) ( ) rA/C = 3.1 m i + 1.2 m k
Have MC = rA/C × RA
i j k
= ⋅
0 3.1 1.2 Nm
216 − 1017 −
108
= (885.6 N⋅m)i + (259.2 N⋅m) j − (669.6 N⋅m)k
MC = (886 N⋅m)i + (259 N⋅m) j − (670 N⋅m)k W

Have MA = rC/A × F
where ( ) ( ) ( ) rC/A = 215 mm i − 50 mm j + 140 mm k
Fx = −(36 N)cos 45°sin12°
Fy = −(36 N)sin 45°
Fz = −(36 N)cos 45°cos12°
∴ F = −(5.2926 N)i − (25.456 N) j − (24.900 N)k
i j k
A = − ⋅
M
and 0.215 0.050 0.140 N m
− − −
5.2926 25.456 24.900
= (4.8088 N⋅m)i + (4.6125 N⋅m) j − (5.7377 N⋅m)k
MA = (4.81 N⋅m)i + (4.61 N⋅m) j − (5.74 N⋅m)k W

Have MO = rA/O × R
where ( ) ( ) rA/D = 30 ft j + 3 ft k
( ) ( ) T1 = −  62 lb cos10° i −  62 lb sin10° j
= −(61.058 lb)i − (10.766 lb) j
JJJG
T AB
AB
T =
2 2
( ) ( ) ( ) ( )
i − j +
k
5 ft 30 ft 6 ft
( )2 ( )2 ( )2
62 lb
+ − +
5 ft 30 ft 6 ft
=
= (10 lb)i − (60 lb) j + (12 lb)k
∴ R = −(51.058 lb)i − (70.766 lb) j + (12 lb)k
i j k
O = ⋅
0 30 3lbft
51.058 70.766 12
− −
M
= (572.30 lb⋅ft)i − (153.17 lb⋅ft) j + (1531.74 lb⋅ft)k
MO = (572 lb⋅ft)i − (153.2 lb⋅ft) j + (1532 lb⋅ft)k W

(a) Have MO = rB/O × TBD
where rB/O = (2.5 m)i + (2 m) j
JJJG
T BD
T =
BD BD
BD
( ) ( ) ( ) ( )
− 1 m i − 2 m j + 2 m
k
=
 ( )2 ( )2 ( )2
900 N
1m 2 m 2 m
− + − +
= −(300 N)i − (600 N) j + (600 N)k
Then
i j k
O= ⋅
2.5 2 0 N m
300 600 600
− −
M
MO = (1200 N⋅m)i − (1500 N⋅m) j − (900 N⋅m)k W
continued

(b) Have MO = rB/O × TBE
where rB/O = (2.5 m)i + (2 m) j
JJJG
T BE
T =
BE BE
BE
( ) ( ) ( ) ( )
− 0.5 m i − 2 m j − 4 m
k
=
 ( )2 ( )2 ( )2
675 N
0.5 m + − 2 m + −
4 m
= −(75 N)i − (300 N) j − (600 N)k
Then
i j k
O= ⋅
2.5 2 0 N m
75 300 600
− − −
M
MO = −(1200 N⋅m)i + (1500 N⋅m) j − (600 N⋅m)k W

Have MC = rA/C × P
where
rA/C = rB/C + rA/B
= (16 in.)(−cos80°cos15°i − sin80°j − cos80°sin15°k)
+(15.2 in.)(−sin 20°cos15°i + cos 20°j − sin 20°sin15°k)
= −(7.7053 in.)i − (1.47360 in.) j − (2.0646 in.)k
and P = (150 lb)(cos5°cos70°i + sin 5°j − cos5°sin 70°k)
= (51.108 lb)i + (13.0734 lb) j − (140.418 lb)k
Then
i j k
C = − − − ⋅
7.7053 1.47360 2.0646 lb in.
51.108 13.0734 −
140.418
M
= (233.91 lb⋅in.)i − (1187.48 lb⋅in.) j − (25.422 lb⋅in.)k
or MC = (19.49 lb⋅ft)i − (99.0 lb⋅ft) j − (2.12 lb⋅ft)k W

Have MC = rA/C × FBA
where ( ) ( ) ( ) rA/C = 0.96 m i − 0.12 m j + 0.72 m k
and FBA = λBAFBA
  − ( 0.1 m ) i + ( 1.8 m ) j − ( 0.6 m
)
k
 =   
 ( ) + ( ) + ( )

( ) 2 2 2
i j k
∴ C = − ⋅
228 N
0.1 1.8 0.6 m
= −(12.0 N)i + (216 N) j − (72 N)k
0.96 0.12 0.72 N m
12.0 216 72
− −
M
= −(146.88 N⋅m)i + (60.480 N⋅m) j + (205.92 N⋅m)k
or MC = −(146.9 N⋅m)i + (60.5 N⋅m) j + (206 N⋅m)k W

Have MC = TAD d
where d = Perpendicular distance from C to line AD
T AD
i j k
2.4 m 3.1 m 1.2 m
i j k
JJJG
with MC = rA/C TAD
and ( ) ( ) rA/C = 3.1 m j + 1.2 m k
AD AD
AD
T =
JJJG
( ) ( ) ( ) ( )
( )2 ( )2 ( )2
369 N
2.4 m 3.1 m 1.2 m
AD
 − −  =
+ − + −
T
= (216 N)i − (279 N) j − (108 N)k
Then
0 3.1 1.2 Nm
216 279 108
C= ⋅
− −
M
= (259.2 N⋅m) j − (669.6 N⋅m)k
and ( )2 ( )2 MC = 259.2 N⋅m + −669.6 N⋅m
= 718.02 N⋅m
∴ 718.02 N⋅m = (369 N)d
or d = 1.946 m W

M = T
Have O AC d
where d = Perpendicular distance from O to rope AC
with MO = rA/O × TAC
and ( ) ( ) rA/O = 30 ft j + 3 ft k
TAC = − (62 lb)cos10° i − (62 lb)sin10° j
= −(61.058 lb)i − (10.766 lb) j
Then
i j k
0 30 3lbft
61.058 10.766 0
O= ⋅
− −
M
= (32.298 lb⋅ft)i − (183.174 lb⋅ft) j + (1831.74 lb⋅ft)k
and ( )2 ( )2 ( )2 MO = 32.298 lb⋅ft + −183.174 lb⋅ft + 1831.74 lb⋅ft
= 1841.16 lb⋅ft
∴1841.16 lb⋅ft = (62 lb)d
or d = 29.7 ft W

Have MO = TAB d
where d = Perpendicular distance from O to rope AB
with MO = rA/O × TAB
and ( ) ( ) rA/O = 30 ft j + 3 ft k
JJJG
T AB
T =
AB AB
AB
( ) ( ) ( ) ( )
 5 ft i − 30 ft j + 6 ft
k
=
 ( )2 ( )2 ( )2
62 lb
5 ft + − 30 ft +
6 ft
= (10 lb)i − (60 lb) j + (12 lb)k
Then
i j k
0 30 3 lbft
10 60 12
O = ⋅
−
M
= (540 lb⋅ft)i + (30 lb⋅ft) j − (300 lb⋅ft)k
and ( )2 ( )2 ( )2 MO = 540 lb⋅ft + 30 lb⋅ft + −300 lb⋅ft
= 618.47 lb⋅ft
∴ 618.47 lb⋅ft = (62 lb)d
or d = 9.98 ft W

Have MC = TBD d
where d = Perpendicular distance from C to cable BD
with MC = rB/C × TB/D
and ( ) rB/C = 2 m j
JJJG
T BD
T =
BD BD
BD
( ) ( ) ( ) ( )
− 1m i − 2 m j + 2 m
k
=
 ( )2 ( )2 ( )2
900 N
1m 2 m 2 m
− + − +
= −(300 N)i − (600 N) j + (600 N)k
Then
i j k
0 2 0 Nm
300 600 600
C= ⋅
− −
M
= (1200 N⋅m)i + (600 N⋅m)k
and ( )2 ( )2 MC = 1200 N⋅m + 600 N⋅m
= 1341.64 N⋅m
∴1341.64 = (900 N)d
or d = 1.491 m W

Have MC = Pd
From the solution of problem 3.25
MC = (233.91 lb⋅in.)i − (1187.48 lb⋅in.) j − (25.422 lb⋅in.)k
Then
( ) ( ) ( ) 2 2 2 MC = 233.91 + −1187.48 + −25.422
= 1210.57 lb⋅in.
d MC
and 1210.57 lb.in.
150 lb
= =
P
or d = 8.07 in.W

Have |MD | = FBAd
where d = perpendicular distance from D to line AB.
MD = rA/D × FBA
( ) ( ) rA/D = − 0.12 m j + 0.72 m k
( ( 0.1m ) i ( 1.8 m ) j ( 0.6 m
) k
)
( ) 2 2 2
( ) ( ) ( )
i j k
228 N
0.1 1.8 0.6 m
BA BAFBA
− + −
= =
+ +
F λ
= −(12.0 N)i + (216 N) j − (72 N)k
0 0.12 0.72 N m
12.0 216 72
∴ D = − ⋅
− −
M
= −(146.88 N⋅m)i − (8.64 N⋅m) j − (1.44 N⋅m)k
and ( )2 ( )2 ( )2 |MD | = 146.88 + 8.64 + 1.44 = 147.141N⋅m
∴ 147.141N⋅m = (228 N)d
d = 0.64536 m
or d = 0.645 mW

Have |MC | = FBAd
where d = perpendicular distance from C to line AB.
MC = rA/C × FBA
( ) ( ) ( ) rA/C = 0.96 m i − 0.12 m j + 0.72 m k
( ( 0.1m ) i ( 1.8 m ) j ( 0.6
) k
)
( ) ( ) ( )
( ) 2 2 2
i j k
228 N
0.1 1.8 0.6 m
BA BAFBA
− + −
= =
+ +
F λ
= −(12.0 N)i + (216 N) j − (72 N)k
0.96 0.12 0.72 N m
12.0 216 72
∴ C = − ⋅
− −
M
= −(146.88 N⋅m)i − (60.48 N⋅m) j + (205.92 N⋅m)k
and ( )2 ( )2 ( )2 |MC | = 146.88 + 60.48 + 205.92 = 260.07 N⋅m
∴ 260.07 N⋅m = (228 N)d
d = 1.14064 m
or d = 1.141mW

