COSMOS Chapter 3 Solutions

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 M B = − rA/BQ

= − (0.225 m )(57.851 N )

= −13.0165 N ⋅ m

M B = 13.02 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.



Fx = ( 90 N ) cos 25°

= 81.568 N

Fy = ( 90 N ) sin 25°

= 38.036 N

x = ( 0.225 m ) cos 65°

= 0.095089 m

y = (0.225 m ) sin 65°

= 0.20392 m

M B = xFy − yFx

= ( 0.095089 m )( 38.036 N ) − ( 0.20392 m )( 81.568 N )

= −13.0165 N ⋅ m

M B = 13.02 N ⋅ m




Px = ( 3 lb ) sin 30°

= 1.5 lb

Py = ( 3 lb ) cos 30°

= 2.5981 lb

M A = xB/ A Py + yB/ A Px

= ( 3.4 in.)( 2.5981 lb ) + ( 4.8 in.)(1.5 lb )

= 16.0335 lb ⋅ in.

M A = 16.03 lb ⋅ in.




For P to be a minimum, it must be perpendicular to the line joining points
A and B

with rAB = ( 3.4 in.)2 + ( 4.8 in.)2
= 5.8822 in.

 y
α = θ = tan −1  
x  

 4.8 in. 
= tan −1  
 3.4 in. 

= 54.689°

Then M A = rAB Pmin

M A 19.5 lb ⋅ in.
or Pmin = =
rAB 5.8822 in.

= 3.3151 lb

∴ Pmin = 3.32 lb 54.7°

or Pmin = 3.32 lb 35.3°




By definition M A = rB/ A P sin θ

where θ = φ + ( 90° − α )
 4.8 in. 
and φ = tan −1  
 3.4 in. 
= 54.689°

Also rB/ A = ( 3.4 in.)2 + ( 4.8 in.)2
= 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°




(a) (a) M A = rB/ A × TBF

M A = xTBFy + yTBFx

= ( 2 m )( 200 N ) sin 60° + ( 0.4 m )( 200 N ) cos 60°

= 386.41 N ⋅ m

or M A = 386 N ⋅ m
(b)
(b) For FC to be a minimum, it must be perpendicular to the line
joining A and C.

∴ M A = d ( FC )min

with d = ( 2 m )2 + (1.35 m )2
= 2.4130 m

Then 386.41 N ⋅ m = ( 2.4130 m ) ( FC )min

( FC )min = 160.137 N

 1.35 m 
and φ = tan −1   = 34.019°
 2m 

θ = 90 − φ = 90° − 34.019° = 55.981°

∴ ( FC )min = 160.1 N 56.0°




(a) M A = xTBFy + yTBFx

= ( 2 m )( 200 N ) sin 60° + ( 0.4 m )( 200 N ) cos 60°
= 386.41 N ⋅ m
or M A = 386 N ⋅ m

(b)
Have M A = xFC
MA 386.41 N ⋅ m
or FC = =
x 2m
= 193.205 N
∴ FC = 193.2 N
(c) For FB to be minimum, it must be perpendicular to the line joining A
and B
∴ M A = d ( FB )min

with d = ( 2 m )2 + ( 0.40 m )2 = 2.0396 m

Then 386.41 N ⋅ m = ( 2.0396 m ) ( FC )min

( FC )min = 189.454 N

 2m 
and θ = tan −1   = 78.690°
 0.4 m 
or ( FC )min = 189.5 N 78.7°




( )
(a)
M B = rA/B cos15° W

= (14 in.)( cos15° )( 5 lb )
= 67.615 lb ⋅ in.
or M B = 67.6 lb ⋅ in.

(b)
M B = rD/B P sin 85°

67.615 lb ⋅ in. = ( 3.2 in.) P sin 85°
or P = 21.2 lb

(c) For ( F )min, F must be perpendicular to BC.

Then, M B = rC/B F

67.615 lb ⋅ in. = (18 in.) F
or F = 3.76 lb 75.0°




35 in. 5
(a) Slope of line EC = =
76 in. + 8 in. 12

12
Then TABx = (TAB )
13

12
= ( 260 lb ) = 240 lb
13

5
and TABy = ( 260 lb ) = 100 lb
13

Then M D = TABx ( 35 in.) − TABy ( 8 in.)

= ( 240 lb )( 35 in.) − (100 lb )( 8 in.)

= 7600 lb ⋅ in.

or M D = 7600 lb ⋅ in.

