Solutions Manual for
Applied Statics and Strength of Materials Seventh Edition George F. Limbrunner (Inactive) Craig T. D’Allaird
Contents Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 Chapter 12 Chapter 13 Chapter 14 Chapter 15 Chapter 16 Chapter 17 Chapter 18 Chapter 19 Chapter 20
1 12 20 49 71 115 158 177 202 221 232 267 285 330 361 398 428 474 502 513
iii Copyright © 2021 Pearson Education, Inc.
INSTRUCTOR’S MANUAL FOR
APPLIED STATICS AND STRENGTH OF MATERIALS Seventh Edition George F. Limbrunner, P.E., (INACTIVE) Craig T. D’Allaird, P.E.
NOTES: 1. The solutions presented herein are, in general, somewhat abbreviated to conserve space. Very little explanation is furnished. Sketches are kept to a minimum. Few checks are shown. 2. The solutions follow the procedures developed in the examples in the text. 3. The solutions are based on the limited tables furnished in the text and/or the appendices. The tables furnished are for the purposes of this text only and should not be used for design. 4. The solutions for the design problems are generally not the only solution nor are they necessarily the most economical solutions. 5. Please note that problem numbers in the solution manual are depicted with both dashes (‐) and periods (.) between the chapter and problem numbers. These are interchangeable. 6. It should be noted that the previous editions of the text used lowercase s to denote stresses. This has changed in the 7th edition, but due to time constraints this solution manual has not been completely updated to reflect such. 7. If you find errors in this manual or in the text, please forward them to me at c.dallaird@hvcc.edu. Craig T. D’Allaird Troy, NY January 2021
iv Copyright © 2021 Pearson Education, Inc.
Prob. 1.1 (a) 𝑐 = √10 + 7 = 12.21 ft (b) 𝑏 = √20 − 16 = 12.00 m -----------------------------------------------------------Prob. 1.2 (a) 𝑎 = 25 sin 48° = 18.58 ft (b) 𝑏 = √25 − 18.58 = 16.73 ft (c) ℎ = b sin 48° = 16.73 sin 48° = 12.43 ft -----------------------------------------------------------Prob. 1.3 𝑎 = √72 − 67.3 = 25.59 ft A = cos–1(67.3/72) = 20.8° B = sin–1(67.3/72) = 69.2° -----------------------------------------------------------Prob. 1.4 AB = 28 sin 70° = 26.3 ft -----------------------------------------------------------Prob. 1.5 𝑐=
10 + 6 = 11.6 ft
𝜃 = tan (6⁄10) = 31.0° -----------------------------------------------------------Prob. 1.6 𝐴𝐵 =
12 + 16 = 20 ft
𝐵𝐶 = 12 + 32 = 34.2 ft 𝐴 = tan (12⁄16) = 36.9° 𝐶 = tan (12⁄32) = 20.6° -----------------------------------------------------------Prob. 1.7 θ = sin (5⁄6) = 56.4° 𝑥=
12 + 10 = 6.63 ft
-----------------------------------------------------------1-1
Prob. 1.8 Assume all angles to be 45° 𝑅 = +2 + 3 cos 45° + 0 − 6 sin 45° = 0.1213 mi 𝑅 = 0 − 3 sin 45° + 6 − 6 sin 45° = −0.364 mi. 𝑅=
(−0.1213) + (−0.364) = 0.384 mi.
-----------------------------------------------------------Prob. 1.9 (a) c2 = 112 + 132 – 2(11)(13)cos 80° = 15.50 ft sin 80° sin A sin B = = 15.50 11 13 →A = 44.3°, B = 55.7° (b) a2 = 782 + 852 – 2(78)(85)cos 72° = 96.0 ft sin 72° sin C sin B = = a 85 78 →A = 57.4°, B = 50.6° (c) Right Triangle Check Check a2 + b2 = c2 𝐴 = tan (7⁄24) = 16.26° 𝐵 = tan (24⁄7) = 73.7° -----------------------------------------------------------Prob. 1.10 C = 180° – 55° – 63° = 62° 𝑎 100 𝑐 = = sin 63° sin 55° sin 62° sin 63° (100) = 108.8 ft ∴𝑎= sin 55° sin 62° (100) = 107.8 ft &𝑐 = sin 55° Perimeter = 𝑎 + 𝑏 + 𝑐 = 317 ft -----------------------------------------------------------1-2
Prob. 1.11 S = weight of one shock P = weight of one set of brake pads Eq1: 8S + 10P = 101.6 lb Eq2: 10S + 6P = 106.2 lb Multiply Eq2 by 8/10: Eq3: 8S + 4.8P = 84.96 lb 8S + 10P = 101.6 lb - 8S – 4.8P = -84.96 lb 5.2P = 16.64 lb P = 3.20 lb ∴
S = 8.70 lb
----------------------------------------------------------------Prob. 1.12
20 26 = → ∴ 𝐵 = 41.77° sin B sin 60° D1= 78.23° and D2= 101.77° 𝐴𝐵 20 = →∴ 𝐴𝐵 = 29.39 ft sin 78.23° sin 41.77° 𝐵𝐶 =
(𝐴𝐵) + 50 − 2(𝐴𝐵)(50) cos 60°
𝐵𝐶 = 43.52 ft 50 43.5 = → ∴ 𝐵 = 84.25° sin B sin 60° C = 180° – 60° – 84.25° = 35.75° ------------------------------------------------------------
1-3
Prob. 1.13