(a) Have d = rC/A sinθ = λAB × rC/A
where d = Perpendicular distance from C to pipe AB
with
λ AB i j k
7 + 4 −
32
7 4 32
( )2 ( )2 ( )2
= =
AB AB
+ + −
1 (7 4 32 )
33
= i + j − k
and ( ) ( ) ( ) rC/A = − 14 ft i + 5 ft j +  L − 22 ftk
λ r
Then /
i j k
1 7 4 32 ft
33
× = −
L
14 5 22
AB C A
− −
1 { 4( 22) 32(5) 32(14) 7( 22) 7(5) 4(14) }ft
33
=  L − +  i +  − L −  j +  + k
= 1  (4 L + 72) i + ( − 7 L + 602) j + 91 kft
33

and d = 1 (4 L + 72)2 + ( − 7 L + 602)2 +
(91)2
33
2
d dd L L
( ) , 1 2 4 4 72 2 7 7 602 0
For ( )( ) ( )( )
=  + + − − +  =
min 2
L
d 33
or 65L − 3926 = 0
or L = 60.400 ft
But L Lgreenhouse so L = 30.0 ft W
(b) with L = 30 ft, d = 1 (4 × 30 + 72)2 + ( − 7 × 30 + 602)2 +
(91)2
33
or d = 13.51 ft W
Note: with L = 60.4 ft,
1 (4 60.4 72)2 ( 7 60.4 602)2 (91)2 11.29 ft
33
d= × + + − × + + =

P⋅Q = (−4i + 8j − 3k) ⋅ (9i − j − 7k)
= (−4)(9) + (8)(−1) + (−3)(−7)
= −23
or P⋅Q = −23W
P⋅S = (−4i + 8j − 3k) ⋅ (5i − 6j + 2k)
= (−4)(5) + (8)(−6) + (−3)(2)
= −74
or P⋅S = −74 W
Q⋅S = (9i − j − 7k) ⋅ (5i − 6j + 2k)
= (9)(5) + (−1)(−6) + (−7)(2)
= 37
or Q⋅S = 37 W

By definition
B⋅C = BCcos(α − β )
where B = B (cosβ )i + (sinβ ) j
C = C (cosα )i + (sinα ) j
∴ (Bcosβ )(C cosα ) + (Bsinβ )(Csinα ) = BC cos(α − β )
or cosβ cosα + sinβ sinα = cos(α − β ) (1)
By definition
B′⋅C = BC cos(α + β )
where B′ = (cosβ )i − (sinβ ) j
∴ (Bcosβ )(Ccosα ) + (−Bsinβ )(Csinα ) = BCcos(α + β )
or cosβ cosα − sinβ sinα = cos(α + β ) (2)
Adding Equations (1) and (2),
2 cosβ cosα = cos(α − β ) + cos(α + β )
or cos cos 1 cos( ) 1 cos( )
α β = α + β + α − β W
2 2

First note:
rB/A = (0.56 m)i + (0.9 m) j
( ) ( ) rC/A = 0.9 m j − 0.48 m k
( ) ( ) ( ) rD/A = − 0.52 m i + 0.9 m j + 0.36 m k
( )2 ( )2
rB/A = 0.56 m + 0.9 m = 1.06 m
( )2 ( )2
rC/A = 0.9 m + −0.48 m = 1.02 m
( )2 ( )2 ( )2
rD/A = −0.52 m + 0.9 m + 0.36 m = 1.10 m
By definition rB/A ⋅rD/A = rB/A rD/A cosθ
or (0.56i + 0.9j) ⋅ (−0.52i + 0.9j + 0.36k) = (1.06)(1.10)cosθ
(0.56)(−0.52) + (0.9)(0.9) + (0)(0.36) = 1.166cosθ
cosθ = 0.44494
θ = 63.6° W

From the solution to problem 3.37
rC/A = 1.02 m with rC/A = (0.9 m)i − (0.48 m) j
( ) ( ) ( ) rD/A = 1.10 m with rD/A = − 0.52 m i + 0.9 m j + 0.36 m k
Now by definition
rC/A ⋅ rD/A = rC/A rD/A cosθ
or (0.9j − 0.48k) ⋅ (−0.52i + 0.9j + 0.36k) = (1.02)(1.10)cosθ
0(−0.52) + (0.9)(0.9) + (−0.48)(0.36) = 1.122cosθ
cosθ = 0.56791
or θ = 55.4° W

(a) By definition
λBC + λEF = (1)(1) cosθ
where
( ) ( ) ( )
( )2 ( )2 ( )2
i j k
32 ft 9 ft 24 ft
32 9 24 ft
BC
− −
=
+ − + −
λ
1 (32 9 24 )
41
= i − j − k
( ) ( ) ( )
( )2 ( )2 ( )2
i j k
14 ft 12 ft 12 ft
14 12 12 ft
EF
− − +
=
− + − +
λ
1 ( 7 6 6 )
11
= − i − j + k
(32 i − 9 j − 24 k ) ( − 7 i − 6 j +
6 k
)
Therefore ⋅ =
cos
41 11
θ
(32)(−7) + (−9)(−6) + (−24)(6) = (41)(11)cosθ
cosθ = −0.69623
or θ = 134.1° W
(b) By definition ( EG )BC ( EF )cos T = T θ
= (110 lb)(−0.69623)
= −76.585 lb
or ( EF )BC 76.6 lb T = − W

(a) By definition
λBC ⋅ λEG = (1) (1) cosθ
where
( ) ( ) ( )
( )2 ( )2 ( )2
i j k
32 ft 9 ft 24 ft
32 9 24 ft
BC
− −
=
+ − + −
λ
1 (32 9 24 )
41
= i − j − k
( ) ( ) ( )
i j k
16 ft − 12 ft +
9.75
16 12 9.75 ft
( )2 ( )2 ( )2
EG
=
+ − +
λ
1 (16 12 9.75 )
22.25
= i − j + k
(32 i − 9 j − 24 k ) (16 i − 12 j +
9.75 k
)
Therefore ⋅ =
cos
41 22.25
θ
(32)(16) + (−9)(−12) + (−24)(9.75) = (41)(22.25)cosθ
cosθ = 0.42313
or θ = 65.0° W
(b) By definition ( EG )BC ( EG )cos T = T θ
= (178 lb)(0.42313)
= 75.317 lb
or ( EG )BC 75.3 lb T = W

First locate point B:
3.5
d =
22 14
or d = 5.5 m
(a) ( )2 ( )2 ( )2 dBA = 5.5 + 0.5 + −22 + −3 = 23 m
Locate point D:
(−3.5 − 7.5sin 45°cos15°), (14 + 7.5cos 45°),
(0 + 7.5sin 45°sin15°)m
or (−8.6226 m, 19.3033 m, 1.37260 m)
Then
( )2 ( )2 ( )2 dBD = −8.6226 + 5.5 + 19.3033 − 22 + 1.37260 − 0 m
= 4.3482 m
⋅ ( 6 i − 22 j − 3 k ) ⋅ ( − 3.1226 i − 2.6967 j +
1.37260
k
)
and ( )( )
cos
23 4.3482
d d
d d
BA BD
θ
ABD
= =
BA BD
= 0.36471
or θ ABD = 68.6° W
(b) ( BA )BD BA cos ABD T = T θ
= (230 N)(0.36471)
or ( BA )BD 83.9 N T = W

First locate point B:
3.5
d =
22 14
or d = 5.5 m
(a) Locate point D:
(−3.5 − 7.5sin 45°cos10°), (14 + 7.5cos 45°),
3.2227 2.6967 0.92091 5.2227 5.3033 0.92091
d d
d d
(0 + 7.5sin 45°sin10°)m
or (−8.7227 m, 19.3033 m, 0.92091 m)
Then
dDC = (5.2227 m)i − (5.3033 m) j − (0.92091 m)k
and
( )2 ( )2 ( )2 dDB = −5.5 + 8.7227 + 22 − 19.3033 + 0 − 0.92091 m
= 4.3019 m
and ( ) ( )
( )( )
cos
4.3019 7.5
DB DC
BDC
DB DC
θ
⋅ + − ⋅ − −
= =
i j k i j k
= 0.104694
or θ BDC = 84.0° W
(b) ( BD )DC BD cos BDC (250 N)(0.104694) T = T θ =
or ( BD )DC 26.2 N T = W

Volume of parallelopiped is found using the mixed triple product
(a) Vol = P⋅(Q × S)
3 4 1
7 6 8 in.
9 2 3
3
−
= − −
− −
= (−54 + 288 + 14 − 48 + 84 − 54)in.3
= 230 in.3
5 7 4
6 2 5 in.
4 8 9
− −
= −
− −
= (−90 + 140 + 192 + 200 − 378 − 32) in.3
= 32 in.3
or Volume = 230 in.3 W
(b) Vol = P⋅(Q × S)
3
or Volume = 32 in.3 W

For the vectors to all be in the same plane, the mixed triple product is zero.
P⋅(Q × S) = 0
3 7 5
− −
0 2 1 4
∴ = − −
8 Sy −
6
0 = 18 + 224 − 10Sy − 12Sy + 84 − 40
So that 22 Sy = 286
Sy = 13
or Sy = 13.00 W

Have rC = (2.25 m)k
JJJG
T CE
T =
CE CE
CE
( ) ( ) ( ) ( )
 0.90 m i + 1.50 m j − 2.25 m
k
=
 ( )2 ( )2 ( )2
1349 N
0.90 1.50 2.25 m
T
CE
+ + −
= (426 N)i + (710 N) j − (1065 N)k
Now MO = rC × TCE
i j k
0 0 2.25 Nm
426 710 1065
= ⋅
−
= − (1597.5 N⋅m)i + (958.5 N⋅m) j
∴ Mx = −1598 N⋅m, My = 959 N⋅m, Mz = 0 W