(b) Have M D = TABx ( y ) + TABy ( x )

= ( 240 lb )( 0 ) + (100 lb )( 76 in.)

= 7600 lb ⋅ in.

or M D = 7600 lb ⋅ in.




35 in. 7
Slope of line EC = =
112 in. + 8 in. 24
24
Then TABx = TAB
25
7
and TABy = TAB
25
Have M D = TABx ( y ) + TABy ( x )

24 7
∴ 7840 lb ⋅ in. = TAB ( 0 ) + TAB (112 in.)
25 25
TAB = 250 lb

or TAB = 250 lb




The minimum value of d can be found based on the equation relating the moment of the force TAB about D:

M D = (TAB max ) y ( d )

where M D = 1152 N ⋅ m

(TAB max ) y = TAB max sin θ = ( 2880 N ) sin θ

1.05 m
Now sin θ =
(d + 0.24 ) + (1.05 ) m
2 2

 
1.05
∴ 1152 N ⋅ m = 2880 N   (d )
 

 (d + 0.24 ) + (1.05 )
2 2



or ( d + 0.24 )2 + (1.05)2 = 2.625d

or (d + 0.24 ) + (1.05 ) = 6.8906d 2
2 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




with d AB = ( 42 mm )2 + (144 mm )2
= 150 mm

42 mm
sin θ =
150 mm

144 mm
cosθ =
150 mm

and FAB = − FAB sin θ i − FAB cosθ j

2.5 kN
= ( − 42 mm ) i − (144 mm ) j
150 mm  

= − ( 700 N ) i − ( 2400 N ) j

Also rB/C = − ( 0.042 m ) i + ( 0.056 m ) j

Now M C = rB/C × FAB

= ( − 0.042 i + 0.056 j) × ( − 700 i − 2400 j) N ⋅ m

= (140.0 N ⋅ m ) k

or M C = 140.0 N ⋅ m




with d AB = ( 42 mm )2 + (144 mm )2
= 150 mm

42 mm
sin θ =
150 mm

144 mm
cosθ =
150 mm

FAB = − FAB sin θ i − FAB cosθ j

2.5 kN
= ( − 42 mm ) i − (144 mm ) j
150 mm  

= − ( 700 N ) i − ( 2400 N ) j

Also rB/C = − ( 0.042 m ) i − ( 0.056 m ) j

Now M C = rB/C × FAB

= ( − 0.042 i − 0.056 j) × ( − 700i − 2400 j) N ⋅ m

= ( 61.6 N ⋅ m ) k

or M C = 61.6 N ⋅ m




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




Note: B = B ( cos β i + sin β j)

B′ = B ( cos β i − sin β j)

C = C ( cos α i + sin α j)

By definition: B × C = BC sin (α − β ) (1)

B′ × C = BC sin (α + β ) (2)

Now ... B × C = B ( cos β i + sin β j) × C ( cos α i + sin α j)

= BC ( cos β sin α − sin β cos α ) k (3)

and B′ × C = B ( cos β i − sin β j) × C ( cos α i + sin α j)

= BC ( cos β sin α + sin β cos α ) k (4)

Equating the magnitudes of B × C from equations (1) and (3) yields:

BC sin (α − β ) = BC ( cos β sin α − sin β cos α ) (5)

Similarly, equating the magnitudes of B′ × C from equations (2) and (4) yields:

BC sin (α + β ) = BC ( cos β sin α + sin β cos α ) (6)

Adding equations (5) and (6) gives:

sin (α − β ) + sin (α + β ) = 2cos β sin α

1 1
or sin α cos β = sin (α + β ) + sin (α − β )
2 2




Have d = λ AB × rO/ A

rB/ A
where λ AB =
rB/ A

and rB/ A = ( −210 mm − 630 mm ) i

+ ( 270 mm − ( −225 mm ) ) j

= − ( 840 mm ) i + ( 495 mm ) j

rB/ A = ( −840 mm )2 + ( 495 mm )2
= 975 mm

− ( 840 mm ) i + ( 495 mm ) j
Then λ AB =
975 mm
1
= ( −56i + 33j)
65
Also rO/ A = ( 0 − 630 ) i + ( 0 − (−225) ) j

= − ( 630 mm ) i + ( 225 mm ) j

1
∴d = ( −56i + 33j) × − ( 630 mm ) i + ( 225 mm ) j
 
65
= 126.0 mm
d = 126.0 mm




A×B
(a) λ =
A×B
where A = 12i − 6 j + 9k
B = − 3i + 9 j − 7.5k
Then
i j k
A × B = 12 − 6 9
− 3 9 − 7.5

