Title | Sol Cap 03 - Edicion 8 - Solucionario |
---|---|
Author | HENRY SANTAMARIA |
Course | Estática |
Institution | Universidad del Atlántico |
Pages | 180 |
File Size | 9.5 MB |
File Type | |
Total Downloads | 34 |
Total Views | 137 |
Solucionario...
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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.
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Chapter 3, Solution 2.
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
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.
COSMOS: Complete Online Solutions Manual Organization System
Chapter 3, Solution 3.
Px = ( 3 lb ) sin 30° = 1.5 lb
Py = ( 3 lb ) cos30 ° = 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.
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.
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Chapter 3, Solution 4.
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 x
α = θ = tan −1
4.8 in. = tan−1 3.4 in.
= 54.689° Then
M A = rAB Pmin
or
Pmin =
M A 19.5 lb⋅ in. = 5.8822 in. rAB
= 3.3151 lb ∴ Pmin = 3.32 lb
54.7° or
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.
Pmin = 3.32 lb
35.3°
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Chapter 3, Solution 5.
M A = rB /A P sin θ
By definition where
θ = φ + ( 90° − α )
and
φ = tan −1
4.8 in. 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°
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.
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Chapter 3, Solution 6.
(a)
(a) M A = rB/ A × TBF
M A = xT BFy + yT BFx = ( 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 d =
with
( 2 m) 2 + ( 1.35 m) 2
= 2.4130 m Then 386.41 N ⋅m = (2.4130 m )( FC )min
( FC )min and
= 160.137 N 1.35 m = 34.019° 2m
φ = tan −1
θ = 90 − φ = 90° − 34.019° = 55.981° ∴ (FC )min = 160.1 N
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.
56.0°
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Chapter 3, Solution 7.
(a)
M A = xTBF + yTBF y x
= ( 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 or
M A = xF C
FC =
M A 386.41 N ⋅ m = 2m x
= 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 Then
d =
= 2.0396 m
386.41 N ⋅m = (2.0396 m ) ( FC )min
( FC ) min and
( 2 m )2 + (0.40 m )2
= 189.454 N
2m = 78.690° 0.4 m
θ = tan−1
or
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.
( FC ) min
= 189.5 N
78.7° ⊳
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Chapter 3, Solution 8.
(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
(c)
P = 21.2 lb ⊳
For ( F) min , F must be perpendicular to BC. Then,
M B = rC/B F 67.615 lb ⋅in. = (18 in. ) F or
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.
F = 3.76 lb
75.0 ° ⊳
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Chapter 3, Solution 9.
Slope of line EC =
(a)
Then
and Then
TABx =
35 in. 5 = 76 in. + 8 in. 12
12 (T ) 13 AB
=
12 ( 260 lb ) = 240 lb 13
TABy =
5 ( 260 lb ) = 100 lb 13
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 = T ABx ( y ) + T ABy (x ) = ( 240 lb )( 0) + (100 lb)( 76 in.) = 7600 lb ⋅in. or M D = 7600 lb ⋅in.
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.
⊳
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Chapter 3, Solution 10.
Slope of line EC =
35 in. 7 = 112 in. + 8 in. 24
Then
TABx =
24 T AB 25
and
TABy =
7 T 25 AB
M D = TABx ( y) + T ABy ( x )
Have
∴ 7840 lb ⋅in. =
24 7 TAB ( 0 ) + TAB (112 in. ) 25 25 TAB = 250 lb or T AB = 250 lb ⊳
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.
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Chapter 3, Solution 11.
The minimum value of d can be found based on the equation relating the moment of the forceT
ABabout
D:
M D = ( TAB max ) y ( d ) M D = 1152 N ⋅ m
where
( TAB max ) y Now
sin θ =
= TAB max sinθ = ( 2880 N ) sinθ
1.05 m
( d + 0.24) 2 + (1.05) 2 m
∴ 1152 N ⋅m = 2880 N
1.05
( d + 0.24) 2
(d ) 2 + (1.05)
( d + 0.24)2 + ( 1.05)2
or or
(d
or
= 2.625d
+ 0.24 ) + (1.05 ) = 6.8906d2 2
2
2
5.8906 d − 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
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.
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Chapter 3, Solution 12.
with d AB =
( 42 mm )2 + (144 mm )2
= 150 mm sinθ =
42 mm 150 mm
cos θ =
144 mm 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 MC = rB/C × FAB = ( −0.042 i + 0.056 j) × (− 700 i − 2400 j) N ⋅m = (140.0 N ⋅ m ) k or
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.