Have rE = (0.90 m)i + (1.50 m) j
JJJG
T DE
T =
DE DE
DE
( ) ( ) ( ) ( )
− 2.30 m i + 1.50 m j − 2.25 m
k
=
 ( )2 ( )2 ( )2
1349 N
2.30 1.50 2.25 m
− + + −
= −(874 N)i + (570 N) j − (855 N)k
Now MO = rE × TDE
i j k
0.90 1.50 0 N m
874 570 855
= ⋅
− −
= −(1282.5 N⋅m)i + (769.5 N⋅m) j + (1824 N⋅m)k
∴Mx = −1283 N⋅m, My = 770 N⋅m, Mz = 1824 N⋅m W

Have z ( B )y BA ( C )y CD
= ⋅ ×  +  ×  M k  r T  k  r T 
where Mz = −(48 lb⋅ft)k
( B )y ( C )y (3 ft) r = r = j
( ) ( ) ( ) ( )
JJJG
 4.5 ft i − 3 ft j + 9 ft
k
= =
 ( )2 ( )2 ( )2
14 lb
4.5 3 9 ft
T BA
T
BA BA
BA
+ − +
= (6 lb)i − (4 lb) j + (12 lb)k
( ) ( ) ( )
( )2 ( )2 ( )2
JJJG
 i − j − k
= =
 6 ft 3 ft 6 ft
6 3 6 ft
T CD T
T
CD CD CD
CD
+ − + −
= TCD i − j − k
(2 2 )
3
Then −(48 lb⋅ft) = k⋅{(3 ft) j × (6 lb)i − (4 lb) j + (12 lb)k}
 + k ⋅(3 ft) j × TCD (2 i − j − 2 k
)
   3

or −48 = −18 − 2TCD
TCD = 15.00 lb W

Have y ( B )z BA ( C )z CD
M = j⋅ r × T  × j⋅ r × T 
where My = 156 lb⋅ft
( B )z (24 ft) ; ( C )z (6 ft) r = k r = k
( ) ( ) ( )
( )2 ( )2 ( )2
JJJG
 i − j + k
= =
 4.5 ft 3 ft 9 ft
4.5 3 9 ft
T BA T
T
BA BA BA
BA
+ − +
= TBA i − j + k
(4.5 3 9 )
10.5
( ) ( ) ( ) ( )
JJJG
 6 ft i − 3 ft j + 9 ft
k
= =
 ( )2 ( )2 ( )2
7.5 lb
6 3 9 ft
T CD
T
CD CD
CD
+ − +
= (5 lb)i − (2.5 lb) j − (5 lb)k
⋅ = ⋅ × TBA − + 
Then (156 lb ft) j (24 ft) k (4.5 i 3 j 9 k
)
10.5
 
+ j⋅{(6 ft)k × (5 lb)i − (2.5 lb) j − (5 lb)k}
or 156 108 30
10.5 BA = T +
TBA = 12.25 lb W

Based on Mx = (Pcosφ )(0.225 m)sinθ  − (Psinφ )(0.225 m)cosθ  (1)
My = −(Pcosφ )(0.125 m) (2)
Mz = −(Psinφ )(0.125 m) (3)
By ( )
M P
M P
Equation 3 sin 0.125
( )
( )( )
( )( )
:
z
y
φ
φ
−
=
−
Equation 2 cos 0.125
−
or 4 tan 9.8658
23
= ∴ = °
φ φ
−
or φ = 9.87° W
From Equation (2)
−23 N⋅m = −(Pcos9.8658°)(0.125 m)
P = 186.762 N
or P = 186.8 N W
From Equation (1)
26 N⋅m = (186.726 N)cos9.8658° (0.225 m)sinθ 
− (186.726 N)sin 9.8658° (0.225 m)cosθ 
or 0.98521sinθ − 0.171341cosθ = 0.61885
Solving numerically,
θ = 48.1° W

Based on Mx = (Pcosφ )(0.225 m)sinθ  − (Psinφ )(0.225 m)cosθ  (1)
My = −(Pcosφ )(0.125 m) (2)
Mz = −(Psinφ )(0.125 m)
By ( )
M P
M P
Equation 3 sin 0.125
( )
( )( )
( )( )
:
z
y
φ
φ
−
=
−
Equation 2 cos 0.125
or 3.5 tan ; 9.9262
20
φ φ
−
= = °
−
From Equation (3):
−3.5 N⋅m = −(Psin 9.9262°)(0.125 m)
P = 162.432 N
From Equation (1):
Mx = (162.432 N)(0.225 m)(cos9.9262°sin 60° − sin 9.9262°cos60°)
= 28.027 N⋅m
or Mx = 28.0 N⋅m W

First note:
JJJG
T BA
T =
BA BA
BA
( ) ( ) ( ) ( )
+  − +  + − =
i j k
L
L
4 1.5 1 6
BC
( ) ( ) ( ) 2 2 2
70 lb
+  − +  + −
4 1.5 1 6
BC
( ) ( )
i L
j k
4 + 0.5 − −
6
BC
( )2
70 lb
52 0.5
L
BC
=
+ −
rA = (4 ft)i + (1.5 ft) j − (12 ft)k
Have MO = rA × TBA
70 lb 4 ft 1.5 ft 12 ft
= −
( ) ( ) 2
52 + 0.5 − L BC 4 0.5 − L
BC −
6 763 lb ft 70 1.5 6 12 0.5 lb ft
− ⋅ =  − + −  ⋅
2 190.81 190.81 4 25.19 6198.8225
− ± − −
=
BC 2 25.19 L
i j k
For the i components:
( )
( ) ( ) 2
52 + 0.5
BC
BC
L
L
−
or ( )2 10.9 52 + 0.5 − LBC = 3 + 12LBC
or (10.9)2 52 + (0.5 − LBC )2  = 9 + 72LBC + 144L2BC
or 25.19L2BC + 190.81LBC − 6198.8225 = 0
Then
( ) ( )( )
( )
Taking the positive root LBC = 12.35 ft W

First note:
JJJG
T BA
T =
BA BA
BA
( ) ( ) ( ) ( )
+  − +  + − =
i j k
L
L
4 1.5 1 6
BC
( ) ( ) ( ) 2 2 2
70 lb
+  − +  + −
4 1.5 1 6
BC
( ) ( )
i L
j k
4 + 0.5 − −
6
BC
( )2
70 lb
52 0.5
L
BC
=
+ −
rA = (4 ft)i + (1.5 ft) j − (12 ft)k
Have MO = rA × TBA
BA
i j k
= −
( ) ( ) 2
T BA
L
− ⋅ =  − + −  ⋅
L
T BA
L
= +
L
T BA
L
− ⋅ =  − −  ⋅
L
T L
315 4 BA
1
= +
L
1 300 1 4
2 315 41 BC
L
L
4 ft 1.5 ft 12 ft
52 + 0.5 4 0.5 6
BC BC
T
L L
− − −
For the i components:
( )
( ) ( ) 2
900 lb ft 1.5 6 12 0.5 lb ft
52 + 0.5
BC
BC
−
or
( )
( ) 2
300 1 4
52 + 0.5
BC
BC
−
(1)
For the k components:
( )
( ) ( ) 2
315 lb ft 4 0.5 1.5 4 lb ft
52 + 0.5
BC
BC
−
or
( )
( ) 2
52 + 0.5
BC
BC
−
(2)
Then, ( )
( ) ( )
BC
+
⇒ =
+
or 59 ft
BC 4 L =
LBC = 14.75 ft W

Have MAD = λ AD ⋅(rB/A × TBH )
where ( ) ( )
i k
0.8 m −
0.6 m
λ = = i −
k
( )2 ( )2
0.8 0.6
0.8 m 0.6 m
AD
+ −
( ) rB/A = 0.4 m i
( ) ( ) ( ) ( )
JJJJG
 0.3 m i + 0.6 m j − 0.6 m
k
= =
 ( )2 ( )2 ( )2
1125 N
0.3 0.6 0.6 m
T BH
T
BH BH
BH
+ + −
Then
0.8 0 0.6
0.4 0 0 180 N m
375 750 750
MAD
−
= =− ⋅
−
or MAD = −180.0 N⋅m W

Have MAD = λ AD ⋅(rB/A × TBG )
where λ AD = (0.8 m)i − (0.6 m)k
( ) rB/A = 0.4 m i
( ) ( ) ( ) ( )
JJJG
− 0.4 m i + 0.74 j − 0.32 m
k
 = =
( )2 ( )2 ( )2
1125 N
0.4 m 0.74 m 0.32 m
T BG
T
BG BG
BG
− + + −
= −(500 N)i + (925 N) j − (400 N)k
Then
0.8 0 0.6
0.4 0 0
500 925 400
MAD
−
=
− −
or MAD = −222 N⋅m W

Have
MAD = λ AD ⋅(rE/A × FEF )
λ =
where AD
JJJG
AD
AD
( 7.2 m ) i ( 0.9 m
)
j
( 7.2 m )2 ( 0.9 m
)2
AD
+
=
+
λ
= 0.99228i + 0.124035 j
( ) ( ) rE/A = 2.1 m i − 0.9 m j
( ) ( ) ( ) ( )
JJJG
 0.3 m i + 1.2 m j + 2.4 m
k
= =
 ( )2 ( )2 ( )2
24.3 kN
0.3 m 1.2 m 2.4 m
F EF
F
EF EF
EF
+ +
= (2.7 kN)i + (10.8 kN) j + (21.6 kN)k
Then
0.99228 0.124035 0
2.1 0.9 0 kN m
2.7 10.8 21.6
MAD = − ⋅
= −19.2899 − 5.6262
= −24.916 kN⋅m
or MAD = −24.9 kN⋅m W

Have MAD = λ AD ⋅(rG/A × EEF )
Where ( ) ( )
i j
7.2 m +
0.9 m
7.2 m 0.9 m
( )2 ( )2
AD
=
+
λ
= 0.99228i + 0.124035 j
( ) ( ) rG/A = 6 m i − 1.8 m j
( ) ( ) ( ) ( )
JJJJG
 − 1.2 m i + 2.4 m j + 2.4 m
k
= =
 ( )2 ( )2 ( )2
21.3 kN
1.2 m 2.4 m 2.4 m
F GH
F
GH GH
GH
− + +
= −(7.1 kN)i + (14.2 kN) j + (14.2 kN)k
Then
0.99228 0.124035 0
MAD = − ⋅
6 1.8 0 kNm
7.1 14.2 14.2
−
= −25.363 − 10.5678
= −35.931 kN⋅m
or MAD = −35.9 kN⋅m W