= ( 45 − 81) i + ( −27 + 90 ) j + (108 − 18 ) k

= 9 ( − 4i + 7 j + 10k )

And A × B = 9 (− 4) 2 + (7)2 + (10)2 = 9 165
9 ( − 4i + 7 j + 10k )
∴λ =
9 165
1
or λ = ( − 4i + 7 j + 10k )
165
A×B
(b) λ =
A×B

where A = −14i − 2 j + 8k

B = 3i + 1.5j − k

Then

i j k
A × B = −14 − 2 8
3 1.5 −1

= ( 2 − 12 ) i + ( 24 − 14 ) j + ( −21 + 6 ) k
= 5 ( −2i + 2 j − 3k )

and A × B = 5 (−2)2 + (2)2 + (−3)2 = 5 17
5 ( −2i + 2 j − 3k )
∴λ =
5 17
1
or λ = ( − 2i + 2 j − 3k )
17




(a) Have A = P × Q

i j k
P × Q = 3 7 −2 in.2
−5 1 3

= [ (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

or A = 44.4 in.2

(b) A = P×Q

i j k
P × Q = 2 − 4 3 in.2
6 −1 5

= [ (−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




(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 − 15.75 j − 40.5k ) N ⋅ m
or M O = − (18.00 N ⋅ m ) i − (15.75 Ν ⋅ m ) j − ( 40.5 N ⋅ m ) k
(b) Have MO = r × F
i j k
= 2 − 0.75 −1 N ⋅ m
7.5 3 − 4.5

= [ (3.375 + 3)i + (−7.5 + 9) j + (6 + 5.625)k ] N ⋅ m
= ( 6.375i + 1.500 j + 11.625k ) N ⋅ m
or M O = ( 6.38 N ⋅ m ) i + (1.500 Ν ⋅ m ) j + (11.63 Ν ⋅ m ) k
(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 M O = 0
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 M O = − ( 24.0 lb ⋅ ft ) i + (12.00 lb ⋅ ft ) j + ( 36.0 lb ⋅ ft ) k

(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 M O = 0

(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 M O = − ( 76.0 lb ⋅ ft ) i − (10.00 lb ⋅ ft ) j + ( 42.0 lb ⋅ ft ) k




With TAB = − ( 369 N ) j

TAB = TAD
AD
= ( 369 N )
( 2.4 m ) i − ( 3.1 m ) j − (1.2 m ) k
AD ( 2.4 m )2 + ( −3.1 m )2 + ( −1.2 m )2

TAD = ( 216 N ) i − ( 279 N ) j − (108 N ) k

Then R A = 2 TAB + TAD

= ( 216 N ) i − (1017 N ) j − (108 N ) k

Also rA/C = ( 3.1 m ) i + (1.2 m ) k

Have M C = rA/C × R A

i j k
= 0 3.1 1.2 N ⋅ m
216 −1017 −108

= ( 885.6 N ⋅ m ) i + ( 259.2 N ⋅ m ) j − ( 669.6 N ⋅ m ) k

M C = ( 886 N ⋅ m ) i + ( 259 N ⋅ m ) j − ( 670 N ⋅ m ) k




Have M A = 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
and M A = 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

M A = ( 4.81 N ⋅ m ) i + ( 4.61 N ⋅ m ) j − ( 5.74 N ⋅ m ) k




Have M O = 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

AB
T2 = T2
AB

= ( 62 lb )
( 5 ft ) i − ( 30 ft ) j + ( 6 ft ) k
( 5 ft )2 + ( − 30 ft )2 + ( 6 ft )2
= (10 lb ) i − ( 60 lb ) j + (12 lb ) k

∴ R = − ( 51.058 lb ) i − ( 70.766 lb ) j + (12 lb ) k

i j k
MO = 0 30 3 lb ⋅ ft
− 51.058 −70.766 12

= ( 572.30 lb ⋅ ft ) i − (153.17 lb ⋅ ft ) j + (1531.74 lb ⋅ ft ) k

M O = ( 572 lb ⋅ ft ) i − (153.2 lb ⋅ ft ) j + (1532 lb ⋅ ft ) k




(a) Have M O = rB/O × TBD

where rB/O = ( 2.5 m ) i + ( 2 m ) j

BD
TBD = TBD
BD

 − (1 m ) i − ( 2 m ) j + ( 2 m ) k 
= ( 900 N )  
( −1 m ) + ( − 2 m ) + ( 2 m )
2 2 2