MC = 140.0 N⋅ m
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Chapter 3, Solution 13.
( 42 mm )2 + (144 mm )2
with d AB =
= 150 mm sinθ =
42 mm 150 mm
cosθ =
144 mm 150 mm
FAB = − FAB sin θi − 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 MC = rB/C × FAB = ( −0.042 i − 0.056 j ) × ( −700 i − 2400 j) N ⋅m = ( 61.6 N ⋅ m) k or
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.
MC = 61.6 N ⋅ m
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Chapter 3, Solution 14.
ΣM D :
88 80 N 105 M D = ( 0.090 m ) × − ( 0.280 m ) 137 × 80 N 137 = −12.5431 N ⋅ m or
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.
M D = 12.54 N ⋅ m
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Chapter 3, Solution 15.
Note: B = B ( cos β i + sin β j )
B′ = B ( cos β i − sin β j ) C = C ( cosα i + sinα 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 and
(3)
B′ × C = B ( cos β i − sin β j) × C (cos α i + sin α j) = BC( cos β sinα + sin β cosα )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 cosα )
(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 cosα )
(6)
Adding equations (5) and (6) gives: sin (α − β ) + sin ( α + β ) = 2cos β sinα or
sin α cos β =
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.
1 1 sin ( α + β ) + sin ( α − β ) 2 2
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Chapter 3, Solution 16.
Have d = λ AB × rO/ A where λ AB =
rB/ A 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 λAB =
975 mm
=
1 ( −56i + 33j) 65
Also rO /A = ( 0 − 630) i + ( 0 − (− 225)) j
= − ( 630 mm ) i + ( 225 mm ) j
∴d =
1 (− 56i + 33j ) × − ( 630 mm ) i + ( 225 mm ) j 65
= 126.0 mm d = 126.0 mm
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.
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Chapter 3, Solution 17.
(a) where
λ =
A×B A ×B
A = 12i − 6 j + 9k B = − 3i + 9 j − 7.5k
Then i
j
k
A × B = 12 − 6 −3 9
9 − 7.5
= ( 45 − 81) i + ( −27 + 90 ) j + (108 − 18 )k = 9 (− 4i + 7 j + 10 k ) And A × B = 9 (− 4) 2 + (7) 2 + (10) 2 = 9 165
∴λ =
9 ( − 4i + 7j + 10k ) 9 165 or λ =
(b) where
λ =
A×B A ×B
A = −14i − 2j + 8k B = 3i + 1.5 j − k
Then
j k i A × B = −14 − 2 8 1.5 −1
3
= ( 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
∴λ =
5( − 2 i + 2 j − 3k) 5 17
or λλ =
1 17
( − 2i + 2j − 3k ) ⊳
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.
1 165
( − 4i + 7j + 10k ) ⊳
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Chapter 3, Solution 18.
(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
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.
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Chapter 3, Solution 19.
(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 MO = − (18.00 N⋅ m )i − (15.75 Ν ⋅ m) j − ( 40.5 N⋅ m) k (b)
Have
MO = r × F i = 2 7.5
j − 0.75 3
k −1 N ⋅ m − 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 MO = ( 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.
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.
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Chapter 3, Solution 20.
(a)
Have
MO = r × F i j = −7.5 3
k −6 lb ⋅ft
−6
3
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 (b)
Have
MO = r × F i j k = −7.5 1.5 −1 lb ⋅ft −6 4
3
= [ (6 − 6)i + (−3 + 3) j + (4.5 − 4.5) k] lb ⋅ ft or M O = 0 (c)
Have
MO = r × F i j = −8 2 3
−6
k −14 lb ⋅ft 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
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.
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Chapter 3, Solution 21.
With
TAB = − ( 369 N) j TAB = TAD
JJJG AD = ( 369 N ) AD
( 2.4 m) i − ( 3.1 m) j − ( 1.2 m) k ( 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 = r A/C × R A
=
i 0
j 3.1
k 1.2 N ⋅ m
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
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.
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Chapter 3, Solution 22.
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 and
MA =
i 0.215
j − 0.050
k 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
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.
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Chapter 3, Solution 23.
Have
M O = r A/O × R
where
rA/ D = ( 30 ft ) j + (3 ft ) k T1 = −( 62 lb) cos10 ° i − (62 lb ) sin10 ° j