Have M = λ ⋅ ( r × P
)
OA OA C/O where
From triangle OBC
a
( ) 2 x
OA =
( ) ( ) 1
tan30
2 3 2 3 z x
OA = OA + OA + OA
2 2
=  a      + + a
 
   
a OA
∴ OA = a − − = a
OAλ = i + j + k
P = λ P
° − °
P = i − k
  ∴ =  
aP 
= −  −
aP =
a a
OA OA
 
= ° =   =
 
Since ( ) ( ) ( ) ( ) 2 2 2 2
x y z
or ( )
2 2
2 y 2 3
( )
2 2
a a
2 2
y 4 12 3
Then /
2
2 3 2 3
AO
a a
r = i + a j + k
and
1 2 1
2 3 2 3
BC
( asin30 ) ( acos30
)( P
) a
=
i k
( 3 )
2
C/O r = ai
( )
1 2 1
2 3 2 3
1 0 0 2
1 0 3
OA
P
M a
 
−
2
( 1 )( 3
) 2 3
 
2
2
OA
aP
M =

(a) For edge OA to be perpendicular to edge BC,
OA BC = 0 uuur uuur ⋅⋅⋅⋅
where
From triangle OBC
a
( OA ) =
x
2 ( ) ( ) 1
tan30
2 3 2 3 z x
    ∴ =   + +  
i j k
OA OA
= a a i − k
= a ( i − 3 k
)
 a   
 i + a a
OA
j +   k  ⋅⋅⋅⋅ i − k
=
   
2 2
a a
+ OA − =
M = Pd with P acting along BC and d the
uuur uuur
Pa
∴ = Pd
a a
OA OA
 
= ° =   =
 
a a
( )
 2  y  2 3

uuur
uuur
and BC = (asin 30°)i − (acos30°)k
3
2 2
2
Then ( ) ( 3 ) 0
2 y 2 3 2
or ( ) ( )
0 0
4 y 4
∴ OA BC = 0 uuur uuur ⋅⋅⋅⋅
uuur
so that OA
uuur
is perpendicular to . BC

(b) Have OA ,
perpendicular distance from OA to BC.
From the results of Problem 3.57,
2
OA
Pa
M =
2
or
a
d =
2

Have MDI = λDI ⋅(rF/I × TEF )
where ( ) ( )
i j
( )2 ( )2
T EF
 3.6 ft i − 10.8 ft j + 16.2 ft
k
=
 Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
4.8 ft 1.2 ft
4.8 ft 1.2 ft
DI
DI
DI
−
= =
+ −
λ
JJJG
= 0.97014i − 0.24254 j
( ) rF/I = 16.2 ft k
EF EF
EF
T =
JJJG
( ) ( ) ( ) ( )
( )2 ( )2 ( )2
29.7 lb
3.6 ft + − 10.8 ft +
16.2 ft
= (5.4 lb)i − (16.2 lb) j + (24.3 lb)k
Then
0.97014 0.24254 0
0 0 16.2lbft
5.4 16.2 24.3
MDI
−
= ⋅
−
= −21.217 + 254.60
= 233.39 lb⋅ft
or MDI = 233 lb⋅ft W

Have MDI = λDI ⋅(rG/I × TEG )
where ( ) ( )
i j
4.8 ft 1.2 ft
4.8 ft 1.2 ft
( )2 ( )2
DI
DI
DI
T EG
 3.6 ft i − 10.8 ft j − 35.1 ft
k
=
 Vector Mechanics for Engineers: Statics and Dynamics, 8/e, Ferdinand P. Beer, E. Russell Johnston, Jr.,
−
= =
+ −
λ
JJJG
= 0.97014i − 0.24254 j
( ) rG/I = − 35.1 ft k
EG EG
EG
T =
JJJG
( ) ( ) ( ) ( )
( )2 ( )2 ( )2
24.6 lb
3.6 ft + − 10.8 ft + −
35.1 ft
= (2.4 lb)i − (7.2 lb) j − (23.4 lb)k
Then
0.97014 0.24254 0
0 0 35.1lbft
2.4 7.2 23.4
MDI
−
= − ⋅
− −
= 20.432 − 245.17
= −224.74 lb⋅ft
or MDI = −225 lb⋅ft W

First note that F1 = F1λ1 and F2 = F2λ2
Let M1 = moment of F2 about the line of action of M1
and M2 = moment of F1 about the line of action of M2
Now, by definition
( ) ( ) M1 = λ1 ⋅ rB/A × F2 = λ1 ⋅ rB/A × λ2 F2
( ) ( ) M2 = λ2 ⋅ rA/B × F1 = λ2 ⋅ rA/B × λ1 F1
Since F1 = F2 = F and rA/B = −rB/A
( ) M1 = λ1 ⋅ rB/A × λ2 F
( ) M2 = λ2 ⋅ −rB/A × λ1 F
Using Equation (3.39)
( ) ( ) λ1 ⋅ rB/A × λ2 = λ2 ⋅ −rB/A × λ1
so that ( ) M2 = λ1 ⋅ rB/A × λ2 F
∴ M12 = M21W

From the solution of Problem 3.53:
λAD = 0.8i − 0.6k
TBH = (375 N)i + (750 N) j − (750 N)k; TBH = 1125 N
MAD = −180 N⋅m
Only the perpendicular component of TBH contributes to the moment of TBH about line AD. The
parallel component of TBH will be used to find the perpendicular component.
Have ( )TBH Parallel = λAD ⋅TBH
= [0.8i − 0.6k] ⋅ (375 N)i + (750 N) j − (750 N)k
= (300 + 450)N
= 750 N
Since ( ) ( ) TBH = TBH Perpendicular + TBH Parallel
Then ( ) ( )2 ( )2
TBH Perpendicular = TBH − TBH Parallel
( )2 ( )2 = 1125 N − 750 N
= 838.53 N
and ( )MAD = TBH Perpendicular d
180 N⋅m = (838.53 N)d
d = 0.21466 m
or d = 215 mm W

λAD = 0.8i − 0.6k
(500 N) (925 N) (400 N)
1125 N
T i j k
BG
TBG
= − + −
=
MAD = −222 N⋅m
Only the perpendicular component of TBG contributes to the moment of TBG about line AD. The
parallel component of TBG will be used to find the perpendicular component.
Have ( )TBG Parallel = λAD ⋅TBG
= [0.8i − 0.6k] ⋅ −(500 N)i + (925 N) j − (400 N)k
= (−400 + 240)N
= −160 N
Since ( ) ( ) TBG = TBG Perpendicular + TBG Parallel
Then ( ) ( )2 ( )2
TBG Perpendicular = TBG − TBG Parallel
( )2 ( )2 = 1125 N − −160 N
= 1113.56 N
and ( )MAD = TBG Perpendicular d
222 N⋅m = (1113.56 N)d
d = 0.199361 m
or d = 199.4 mm W

λDI = 0.97014i − 0.24254 j
(5.4 lb) (16.2 lb) (24.3 lb)
29.7 lb
T i j k
EF
TEF
= − +
=
MDI = 233.39 lb⋅ft
Only the perpendicular component of TEF contributes to the moment of TEF about line DI. The
parallel component of TEF will be used to find the perpendicular component.
Have ( )TEF Parallel = λDI ⋅TEF
= [0.97014i − 0.24254 j] ⋅ (5.4 lb)i − (16.2 lb) j + (24.3 lb)k
= (5.2388 + 3.9291)
= 9.1679 lb
Since ( ) ( ) TEF = TEF Perpendicular + TEF Parallel
Then ( ) ( )2 ( )2
TEF Perpendicular = TEF − TEF Parallel
( )2 ( )2 = 29.7 − 9.1679
= 28.250 lb
and ( )MDI = TEF Perpendicular d
233.39 lb⋅ft = (28.250 lb)d
d = 8.2616 ft
or d = 8.26 ft W

λDI = 0.97014i − 0.24254 j
(2.4 lb) (7.2 lb) (23.4 lb)
24.6 lb
T i j k
EG
TEG
= − −
=
MDI = − 224.74 lb⋅ft
Only the perpendicular component of TEG contributes to the moment of TEG about line DI. The
parallel component of TEG will be used to find the perpendicular component.
Have ( )TEG Parallel = λDI ⋅TEG
= [0.97014 i − 0.24254 j] ⋅ (2.4 lb) i − (7.2 lb) j − (23.4 lb)k
= (2.3283 + 1.74629)
= 4.0746 lb
Since ( ) ( ) TEG = TEG Perpendicular + TEG Parallel
Then ( ) ( )2 ( )2
TEG Perpendicular = TEG − TEG Parallel
( )2 ( )2 = 24.6 − 4.0746
= 24.260 lb
and ( )MDI = TEG Perpendicular d
224.74 lb⋅ft = (24.260 lb)d
d = 9.2638 ft
or d = 9.26 ft W

From the solution of Prob. 3.55:
λAD = 0.99228i + 0.124035 j
FEF = (2.7 kN)i + (10.8 kN) j + (21.6 kN)k
FEF = 24.3 kN
MAD = −24.916 kN⋅m
Only the perpendicular component of FEF contributes to the moment of FEF about edge AD. The
parallel component of FEF will be used to find the perpendicular component.
Have ( )FEF Parallel = λAD ⋅FEF
= [0.99228i + 0.124035 j] ⋅ (2.7 kN)i + (10.8 kN) j + (21.6 kN)k
= 4.0187 kN
Since ( ) ( ) FEF = FEF Perpendicular + FEF Parallel
Then ( ) ( )2 ( )2
FEF Perpendicular = FEF − FEF Parallel
( )2 ( )2 = 24.3 − 4.0187
= 23.965 kN
and ( )MAD = FEF Perpendicular d
24.916 kN⋅m = (23.965 kN)d
d =1.039683m or d = 1.040 m W