= − ( 300 N ) i − ( 600 N ) j + ( 600 N ) k

Then

i j k
MO = 2.5 2 0 N⋅m
− 300 − 600 600

M O = (1200 N ⋅ m ) i − (1500 N ⋅ m ) j − ( 900 N ⋅ m ) k

continued



(b) Have M O = rB/O × TBE

where rB/O = ( 2.5 m ) i + ( 2 m ) j

BE
TBE = TBE
BE

 − ( 0.5 m ) i − ( 2 m ) j − ( 4 m ) k 
= ( 675 N )  
( 0.5 m ) + ( −2 m ) + ( − 4 m )2
2 2

= − ( 75 N ) i − ( 300 N ) j − ( 600 N ) k

Then

i j k
MO = 2.5 2 0 N⋅m
− 75 − 300 − 600

M O = − (1200 N ⋅ m ) i + (1500 N ⋅ m ) j − ( 600 N ⋅ m ) k




Have M C = rA/C × P

where
rA/C = rB/C + rA/B

= (16 in.)( − cos80° cos15°i − sin 80° 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 )( cos 5° cos 70°i + sin 5° j − cos 5° sin 70°k )

= ( 51.108 lb ) i + (13.0734 lb ) j − (140.418 lb ) k

Then
i j k
M C = −7.7053 −1.47360 −2.0646 lb ⋅ in.
51.108 13.0734 −140.418

= ( 233.91 lb ⋅ in.) i − (1187.48 lb ⋅ in.) j − ( 25.422 lb ⋅ in.) k

or M C = (19.49 lb ⋅ ft ) i − ( 99.0 lb ⋅ ft ) j − ( 2.12 lb ⋅ ft ) k




Have M C = rA/C × FBA

where rA/C = ( 0.96 m ) i − ( 0.12 m ) j + ( 0.72 m ) k

and FBA = λ BA FBA

 
=  − ( 0.1 m ) i + (1.8 m ) j − ( 0.6 m ) k  ( 228 N )
 

 ( 0.1)2 + (1.8)2 + ( 0.6 )2 m  

= − (12.0 N ) i + ( 216 N ) j − ( 72 N ) k

i j k
∴ M C = 0.96 −0.12 0.72 N ⋅ m
−12.0 216 −72

= − (146.88 N ⋅ m ) i + ( 60.480 N ⋅ m ) j + ( 205.92 N ⋅ m ) k

or M C = − (146.9 N ⋅ m ) i + ( 60.5 N ⋅ m ) j + ( 206 N ⋅ m ) k




Have M C = TAD d

where d = Perpendicular distance from C to line AD
with M C = rA/C TAD

and rA/C = ( 3.1 m ) j + (1.2 m ) k

AD
TAD = TAD
AD
( 2.4 m ) i − ( 3.1 m ) j − (1.2 m ) k 
TAD = ( 369 N )  
( 2.4 m ) + ( − 3.1 m ) + ( −1.2 m )2
2 2

= ( 216 N ) i − ( 279 N ) j − (108 N ) k

Then
i j k
MC = 0 3.1 1.2 N ⋅ m
216 − 279 −108

= ( 259.2 N ⋅ m ) j − ( 669.6 N ⋅ m ) k

and MC = ( 259.2 N ⋅ m )2 + ( −669.6 N ⋅ m )2
= 718.02 N ⋅ m

∴ 718.02 N ⋅ m = ( 369 N ) d

or d = 1.946 m




Have M O = TAC d
where d = Perpendicular distance from O to rope AC
with M O = 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
MO = 0 30 3 lb ⋅ ft
− 61.058 −10.766 0

= ( 32.298 lb ⋅ ft ) i − (183.174 lb ⋅ ft ) j + (1831.74 lb ⋅ ft ) k

and MO = ( 32.298 lb ⋅ ft )2 + ( −183.174 lb ⋅ ft )2 + (1831.74 lb ⋅ ft )2
= 1841.16 lb ⋅ ft

∴ 1841.16 lb ⋅ ft = ( 62 lb ) d

or d = 29.7 ft




Have M O = TAB d
where d = Perpendicular distance from O to rope AB
with M O = rA/O × TAB

and rA/O = ( 30 ft ) j + ( 3 ft ) k

AB
TAB = TAB
AB
( 5 ft ) i − ( 30 ft ) j + ( 6 ft ) k 
= ( 62 lb )  
( 5 ft ) + ( − 30 ft ) + ( 6 ft )2
2 2