From the solution of Prob. 3.56:
λAD = 0.99228i + 0.124035 j
FGH = −(7.1 kN)i + (14.2 kN) j + (14.2 kN)k
FGH = 21.3 kN
MAD = −35.931 kN⋅m
Only the perpendicular component of FGH contributes to the moment of FGH about edge AD. The
parallel component of FGH will be used to find the perpendicular component.
Have ( )FGH Parallel = λAD ⋅FGH
= (0.99228i + 0.124035 j) ⋅ −(7.1 kN)i + (14.2 kN) j + (14.2 kN)k
= −5.2839 kN
Since ( ) ( ) FGH = FGH Perpendicular + FGH Parallel
Then ( ) ( )2 ( )2
FGH Perpendicular = FGH − FGH Parallel
( )2 ( )2 = 21.3 − 5.2839
= 20.634 kN
and ( )MAD = FGH Perpendicular d
35.931 kN⋅m = (20.634 kN)d
d = 1.741349m or d = 1.741 m W

(a) Have M1 = d1F1
Where d1 = 0.6 m and F1 = 40 N
( )( ) ∴ M1 = 0.6 m 40 N
or M1 = 24.0 N⋅m W
(b) Have MTotal = M1 + M2
8 N⋅m = 24.0 N⋅m − (0.820 m)(cosα )(24 N)
∴ cosα = 0.81301
or α = 35.6° W
(c) Have M1 + M2 = 0
( ) 24 N⋅m − d2 24 N = 0
or d2 = 1.000 m W

(a) M = Fd
12 N⋅m = F (0.45 m)
or F = 26.7 N W
(b) M = Fd
12 N⋅m = F (0.24 m)
or F = 50.0 N W
(c) M = Fd
( )2 ( )2 Where d = 0.45 m + 0.24m
= 0.51 m
12 N⋅m = F (0.51 m)
or F = 23.5 N W

(a) Note when a = 8 in., rC/F is perpendicular to the inclined 10 lb forces.
Have M = ΣFd ( )
= −(10 lb)a + 8 in. + 2(1 in.) − (10 lb)2a 2 + 2(1 in.)
For a = 8 in.,
M = −(10 lb)(18 in. + 24.627 in.)
= −426.27 lb⋅in.
or M = 426 lb⋅in. W
(b) Have M = 480 lb⋅in.
Also M = Σ(M + Fd ) ( )
= Moment of couple due to horizontal forces at A and D +
Moment of force-couple systems at C and F about C.
Then −480 lb⋅in. = −10 lb a + 8 in. + 2(1 in.) + MC + MF + FX (a + 8 in.) + Fy (2a)
Where MC = −(10 lb)(1 in.) = −10 lb⋅in.
MF = MC = −10 lb⋅in.
Fx −
=
10 lb
2 Fy −
=
10 lb
2 ∴ −480 lb⋅in. = −10 lb(a + 10 in.) − 10 lb⋅in. − 10 lb⋅in.
10 lb ( 8 in.) 10 lb (2 )
2 2
− a + − a
303.43 = 31.213 a
or a = 9.72in.W

(a) Have M = ΣFd ( )
= (9 lb)(13.8 in.) − (2.5 lb)(15.2 in.)
= (86.2 lb⋅in.)
M = 86.2 lb⋅in. W
(b) Have M = Td = 86.2 lb⋅in.
For T to be a minimum, d must be maximum.
∴Tmin must be perpendicular to line AC.
tan 15.2 in.
11.4 in.
θ =
θ = 53.130°
or θ = 53.1° W
(c) Have M = Tmindmax Where M = 86.2 lb⋅in.
( )2 ( )2 ( )
dmax = 15.2 in. + 11.4 in. + 2 1.2 in.
= 21.4 in.
( ) ∴86.2 lb⋅ in. = Tmin 21.4 in.
Tmin = 4.0280 lb
or Tmin = 4.03 lb W

Based on M = M1 + M2
( ) M1 = 18 N⋅m k
( ) M2 = 7.5 N⋅m i
∴M = (7.5 N⋅m)i + (18 N⋅m)k
and ( )2 ( )2 M = 7.5 N⋅m + 18 N⋅m
= 19.5 N⋅m
M ⋅ i + ⋅
k λ
= =
or M = 19.50 N⋅m W
With (7.5 N m) (18 N m)
M 19.5 N ⋅
m
5 12
13 13
= i + k
Then cos 5 67.380
x 13 x θ = ∴θ = °
cosθ y = 0 ∴θ y = 90°
cos 12 22.620
z 13 z θ = ∴θ = °
or θ x = 67.4°, θ y = 90.0°, θ z = 22.6° W

Have M = M1 + M2
Where M1 = rC/B × PIC
( ) ( ) rC/B = 38.4 in. i − 16 in. j
PIC = −(25 lb)k
i j k
M
1 38.4 16 0 lb in.
∴ = − ⋅
0 0 −
25
= (400 lb⋅in.)i + (960 lb⋅in.) j
and M2 = rD/A × PZE
( ) ( ) rD/A = 8 in. j − 22 in. k
( ) ( ) ( )
JJJG
− 19.2 in. i + 22 in.
k
 = =
( )2 ( )2
36.5 lb
19.2 in. 22 in.
P ED
P
ZE ZE
ED
− +
= −(24 lb)i + (27.5 lb)k
i j k
2 0 8 22lbin.
M
∴ = − ⋅
24 0 27.5
−
( ) ( ) ( ) M2 = 220 lb⋅in. i + 528 lb⋅in. j + 192 lb⋅in. k
and M = M1 + M2
= (400 lb⋅in.)i + (960 lb⋅in.) j + (220 lb⋅in.)i + (528 lb⋅in.) j + (192 lb⋅in.)k
= (620 lb⋅in.)i + (1488 lb⋅in.) j + (192 lb⋅in.)k
continued

( )2 ( )2 ( )2 M =  620 + 1488 + 192  lb⋅in.  
= 1623.39 lb⋅in.
M ⋅ i + ⋅ j + ⋅
k λ
= =
or M = 1.623 kip⋅in.W
(620 lb in.) (1488 lb in.) (192 lb in.)
M 1623.39 lb ⋅
in.
= 0.38192i + 0.91660 j + 0.118271k
cosθ x = 0.38192 or θ x = 67.5° W
cosθ y = 0.91660 or θ y = 23.6° W
cosθ z = 0.118271 or θ z = 83.2°W

Have M = M1 + M2
Where M1 = rE/D × FD
= −(0.7 m)k × (80 N) j
= (56.0 N⋅m)i
And M2 = rG/F × FB
Now ( )2 ( )2 ( )2 dBF = −0.300 m + 0.540 m + 0.350 m
= 0.710m
Then FB = λBFFB
( − 0.300 m) i + (0.540 m) j +
(0.350 m) k
( 71 N
)
0.710 m
=
= −(30 N)i + (54 N) j + (35 N)k
∴ ( ) ( ) ( ) ( ) M2 = 0.54 m j × − 30 N i + 54 N j + 35 N k
= (18.90 N⋅m)i + (16.20 N⋅m)k
Finally M = (56.0 N⋅m)i + (18.90 N⋅m)i + (16.20 N⋅m)k
= (74.9 N⋅m)i + (16.20 N⋅m)k
and ( )2 ( )2 M = 74.9 N⋅m + 16.20 N⋅m
= 76.632 N⋅m or M = 76.6 N⋅m W
cos 74.9 cos 0 cos 16.20
x 76.632 y 76.632 z 76.632 θ = θ = θ =
or θ x = 12.20° θ y = 90.0° θ z = 77.8° W

Have M = (M1 + M2 ) + MP
From the solution to Problem 3.74
( ) ( ) ( ) M1 + M2 = 74.9 N⋅m i + 16.20 N⋅m k
Now MP = rD / E × PE
= (0.54 m) j + (0.70 m)k × (90 N)i
= (63.0 N⋅m) j − (48.6 N⋅m)k
∴ M = (74.9i + 16.20k) + (63.0 j − 48.6 k)
= (74.9 N⋅m)i + (63.0 N⋅m) j − (32.4 N⋅m)k
and ( )2 ( )2 ( )2 M = 74.9 N⋅m + 63.0 N⋅m + −32.4 N⋅m
= 103.096 N⋅m
−
or M = 103.1 N⋅m W
and cos 74.9 cos 63.0 cos 32.4
θ x = θ 103.096 y = θ
=
103.096 z 103.096 or θ x = 43.4° θ y = 52.3° θ z = 108.3° W

Have M = M1 + M2 + MP
From Problem 3.73 solution:
( ) ( ) M1 = 400 lb⋅in. i + 960 lb⋅in. j
( ) ( ) ( ) M2 = 220 lb⋅in. i + 528 lb⋅in. j + 192 lb⋅in. k
Now MP = rE/A × PE
( ) ( ) ( ) rE/A = 19.2 in. i + 8 in. j − 44 in. k
PE = (52.5 lb) j
Therefore
i j k
19.2 8 44
0 52.5 0
P = −
M
= (2310 lb. in.)i + (1008 lb. in.)k
and M = M1 + M2 + MP
= [(400 + 220 + 2310)i + (960 + 528)j + (192 +1008)k] lb⋅in.
= (2930 lb⋅in.)i + (1488 lb⋅in.) j + (1200 lb⋅in.)k
( )2 ( )2 ( )2 M = 2930 + 1488 + 1200
= 3498.4 lb⋅in.
or M = 3.50 kip⋅in.W
continued

M 2930 i + 1488 j +
= = 1200
k
M
3498.4
λ
= 0.83753i + 0.42534j + 0.34301k
cosθ x = 0.83753
or θ x = 33.1° W
cosθ y = 0.42534
or θ y = 64.8° W
cosθ z = 0.34301
or θ z = 69.9° W

Have M = M1 + M2 + M3
Where M1 = −(1.2 lb⋅ft)cos 25°j − (1.2 lb⋅ft)sin 25°k
M2 = −(1.3 lb⋅ft) j
M3 = −(1.4 lb⋅ft)cos 20°j + (1.4 lb⋅ft)sin 20°k
∴ M = (−1.08757 − 1.3 − 1.31557) j + (−0.507142 + 0.478828)k
= −(3.7031 lb⋅ft) j − (0.028314 lb⋅ft)k
and ( )2 ( )2 M = −3.7031 + −0.028314 = 3.7032 lb⋅ft
− −
= M = j k
M
or M = 3.70 lb⋅ft W
3.7031 0.028314
3.7032
λ
= −0.99997j − 0.0076458k
cosθ x = 0
or θ x = 90°W
cosθ y = −0.99997
or θ y = 179.6° W
cosθ z = −0.0076458
or θ z = 90.4° W

(a) FB = P: ∴ FB = 160.0 N 50.0° W
MB = −rBAPcos10°
= −(0.355 m)(160 N)cos10°
= −55.937 N⋅m
or MB = 55.9 N⋅m W
(b) FC = P: ∴ FC = 160.0 N 50.0° W
( ) C B CB B M M r F⊥
= −
= MB − rCBFB sin 55°
= −55.937 N⋅m − (0.305 m)(160 N)sin 55°
or MC = 95.9 N⋅m W

(a) ΣF: FB = 135 N
or FB = 135 N W
ΣM: MB = P dB
= (135 N)(0.125 m)
= 16.875 N⋅m
or MB = 16.88 N⋅m W
(b) ΣMB : MB = FC d
16.875 N⋅m = FC (0.075 m)
FC = 225 N
or FC = 225 N W
ΣF: 0 = − FB + FC
FB = FC = 225 N
or FB = 225 N W

(a) Based on ΣF: PC = P = 700 N
or PC = 700 N 60°W
ΣMC : MC = −PxdCy + PydCx
where Px = (700 N)cos60° = 350 N
Py = (700 N)sin 60° = 606.22 N
dCx = 1.6 m
dCy = 1.1 m
∴ MC = −(350 N)(1.1 m) + (606.22 N)(1.6 m)
= −385 N⋅m + 969.95 N⋅m
= 584.95 N⋅m
or MC = 585 N⋅m W
(b) Based on ΣFx : PDx = Pcos60°
= (700 N)cos60°
= 350 N
ΣMD : (Pcos60°)(dDA ) = PB (dDB )
(700 N)cos60° (0.6 m) = PB (2.4 m)
PB = 87.5 N
or PB = 87.5 N W

ΣFy : Psin 60° = PB + PDy
(700 N)sin 60° = 87.5 N + PDy
PDy = 518.72 N
( ) ( )2 2
PD = PDx + PDy
( )2 ( )2 = 350 + 518.72 = 625.76 N
P
P
  θ = tan − 1   = tan − 1  518.72    = 55.991
°
350
Dy
Dx
   
or PD = 626 N 56.0°W

ΣFx : 2.8cos65° = FA cosθ + FC cosθ
= (FA + FC )cosθ (1)
ΣFy : 2.8sin 65° = FA sinθ + FC sinθ
= (FA + FC )sinθ (2)
Then (2) tan 65 tan
(1)
⇒ ° = θ
or θ = 65.0°
ΣMA : (27 m)(2.8 kN)sin 65° = (72 m)(FC )sin 65°
or FC = 1.050 kN
From Equation (1): 2.8 kN = FA + 1.050 kN
or FA = 1.750 kN
∴ FA = 1.750 kN 65.0°W
FC = 1.050 kN 65.0° W

Based on
ΣFx : − (54 lb)cos30° = −FB cosα − FC cosα
(FB + FC )cosα = (54 lb)cos30° (1)
ΣFy : (54 lb)sin 30° = FB sinα + FC sinα
or (FB + FC )sinα = (54 lb)sin 30° (2)
From ( )
2
: tan tan30
1
( )
Eq
Eq
α = °
∴α = 30°
Based on ΣMC : (54 lb)cos(30° − 20°) (10 in.) = (FB cos10°)(24 in.)
∴ FB = 22.5 lb
or FB = 22.5 lb 30° W
From Eq. (1), (22.5 + FC )cos30° = (54)cos30°
FC = 31.5 lb
or FC = 31.5 lb 30° W

(a) Based on
ΣFx : −(54 lb)cos30° = − FC cos30°
∴ FC = 54 lb
1 27 : tan 63.006
2 13.7534
27 30.301 lb
B sin 63.006 F = =
or FC = 54.0 lb 30° W
ΣMC : (54 lb)cos10° (10 in.) = MC
∴ MC = 531.80 lb⋅in.
or MC = 532 lb⋅in. W
(b) Based on
ΣFy : (54 lb)sin 30° = FB sinα
or FB sinα = 27 (1)
ΣMB : 531.80 lb⋅in. − (54 lb)cos10° (24 in.)
= − FC (24 in.)cos 20°
FC = 33.012 lb
or FC = 33.0 lb W
And ΣFx : −(54 lb)cos30° = −33.012 lb − FB cosα
FB cosα = 13.7534 (2)
From ( )
( )
Eq
Eq
α = ∴α = °
From Eq. (1), ( °
)
or FB = 30.3 lb 63.0° W

ΣF F + F + F = F
(a) Have :
y C D E
F = −200lb + 150 lb − 150 lb
F = −200 lb
ΣM F d − − F =
(200 lb)(d − 4.5 ft ) − (150 lb)(6 ft ) = 0
ΣM F d − + F =
or F = 200 lb
Have : ( 4.5 ft ) (6 ft ) 0
G C D
d = 9 ft
or d = 9.00 ft
(b) Changing directions of the two 150-lb forces only changes the sign of
the couple.
∴ F = −200 lb
or F = 200 lb
And : ( 4.5 ft ) (6 ft ) 0
G C D
(200 lb)(d − 4.5 ft ) + (150 lb)(6 ft ) = 0
d = 0
or d = 0

(a)
(b)
(c)
Based on ΣFz :
−200 N + 200 N + 240 N = FA
FA = 240 N
or FA = (240 N)k W
Based on ΣMA:
(200 N)(0.7 m) − (200 N)(0.2 m) = MA
MA = 100 N⋅m
or MA = (100.0 N⋅m) jW
Based on ΣFz :
−200 N + 200 N + 240 N = F
F = 240 N
or F = (240 N)k W
Based on ΣMA:
100 N⋅m = (240 N)(x)
x = 0.41667 m
or x = 0.417 m From A along AB W
Based on ΣMB :
−(200 N)(0.3 m) + (200 N)(0.8 m) − P(1 m) = R(0)
P = 100 N
or P = 100.0 N W

Let R be the single equivalent force...
ΣF: R = FA + FC
= (260 N)(cos10°i − sin10°k) + (320 N)(−cos8°i − sin8°k)
= −(60.836 N)i − (89.684 N)k
or R = −(60.8 N)i − (89.7 N)k W
ΣMA : rADRx = rACFC cos8°
rAD (60.836 N) = (0.690 m)(320 N)cos8°
rAD = 3.5941 m
∴R Would have to be applied 3.59 m to the right of A W
on an extension of handle ABC.

(a) Have ΣF: FB + FC + FD = FA
Since FB = −FD
∴ FA = FC = 22 lb 20°
or FA = 22.0 lb 20° W
Have ΣMA : − FBT (r ) − FCT (r ) + FDT (r ) = MA
− (28 lb)sin15° (8 in.) − (22 lb)sin 25° (8 in.)
+ (28 lb)sin 45° (8 in.) = MA
MA = 26.036 lb⋅in.
or MA = 26.0 lb⋅in. W
(b) Have ΣF: FA = FE
or FE = 22.0 lb 20° W
ΣM: MA = [FE cos 20°](a)
∴ 26.036 lb⋅in. = (22 lb)cos 20° (a)
a = 1.25941 in.
or a = 1.259 in. Below AW

(a) Let R be the single equivalent force. Then
R = (120 N)k R = 120 N W
ΣMB : − a(120 N) = −(0.165 m)(90 N)cos15° + (0.201 m)(90 N)sin15°
a = 0.080516 m
∴The line of action is y = 201mm − 80.516 mm =
19.984 mm
2
2 3.1697 3.1697 4 2.48396 0.69263
or y = 19.98 mm W
(b) ΣMB : − (0.201 − 0.040)m (120 N) = −(0.165 m)(90 N)cosθ + (0.201 m)(90 N)sinθ
or cosθ − 1.21818sinθ = 1.30101
or cos2θ = (1.30101 + 1.21818sinθ )2
or 1 − sin2θ = 1.69263 + 3.1697sinθ + 1.48396sin2θ
or 2.48396sin2θ + 3.1697sinθ + 0.69263 = 0
Then
( ) ( )( )
( )
sin
2 2.48396
θ
− ± −
=
or θ = −16.26° and θ = −85.0° W

(a) First note that F = P and that F must be equivalent to (P, MD) at point D,
Where MD = 57.6 N⋅m
For ( )min F = F F must act as far from D as possible
∴ Point of application is at point B W
(b) For ( )min F F must be perpendicular to BD
Now ( )2 ( )2 dDB = 630 mm + −160 mm
= 650 mm
tan 63
16
α =
α = 75.7°
Then MD = dDB F
57.6 N⋅m = (0.650 m)F
F = 88.6 N
or F = 88.6 N 75.7°W

Have ΣF: −(250 kN) j = F
i j k
or F = −(250 kN) jW
Also have ΣMG : rP × P = M
0.030 0 0.060 kN m =
0 250 0
− ⋅
−
M
∴ M = (15 kN⋅m)i + (7.5 kN⋅m)k
or M = (15.00 kN⋅m)i + (7.50 kN⋅m)k W

Have ΣF: TAB = F
JJJG
T AB
T =
where AB AB
AB
i j k
54 lb 2.25 18 9
( )
− +
( 2.25 )2 ( 18 )2 ( 9
)2
=
+ − +
= (6 lb)i − (48 lb) j + (24 lb)k
So that F = (6.00 lb)i − (48.0 lb) j + (24.0 lb)k W
Have ΣME : rA/E × TAB = M
i j k
0 22.5 0 lb ft
6 48 24
⋅ =
−
M
∴M = (540 lb⋅ft)i − (135 lb⋅ft)k
or M = (540 lb⋅ft)i − (135.0 lb⋅ft)k W