= (10 lb ) i − ( 60 lb ) j + (12 lb ) k

Then
i j k
MO = 0 30 3 lb ⋅ ft
10 − 60 12

= ( 540 lb ⋅ ft ) i + ( 30 lb ⋅ ft ) j − ( 300 lb ⋅ ft ) k

and MO = ( 540 lb ⋅ ft )2 + ( 30 lb ⋅ ft )2 + ( −300 lb ⋅ ft )2
= 618.47 lb ⋅ ft

∴ 618.47 lb ⋅ ft = ( 62 lb ) d

or d = 9.98 ft




Have M C = TBD d
where d = Perpendicular distance from C to cable BD
with M C = rB/C × TB/D

and rB/C = ( 2 m ) j

BD
TBD = TBD
BD
 − (1 m ) i − ( 2 m ) j + ( 2 m ) k 
= ( 900 N )  
( −1 m ) + ( − 2 m ) + ( 2 m )2
2 2

= − ( 300 N ) i − ( 600 N ) j + ( 600 N ) k

Then
i j k
MC = 0 2 0 N⋅m
−300 − 600 600

= (1200 N ⋅ m ) i + ( 600 N ⋅ m ) k

and MC = (1200 N ⋅ m )2 + ( 600 N ⋅ m )2
= 1341.64 N ⋅ m

∴ 1341.64 = ( 900 N ) d

or d = 1.491 m




Have M C = Pd

From the solution of problem 3.25

M C = ( 233.91 lb ⋅ in.) i − (1187.48 lb ⋅ in.) j − ( 25.422 lb ⋅ in.) k

Then

MC = ( 233.91)2 + ( −1187.48)2 + ( −25.422 )2
= 1210.57 lb ⋅ in.

MC 1210.57 lb.in.
and d = =
P 150 lb

or d = 8.07 in.




Have | M D | = FBAd
where d = perpendicular distance from D to line AB.

M D = rA/D × FBA

rA/D = − ( 0.12 m ) j + ( 0.72 m ) k

FBA = λ BA FBA =
( − ( 0.1 m ) i + (1.8 m ) j − ( 0.6 m ) k ) ( 228 N )
( 0.1)2 + (1.8)2 + ( 0.6 )2 m

= − (12.0 N ) i + ( 216 N ) j − ( 72 N ) k

i j k
∴ MD = 0 −0.12 0.72 N ⋅ m
−12.0 216 −72

= − (146.88 N ⋅ m ) i − ( 8.64 N ⋅ m ) j − (1.44 N ⋅ m ) k

and |MD | = (146.88)2 + (8.64 )2 + (1.44 )2 = 147.141 N ⋅ m

∴ 147.141 N ⋅ m = ( 228 N ) d

d = 0.64536 m
or d = 0.645 m




Have | M C | = FBAd

where d = perpendicular distance from C to line AB.

M C = rA/C × FBA

rA/C = ( 0.96 m ) i − ( 0.12 m ) j + ( 0.72 m ) k

FBA = λ BA FBA =
( − ( 0.1 m ) i + (1.8 m ) j − ( 0.6 ) k ) ( 228 N )
( 0.1)2 + (1.8)2 + ( 0.6 )2 m

= − (12.0 N ) i + ( 216 N ) j − ( 72 N ) k

i j k
∴ M C = 0.96 −0.12 0.72 N ⋅ m
−12.0 216 −72

= − (146.88 N ⋅ m ) i − ( 60.48 N ⋅ m ) j + ( 205.92 N ⋅ m ) k

and | MC | = (146.88)2 + ( 60.48)2 + ( 205.92 )2 = 260.07 N ⋅ m

∴ 260.07 N ⋅ m = ( 228 N ) d

d = 1.14064 m
or d = 1.141 m




(a) Have d = rC/ A sin θ = λ AB × rC/ A

where d = Perpendicular distance from C to pipe AB
AB 7i + 4 j − 32k
with λ AB = =
AB ( 7 )2 + ( 4 )2 + ( −32 )2
1
= ( 7i + 4 j − 32k )
33
and rC/ A = − (14 ft ) i + ( 5 ft ) j + ( L − 22 ) ft  k
 
i j k
1
Then λ AB × rC/ A = 7 4 − 32 ft
33
−14 5 L − 22

=
1
33 
{      }
 4 ( L − 22 ) + 32 ( 5 )  i + 32 (14 ) − 7 ( L − 22 )  j +  7 ( 5 ) + 4 (14 )  k ft