Have ΣF: TCD = F
JJJG
T CD
T =
where CD CD
CD
i j k
61 lb 0.9 16.8 7.2
( )
− − +
( 0.9 )2 ( 16.8 )2 ( 7.2
)2
=
− + − +
= −(3 lb)i − (56 lb) j + (24 lb)k
So that F = −(3.00 lb)i − (56.0 lb) j + (24.0 lb)k W
Have ΣMO = rC/D × TCD = M
i j k
0 22.5 0 lb ft
3 56 24
⋅ =
− −
M
∴M = (540 lb⋅ft)i + (67.5 lb⋅ft)k
M = (540 lb⋅ft)i + (67.5 lb⋅ft)k W

Have ΣF: TAB = F
JJJG
T AB
T =
where AB AB
AB
i j k
10.5 kN 4.75 2
( )
− − +
( 1 )2 ( 4.75 )2 ( 2
)2
=
− + − +
= −(2 kN)i − (9.5 kN) j + (4 kN)k
So that F = −(2.00 kN)i − (9.50 kN) j + (4.00 kN)k W
Have ΣMO : rA × TAB = M
i j k
3 4.75 0 kNm
2 9.5 4
⋅ =
− −
M
∴M = (19 kN⋅m)i − (12 kN⋅m) j − (19 kN⋅m)k
M = (19.00 kN⋅m)i − (12.00 kN⋅m) j − (19.00 kN⋅m)k W

Let (R, MO ) be the equivalent force-couple system
Then R = (220 N)(−sin 60°j − cos60°k)
= (110 N)(− 3j − k)
or R = −(190.5 N) j − (110 N)k W
Now ΣMO : MO = rOC × R
Where rOC = (0.2 m)i + (0.1 − 0.4sin 20°)m j + (0.4cos 20°m)k
i j k
Then (0.1)(110 N) 2 (1 4sin 20 ) 4cos 20 (m)
O = − − ° °
0 3 1
M
= −(11 N⋅m){(1 − 4sin 20°)(1) − (4cos 20°)( 3) i − 2 j + 2 3 k}
or MO = (75.7 N⋅m)i + (22.0 N⋅m) j − (38.1 N⋅m)k W

Have ΣF: F = FD
JJG
where F AI
AI
F =
i j k
63 lb 14.4 4.8 7.2
( )
− +
( 14.4 )2 ( 4.8 )2 ( 7.2
)2
=
+ − +
So that F = (54.0 lb)i − (18.00 lb) j + (27.0 lb)k W
Have ΣMD : M + rI/O × F = MD
JJJG
where M AC
AC
M =
i k
560 lb in. 9.6 7.2
( )
−
( 9.6 )2 ( 7.2
)2
= ⋅
+ −
= (448 lb⋅in.)i − (336 lb⋅in.)k
i j k
Then (448 lb in.) (336 lb in.) 0 0 14.4 lb in.
D = ⋅ − ⋅ + ⋅
54 −
18 27
M i k
= (448 lb⋅in.)i − (336 lb⋅in.)k + (259.2 lb⋅in.)i + (777.6 lb⋅in.) j
or MD = (707 lb⋅in.)i + (778 lb⋅in.) j − (336 lb⋅in.)k W

First assume that the given force W and couples M1 and M2 act at the
origin.
Now W = −Wj
and M = M1 + M2 = −(M2 cos 25°)i + (M1 − M2 sin 25°)k
Note that since W and M are perpendicular, it follows that they can be
replaced with a single equivalent force.
(a) Have F = W or F = −Wj = −(2.4 N) j
i j k
x z Wz Wx
= = −
W
−
= − ° = − − °   
x = − ° = −
z = − ° = −
or F = −(2.40 N) j W
(b) Assume that the line of action of F passes through point P (x, 0, z).
Then for equivalence
M = rP/O × F
where rP/O = xi + zk
∴ −(M2 cos 25°)i + (M1 − M2 sin 25°)k
0 ( ) ( )
0 0
i k
Equating the i and k coefficients,
z Mz cos 25 and x M1 M2 sin 25
W W
 
(b) For W = 2.4 N, M1 = 0.068 N⋅m, M2 = 0.065 N⋅m
0.068 0.065sin 25 0.0168874 m
2.4
−
or x = −16.89 mmW
0.065cos 25 0.024546 m
2.4
or z = −24.5 mmW

(a) Have ΣMBz : M2z = 0
( ) k⋅ rH/B × F1 + M1z = 0 (1)
where ( ) ( ) rH/B = 31 in. i − 2 in. j
F1 = λEHF1
(6 in.) + (6 in.) −
(7 in.) ( )
i j k
− + −
= ⋅
⋅ − ⋅
− + =
20 lb
11.0 in.
=
i j k
20 lb (6 6 7 )
11.0
= i + j − k
M1z = k⋅M1
M1 = λEJM1
( 3 in. ) ( 7 in.
) ( 480 lb in.
) 2
58 in.
d
d
+
Then from Equation (1),
( )( )
2
0 0 1
20 lb in. 7 480 lb in. 31 2 0 0
11.0 6 6 − 7 d
+
58 continued

Solving for d, Equation (1) reduces to
20 lb ⋅ in. ( ⋅
186 + 12 ) − 3360 lb in. =
0
11.0 d 2
+
58
From which d = 5.3955 in.
or d = 5.40 in.W
20 lb 6 6 7
11.0
(b) F 2 = F 1
= ( i + j − k
) = (10.9091i + 10.9091j − 12.7273k)lb
( ) ( ) ( ) or F2 = 10.91 lb i + 10.91 lb j − 12.73 lb k W
M2 = rH/B × F1 + M1
i j k
⋅ ( − 5.3955) i + 3 j −
7 k
31 2 0 20 lb in. (480 lb in.)
= − + ⋅
11.0 9.3333
6 6 −
7
= (25.455i + 394.55j + 360k)lb⋅in.
+(−277.48i + 154.285j − 360k)lb⋅in.
( ) ( ) M2 = − 252.03 lb⋅in. i + 548.84 lb⋅in. j
( ) ( ) or M2 = − 21.0 lb⋅ft i + 45.7 lb⋅ft j W

(a) a: ΣFy : Ra = −400 N − 600 N
or Ra = 1000 N W
ΣMB : Ma = (2 kN⋅m) + (2 kN⋅m) + (5 m)(400 N)
or Ma = 6.00 kN⋅m W
b: ΣFy : Rb = −1200 N + 200 N
or Rb = 1000 N W
ΣMB : Mb = (0.6 kN⋅m) + (5 m)(1200 N)
or Mb = 6.60 kN⋅m W
c: ΣFy : Rc = 200 N − 1200 N
or Rc = 1000 N W
ΣMB : Mc = −(4 kN⋅m) − (1.6 kN⋅m) − (5 m)(200 N)
or Mc = 6.60 kN⋅m W
d : ΣFy : Rd = −800 N − 200 N
or Rd = 1000 N W
ΣMB : Md = −(1.6 kN⋅m) + (4.2 kN⋅m) + (5 m)(800 N)
or Md = 6.60 kN⋅m W
continued

e: ΣFy : Re = −500 N − 400 N
or Re = 900 N W
ΣMB : Me = (3.8 kN⋅m) + (0.3 kN⋅m) + (5 m)(500 N)
or Me = 6.60 kN⋅m W
f : ΣFy : Rf = 400 N − 1400 N
or R f = 1000 N W
ΣMB : M f = (8.6 kN⋅m) − (0.8 kN⋅m) − (5 m)(400 N)
or Mf = 5.80 kN⋅m W
g: ΣFy : Rg = −1200 N + 300 N
or Rg = 900 N W
ΣMB : Mg = (0.3 kN⋅m) + (0.3 kN⋅m) + (5 m)(1200 N)
or Mg = 6.60 kN⋅m W
h: ΣFy : Rh = −250 N − 750 N
or Rh = 1000 N W
ΣMB : Mh = −(0.65 kN⋅m) + (6 kN⋅m) + (5 m)(250 N)
or Mh = 6.60 kN⋅m W
(b) The equivalent loadings are (b), (d), (h) W

The equivalent force-couple system at B is...
ΣFy : R = −650 N − 350 N
or R = 1000 N
ΣMB : M = (1.6 m)(800 N) + (1.27 kN⋅m) + (5 m)(650 N)
or M = 5.80 kN⋅m
∴ The equivalent loading of Problem 3.98 is (f) W

Equivalent force system...
(a) ΣFy : R = −400 N − 200 N
or R = 600 N W
ΣMA : −d (600 N) = −(200 N⋅m) + (100 N⋅m) − (4 m)(200 N)
or d = 1.500 m W
(b) ΣFy : R = −400 N + 100 N
or R = 300 N W
ΣMA : −d (300 N) = −(200 N⋅m) − (600 N⋅m) + (4 m)(100 N)
or d = 1.333 m W
(c) ΣFy : R = −400 N − 100 N
or R = 500 N W
ΣMA : −d (500 N) = −(200 N⋅m) − (200 N⋅m) − (4 m)(100 N)
or d = 1.600 m W

The equivalent force-couple system at A for each of the five force-couple systems will be determined and
compared to
F = (2 lb) j M = (48 lb⋅in.)i + (32 lb⋅in.)k
To determine if they are equivalent
Force-couple system at B:
Have ΣF: F = (2 lb) j
and ( ) ΣMA : M = ΣMB + rB/A × FB
M = (32 lb⋅in.)i + (16 lb⋅in.)k + (8 in.)i × (2 lb) j
= (32 lb⋅in.)i + (32 lb⋅in.)k
∴ is not equivalent W
Force-couple system at C:
And ( ) ΣMA : M = MC + rC/A × FC
M = (68 lb⋅in.)i + (8 in.)i + (10 in.)k × (2 lb) j
= (48 lb⋅in.)i + (16 lb⋅in.)k
continued

Force-couple system at E:
and ( ) ΣMA : M = ME + rE/A × FE
M = (48 lb⋅in.)i + (16 in.)i − (3.2 in.) j × (2 lb) j
= (48 lb⋅in.)i + (32 lb⋅in.)k
∴ is equivalent W
Force-couple system at G:
Have ΣF: F = (2 lb)i + (2 lb) j
F has two force components
Force-couple system at I:
and ( ) ΣMA : ΣMI + rI/A × FI
M = (80 lb⋅in.)i − (16 in.)k
+ (16 in.)i − (8 in.) j + (16 in.)k × (2 lb) j
M = (48 lb⋅in.)i + (16 lb⋅in.)k