1
= ( 4L + 72 ) i + ( −7 L + 602 ) j + 91k  ft
33  
1
and d = ( 4L + 72 )2 + ( −7 L + 602 )2 + ( 91)2
33
dd 2 1
For (d ) min , = 2  2 ( 4 )( 4L + 72 ) + 2 ( −7 )( −7 L + 602 )  = 0
dL 33  
or 65L − 3926 = 0
or L = 60.400 ft
But L > Lgreenhouse so L = 30.0 ft
1
(b) with L = 30 ft, d = ( 4 × 30 + 72 )2 + ( −7 × 30 + 602 )2 + ( 91)2
33
or d = 13.51 ft
Note: with L = 60.4 ft,
1
d = ( 4 × 60.4 + 72 )2 + ( −7 × 60.4 + 602 )2 + ( 91)2 = 11.29 ft
33




P ⋅ Q = ( − 4i + 8j − 3k ) ⋅ ( 9i − j − 7k )

= ( − 4 )( 9 ) + ( 8 )( −1) + ( − 3)( − 7 )
= − 23
or P ⋅ Q = −23

P ⋅ S = ( − 4i + 8 j − 3k ) ⋅ ( 5i − 6 j + 2k )

= ( − 4 )( 5 ) + ( 8 )( − 6 ) + ( − 3)( 2 )
= − 74
or P ⋅ S = −74

Q ⋅ S = ( 9i − j − 7k ) ⋅ ( 5i − 6 j + 2k )

= ( 9 )( 5 ) + ( −1)( − 6 ) + ( − 7 )( 2 )
= 37
or Q ⋅ S = 37




By definition
B ⋅ C = BC cos (α − β )

where B = B ( cos β ) i + ( sin β ) j
 

C = C ( cos α ) i + ( sin α ) j
 

∴ ( B cos β )( C cos α ) + ( B sin β )( C sin α ) = BC cos (α − β )

or cos β cos α + sin β sin α = cos (α − β ) (1)

By definition
B′⋅ C = BC cos (α + β )

where B′ = ( cos β ) i − ( sin β ) j
 

∴ ( B cos β )( C cos α ) + ( − B sin β )( C sin α ) = BC cos (α + β )

or cos β cos α − sin β sin α = cos (α + β ) (2)

Adding Equations (1) and (2),
2 cos β cos α = cos (α − β ) + cos (α + β )

1 1
or cos α cos β = cos (α + β ) + cos (α − β )
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

rB/ A = ( 0.56 m )2 + ( 0.9 m )2 = 1.06 m

rC/ A = ( 0.9 m )2 + ( − 0.48 m )2 = 1.02 m

rD/ A = ( − 0.52 m )2 + ( 0.9 m )2 + ( 0.36 m )2 = 1.10 m

By definition rB/ A ⋅ rD/ A = rB/ A rD/ A cosθ

or ( 0.56i + 0.9 j) ⋅ ( − 0.52i + 0.9 j + 0.36k ) = (1.06 )(1.10 ) cosθ
( 0.56 )( − 0.52 ) + ( 0.9 )( 0.9 ) + ( 0 )( 0.36 ) = 1.166 cosθ
cosθ = 0.44494
θ = 63.6°




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 ( 0.9 j − 0.48k ) ⋅ ( − 0.52i + 0.9 j + 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°




(a) By definition

λ BC + λ EF = (1) (1) cosθ

where

λ BC =
( 32 ft ) i − ( 9 ft ) j − ( 24 ft ) k
( 32 )2 + ( − 9 )2 + ( − 24 )2 ft
1
= ( 32i − 9 j − 24k )
41

− (14 ft ) i − (12 ft ) j + (12 ft ) k
λ EF =
( −14 )2 + ( −12 )2 + ( 12 )2 ft
1
= ( −7i − 6 j + 6k )
11

Therefore
( 32i − 9 j − 24k ) ⋅ ( −7i − 6 j + 6k ) = cosθ
41 11

( 32 )( −7 ) + ( −9 )( −6 ) + ( −24 )( 6 ) = ( 41)(11) cosθ
cosθ = − 0.69623

or θ = 134.1°

(b) By definition (TEG ) BC = (TEF ) cosθ

= (110 lb )( −0.69623)

= −76.585 lb

or (TEF ) BC = −76.6 lb




(a) By definition

λ BC ⋅ λ EG = (1) (1) cosθ

where

λ BC =
( 32 ft ) i − ( 9 ft ) j − ( 24 ft ) k
( 32 )2 + ( − 9 )2 + ( − 24 )2 ft
1
= ( 32i − 9 j − 24k )
41

λ EG =
(16 ft ) i − (12 ft ) j + ( 9.75) k
(16 )2 + ( −12 )2 + ( 9.75)2 ft
1
= (16i − 12 j + 9.75k )
22.25

Therefore
( 32i − 9 j − 24k ) ⋅ (16i − 12 j + 9.75k ) = cosθ
41 22.25

( 32 )(16 ) + ( −9 )( −12 ) + ( −24 )( 9.75) = ( 41)( 22.25) cosθ
cosθ = 0.42313

or θ = 65.0°

(b) By definition (TEG )BC = (TEG ) cosθ

= (178 lb )( 0.42313)

= 75.317 lb

or (TEG ) BC = 75.3 lb




First locate point B:
d 3.5
=
22 14
or d = 5.5 m

(a) d BA = ( 5.5 + 0.5)2 + ( −22 )2 + ( −3)2 = 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

d BD = ( −8.6226 + 5.5)2 + (19.3033 − 22 )2 + (1.37260 − 0 )2 m

= 4.3482 m

and cosθ ABD =
d BA ⋅ d BD
=
( 6i − 22 j − 3k ) ⋅ ( −3.1226i − 2.6967 j + 1.37260k )
d BA d BD ( 23)( 4.3482 )
= 0.36471
or θ ABD = 68.6°
(b) (TBA )BD = TBA cosθ ABD

= ( 230 N )( 0.36471)

or (TBA ) BD = 83.9 N




First locate point B:
d 3.5
=
22 14
or d = 5.5 m

(a) Locate point D:

( −3.5 − 7.5sin 45° cos10° ) , (14 + 7.5cos 45° ) ,


( 0 + 7.5sin 45° sin10° ) m


or ( −8.7227 m, 19.3033 m, 0.92091 m )

Then

d DC = ( 5.2227 m ) i − ( 5.3033 m ) j − ( 0.92091 m ) k

and

d DB = ( −5.5 + 8.7227 )2 + ( 22 − 19.3033)2 + ( 0 − 0.92091)2 m

= 4.3019 m

and cos θ BDC =
d DB ⋅ d DC
=
( 3.2227i + 2.6967 j − 0.92091k ) ⋅ ( 5.2227i − 5.3033j − 0.92091k )
d DB d DC ( 4.3019 )( 7.5)
= 0.104694

or θ BDC = 84.0°
(b) (TBD )DC = TBD cosθ BDC = ( 250 N )( 0.104694 )

or (TBD )DC = 26.2 N




Volume of parallelopiped is found using the mixed triple product
(a) Vol = P ⋅ ( Q × S )

3 −4 1
= − 7 6 − 8 in.3
9 −2 −3

= ( −54 + 288 + 14 − 48 + 84 − 54 ) in.3

= 230 in.3

or Volume = 230 in.3

(b) Vol = P ⋅ ( Q × S )

−5 −7 4
= 6 − 2 5 in.3
−4 8 −9

= ( −90 + 140 + 192 + 200 − 378 − 32 ) in.3

= 32 in.3

or Volume = 32 in.3




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 − 10S y − 12S y + 84 − 40

So that 22 S y = 286

S y = 13

or S y = 13.00




Have rC = ( 2.25 m ) k

CE
TCE = TCE
CE
( 0.90 m ) i + (1.50 m ) j − ( 2.25 m ) k 
TCE = (1349 N )  
( 0.90 ) + (1.50 ) + ( −2.25 ) m
2 2 2

= ( 426 N ) i + ( 710 N ) j − (1065 N ) k

Now M O = rC × TCE

i j k
= 0 0 2.25 N ⋅ m
426 710 −1065
= − (1597.5 N ⋅ m ) i + ( 958.5 N ⋅ m ) j

∴ M x = −1598 N ⋅ m, M y = 959 N ⋅ m, M z = 0




Have rE = ( 0.90 m ) i + (1.50 m ) j

DE
TDE = TDE
DE

 − ( 2.30 m ) i + (1.50 m ) j − ( 2.25 m ) k 
= (1349 N )  
( − 2.30 ) + (1.50 ) + ( − 2.25) m
2 2 2