First WA = mAg = (38 kg) g
WB = mBg = (29 kg) g
(a) WC = mCg = (27 kg) g
For resultant weight to act at C, ΣMC = 0
Then (38 kg) g (2 m) − (27 kg) g (d ) − (29 kg) g (2 m) = 0
76 58 0.66667 m
27
d −
∴ = =
or d = 0.667 m W
(b) WC = mCg = (24 kg) g
For resultant weight to act at C, ΣMC = 0
Then (38 kg) g (2 m) − (24 kg) g (d ) − (29 kg) g (2 m) = 0
76 58 0.75 m
24
d −
∴ = =
or d = 0.750 m W

ΣF −W − W − W = R
∴R = −200 lb − 175 lb − 135 lb
(a) Have :
C D E
= −510 lb
  Σ =−  
or R = 510 lb
Have :
A
ΣM
−(200 lb)(4.5 ft ) − (175 lb)(7.8 ft ) − (135 lb)(12.75 ft ) = − R(d )
∴−3986.3 lb⋅ft = (−510 lb)d
or d = 7.82 ft
(b) For equal reactions at A and B,
The resultant R must act at midspan.
From
2
A
L
M R
 
∴−(200 lb)(4.5 ft ) − (175 lb)(4.5 ft + a) − (135 lb)(4.5 ft + 2.5 a)
= −(510 lb)(9 ft )
or 2295 + 512.5 a = 4590
and a = 4.48 ft

Have ΣF: −12 kN − WL − 18 kN = −40 kN − 40 kN
WL = 50 kN
or WL = 50.0 kN W
ΣMB : (12 kN)(5 m) + (50 kN)d = (40 kN)(5 m)
d = 2.8 m
or heaviest load (50 kN) is located W
2.80 m from front axle

(a) ΣF: R = (80 N)i − (40 N) j − (60 N) j +(90 N)(−sin 50°i − cos50°j)
= (11.0560 N)i − (157.851 N) j
( )2 ( )2 R = 11.0560 N + −157.851 N
= 158.2 N
−
tan =
157.851
11.0560
θ
θ = 86.0°
or R = 158.2 N 86.0° W
(b)
ΣMF : d −(157.851 N) = (0.32 m)(80 N) − (0.15 m)(40 N) − (0.35 m)(60 N)
− (0.61 m)(90 N)cos50° − (0.16 m)(90 N)sin 50°
or d = 302 mm to the right of F W

(a) : 0 (0.32 m)(80 N) (0.1 m)(40 N) (0.1 m)(60 N) (0.36 m)(90 N)cos
ΣM = + − − α
I
−(0.16 m)(90 N)sinα
or 4sinα + 9cosα = 6.5556
( ) ( ) 2 2
9cosα = 6.5556 − 4sinα
( 2 ) 2 81 1 − sin α = 42.976 − 52.445sinα + 16sin α
2 97sin α − 52.445sinα − 38.024 = 0
Solving by the quadratic formula gives for the positive root
sinα = 0.95230
α = 72.233°
or α = 72.2°
Note: The second root (α = −24.3°) is rejected since 0 α 90°.
(b) ΣF: R = (80 N)i − (40 N) j − (60 N) j
+(90 N)(−sin 72.233°i − cos72.233°j)
= −(5.7075 N)i − (127.463 N) j
( ) ( ) 2 2
R = −5.7075 N + −127.463 N
= 127.6 N
127.463
θ = −
tan
5.7075
−
θ = 87.4°
or R = 127.6 N 87.4°

(a) Have ΣMD : 0 = M − (0.8 in.)(40 lb) − (2.9 in.)(20 lb)cos30° −(3.3 in.)(20 lb)sin 30°
or M = 115.229 lb⋅in.
or M = 115.2 lb⋅in. W
Now, R is oriented at 45° as shown (since its line of action
passes through B and D).
Have ΣFx′ : 0 = (40 lb)cos 45° − (20 lb)cos15°
−(90 lb)cos(α + 45°)
or α = 39.283°
or α = 39.3° W
(b) ΣFx : Rx = 40 − 20sin 30° − 90cos39.283°
= −39.663 lb
Now R = 2Rx or R = 56.1 lb 45.0° W

(a) Reduce system to a force and couple at B:
Have R = ΣF = −(10 lb) j + (25 lb)cos60°i + (25 lb)sin 60°j − (40 lb)i
= −(27.5 lb)i + (11.6506 lb) j
or ( )2 ( )2 R = −27.5 lb + 11.6506 lb = 29.866 lb
tan 1 11.6506 22.960
θ = −   = °
27.5
 
or R = 29.9 lb 23.0° W
Also MB = ΣMB = (80 lb⋅in.)k − (12 in.)i × (−10 lb) j − (8 in.) j × (−40 lb)i
= −(120 lb⋅in.)k
(b)
Have MB = −(120 lb⋅in.)k = −(u)i × (11.6506 lb) j
−(120 lb⋅in.)k = −(11.6506 lb)(u)k
u = 10.2999 in. and x = 12 in. − 10.2999 in.
= 1.7001 in.

Have MB = −(120 lb⋅in.)k = −(v) j × (−27.5 lb)i
−(120 lb⋅in.)k = −(27.5 lb)(v)k
v = 4.3636 in.
and y = 8 in. − 4.3636 in. = 3.6364 in.
or 1.700 in. to the right of A and 3.64 in. above C W

(a) Position origin along centerline of sheet metal at the intersection with
line EF.
(a) Have ΣF = R
R = −0.52 j − 1.05 j − 2.1(sin 45°i + cos 45° j) − 0.64i kips
R = −(2.1249 kips)i − (3.0549 kips) j
( ) ( ) 2 2
R = −2.1249 + −3.0549
= 3.7212 kips
− − 3.0549
θ = 1 tan   = 55.179
°  − 2.1249

or R = 3.72 kips 55.2°
M = ΣM
Have EF EF
Where M = (0.52 kip)(3.6 in.) +
(1.05 kips)(1.6 in.) EF
−(2.1 kips)(0.8 in.) − (0.64 kip)(1.6 in.)sin 45° + 1.6 in.
= 0.123923 kip⋅in.
To obtain distance d left of EF,
Have M = dR = d ( −
3.0549 kips) EF y
0.123923 kip ⋅ in.
d = = −
0.040565 in.
3.0549 kips
−
or d = 0.0406 in. left of EF
(b)
M = ΣM =
Have 0
EF EF
(0.52 kip)(3.6 in.) (1.05 kips)(1.6 in.) EF
M = +
−(2.1 kips)(0.8 in.)
−(0.64 kip)(1.6 in.)sinα + 1.6 in.
∴ (1.024 kip⋅in.)sinα = 0.848 kip⋅in.
or α = 55.9°

(a) Have ΣF = R
R = −0.52j − 1.05j − 2.1(sinαi + cosα j) − 0.64i kips
= − 0.64 kip + (2.1 kips)(sinα )i − 1.57 kips + (2.1 kips)cosα  j
R
R
Then tan 0.64 2.1sin
α
+
1.57 2.1cos
x
y
α
α
= =
+
1.57 tanα + 2.1sinα = 0.64 + 2.1sinα
tan 0.64
1.57
α =
α = 22.178°
or α = 22.2° W
(b) From α = 22.178°
Rx = −0.64 kip − (2.1 kips)sin 22.178°
= −1.43272 kips
Ry = −1.57 kips − (2.1 kips)cos 22.178°
= −3.5146 kips

( )2 ( )2 R = −1.43272 + −3.5146
= 3.7954 kips
or R = 3.80 kips 67.8°W
Then MEF = ΣMEF
Where MEF = (0.52 kip)(3.6 in.) + (1.05 kips)(1.6 in.) − (2.1 kips)(0.8 in.)
−(0.64 kip)(1.6 in.)sin 22.178° + 1.6 in.
= 0.46146 kip⋅in.
To obtain distance d left of EF,
Have MEF = dRy
= d (−3.5146 kips)
0.46146 kip in.
d ⋅
3.5146 kips
=
−
= −0.131298 in.
or d = 0.1313 in. left of EF W

Equivalent force-couple at A due to belts on pulley A
Have ΣF: −120 N − 160 N = RA
∴ RA = 280 N
Have ΣMA : −40 N(0.02 m) = MA
∴ MA = 0.8 N⋅m
Equivalent force-couple at B due to belts on pulley B
Have ΣF: (210 N + 150 N) 25° = RB
∴ RB = 360 N 25°
Have ΣMB : −60 N(0.015 m) = MB
∴ MB = 0.9 N⋅m
Equivalent force-couple at F
Have ΣF: RF = (−280 N) j + (360 N)(cos 25°i + sin 25°j)
= (326.27 N)i − (127.857 N) j
R = RF = RF2x + RF2y = (326.27)2 + (127.857)2 = 350.43 N
R
R
  − θ = tan − 1   = tan − 1  127.857    = − 21.399
°
326.27
Fy
Fx
   
or RF = R = 350 N 21.4°W

Have
ΣMF : MF = −(280 N)(0.06 m) − 0.80 N⋅m
− (360 N)cos 25° (0.010 m)
+ (360 N)sin 25° (0.120 m) − 0.90 N⋅m
MF = −(3.5056 N⋅m)k
To determine where a single resultant force will intersect line FE,
MF = dRy
3.5056 N m 0.027418 m 27.418 mm
127.857 N
F
y
d M
∴ = = = =
R
− ⋅
−
or d = 27.4 mmW

Cap 03 mecanica vectorial para ingenieros estatica 8ed.

Cap 03 mecanica vectorial para ingenieros estatica 8ed.

Recommended

Recommended

More Related Content

What's hot

What's hot (20)

Viewers also liked

Viewers also liked (8)

Similar to Cap 03 mecanica vectorial para ingenieros estatica 8ed.

Similar to Cap 03 mecanica vectorial para ingenieros estatica 8ed. (20)

Cap 03 mecanica vectorial para ingenieros estatica 8ed.