= − ( 874 N ) i + ( 570 N ) j − ( 855 N ) k

Now M O = 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

∴ M x = −1283 N ⋅ m, M y = 770 N ⋅ m, M z = 1824 N ⋅ m




Have M z = k ⋅ ( rB ) y × TBA  + k ( rC ) y × TCD 
   

where M z = − ( 48 lb ⋅ ft ) k

( rB ) y = ( rC ) y = ( 3 ft ) j

BA ( 4.5 ft ) i − ( 3 ft ) j + ( 9 ft ) k 
TBA = TBA = (14 lb )  
BA ( 4.5) + ( − 3) + ( 9 ) ft
2 2 2

= ( 6 lb ) i − ( 4 lb ) j + (12 lb ) k

CD ( 6 ft ) i − ( 3 ft ) j − ( 6 ft ) k 
TCD = TCD = TCD  
CD ( 6 ) + ( − 3) + ( − 6 ) ft
2 2 2

TCD
= (2i − j − 2k )
3

Then {  }
− ( 48 lb ⋅ ft ) = k ⋅ ( 3 ft ) j × ( 6 lb ) i − ( 4 lb ) j + (12 lb ) k 


 T 
+ k ⋅ ( 3 ft ) j ×  CD ( 2 i − j − 2 k )  
  3 

or − 48 = −18 − 2TCD

TCD = 15.00 lb




Have M y = j ⋅ ( rB ) z × TBA  × j⋅ ( rC ) z × TCD 
   

where M y = 156 lb ⋅ ft

( rB ) z = ( 24 ft ) k; ( rC ) z = ( 6 ft ) k

BA ( 4.5 ft ) i − ( 3 ft ) j + ( 9 ft ) k 
TBA = TBA = TBA  
BA ( 4.5) + ( − 3) + ( 9 ) ft
2 2 2

TBA
= ( 4.5i − 3j + 9k )
10.5

CD ( 6 ft ) i − ( 3 ft ) j + ( 9 ft ) k 
TCD = TCD = ( 7.5 lb )  
CD ( 6 ) + ( − 3) + ( 9 ) ft
2 2 2

= ( 5 lb ) i − ( 2.5 lb ) j − ( 5 lb ) k

 T 
Then (156 lb ⋅ ft ) = j ⋅ ( 24 ft ) k × BA ( 4.5i − 3j + 9k ) 
 10.5 

{ }
+ j ⋅ ( 6 ft ) k × ( 5 lb ) i − ( 2.5 lb ) j − ( 5 lb ) k 
 

108
or 156 = TBA + 30
10.5
TBA = 12.25 lb




Based on M x = ( P cos φ ) ( 0.225 m ) sin θ  − ( P sin φ ) ( 0.225 m ) cosθ 
    (1)

M y = − ( P cos φ )( 0.125 m ) (2)

M z = − ( P sin φ )( 0.125 m ) (3)

Equation ( 3) M z − ( P sin φ )( 0.125 )
By : =
Equation ( 2 ) M y − ( P cos φ )( 0.125 )

−4
or = tan φ ∴ φ = 9.8658°
− 23
or φ = 9.87°
From Equation (2)
− 23 N ⋅ m = − ( P cos 9.8658° )( 0.125 m )
P = 186.762 N
or P = 186.8 N
From Equation (1)
26 N ⋅ m = (186.726 N ) cos 9.8658° ( 0.225 m ) sin θ 
  
− (186.726 N ) sin 9.8658°  ( 0.225 m ) cosθ 
  
or 0.98521sin θ − 0.171341cosθ = 0.61885
Solving numerically,
θ = 48.1°




Based on M x = ( P cos φ ) ( 0.225 m ) sin θ  − ( P sin φ ) ( 0.225 m ) cosθ 
    (1)

M y = − ( P cos φ )( 0.125 m ) (2)

M z = − ( P sin φ )( 0.125 m )

Equation ( 3) M z − ( P sin φ )( 0.125 )
By : =
Equation ( 2 ) M y − ( P cos φ )( 0.125 )

− 3.5
or = tan φ ; φ = 9.9262°
− 20
From Equation (3):
− 3.5 N ⋅ m = − ( P sin 9.9262° )( 0.125 m )

P = 162.432 N
From Equation (1):
M x = (162.432 N )( 0.225 m )( cos 9.9262° sin 60° − sin 9.9262° cos 60° )

= 28.027 N ⋅ m
or M x = 28.0 N ⋅ m


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