-----------------------------------------------------------Prob. 1.14
-----------------------------------------------------------Prob. 1.15 AB = 32(tan 55°) = 45.7 m -----------------------------------------------------------Prob. 1.16
------------------------------------------------------------
1-4
Prob. 1.17
--------------------------------------------------------------Prob. 1.18
-----------------------------------------------------------Prob. 1.19
-----------------------------------------------------------1-5
Prob. 1.20
-----------------------------------------------------------Prob. 1.21
------------------------------------------------------------
1-6
Prob. 1.22
-----------------------------------------------------------Prob. 1.23
--------------------------------------------------------------Prob. 1.24 (a)0.015 ton × 2000 lb⁄ton = 30.0 lb (b)30.0 lb × 16 oz.⁄lb = 480 oz. -----------------------------------------------------------Prob. 1.25 (a)5 mi × 5280 ft⁄mi × 1 yd⁄3 ft = 8800 yd (b)5 mi × 5280 ft⁄mi = 26,400 ft -----------------------------------------------------------1-7
Prob. 1.26 60
mi ft 1hr 1 min ft × 5280 × × = 88 hr mi 60 min 60 sec sec
-----------------------------------------------------------Prob. 1.27 43,560
ft 1 yd × acre 3 ft
×
1 rod 5.5 yd
= 160
rod acre
----------------------------------------------------------------Prob. 1.28 (a) 125,000,000,000 gal = 384 × 10 acre − ft gal ft × 43,560 7.481 acre ft (b) 125,000,000,000 gal × 62.4 7.481
gal lb × 2000 ton ft
lb ft = 521 × 10 tons
-----------------------------------------------------------Prob. 1.29 3 7.75 in. (a)27′ − 7 → = 0.646 ft → 27.65 ft in 4 12 ft (b)1.815 ft → in. 0.815 ft × 12
in = 9.78 in. ft
0.78 in.× 32 = 24.96 ∴ 1.815 ft = 1 − 9
25 32
-----------------------------------------------------------Prob. 1.30 Volume=Area × length π(2 in. ) ft (0.25 mi) 5280 mi 4 = in. 1728 ft
12
in. ft
1-8
=28.80 ft3 Flushing Water=2(28.8 ft3)(7.481 gal/ft3) = 431 gal -----------------------------------------------------------Prob. 1.31
-----------------------------------------------------------Prob. 1.32
----------------------------------------------------------------Prob. 1.33
-----------------------------------------------------------Prob. 1.34
------------------------------------------------------------
1-9
Prob. 1.36
-----------------------------------------------------------Prob. 1.37
-----------------------------------------------------------Prob. 1.38
-------------------------------------------------------------------Prob. 1.39
------------------------------------------------------------
1-10
Prob. 1.40
-----------------------------------------------------------Prob. 1.41
-----------------------------------------------------------Prob. 1.42
------------------------------------------------------------
1-11
Prob. 2.1 (a) R = 502 + 602
(b) R = 220 2 + 180 2
= 78.1 lb
= 284 N
θx = tan–1 (60/50)
θx = tan–1 (220/180)
= 50.2°
= 50.7°
(c) R = 4 2 + 1.5 2 = 4.27 k θx = tan–1 (1.5/4) =20.6° -----------------------------------------------------------Prob. 2.3 (a) Fx = +600 cos 30° = +520 lb Fy = −600 sin 30° = −300 lb (b) Fx = −5 cos 45° = −3.54 k Fy = −5 sin 45° = −3.54 k (c) Fx = −1200 sin 30° = −600 lb Fy = +1200 cos 30° = +1039 lb -----------------------------------------------------------Prob. 2.4
(a) Fx = + Fy = +
3 13 2 13
(425) = +354 lb (425) = +236 lb
5 (110) = −42.3 lb 13 12 Fy = − (110) = −101.5 lb 13
(b) Fx = −
2-12
3 (c) Fx = − (3) = −1.80 k 5 4 Fy = + (3) = +2.40 k 5 -----------------------------------------------------------Prob. 2.5 (a) Px = +320 cos 20° = +301 lb Py = −320 sin 20° = −109.4 lb (b) Px = +640 cos 30° = +554 lb Py = −640 sin 30° = −320 lb (c) Px = +320 cos 40° = +245 lb Py = −320 sin 40° = −206 lb (d) Px = +320 cos 88° = +11.17 lb Py = −320 sin 88° = −320 lb -----------------------------------------------------------Prob. 2.6
F = 1.32 + 0.130 2 = 1.304 kN 1.3 θ x = tan −1 0.103 = 85.5° -----------------------------------------------------------Prob. 2.7 R = 175/(cos 45°) = 247 lb θx = tan–1(175/175) = 45° -----------------------------------------------------------Prob. 2.8 Px = 120 cos 50° = 77.1 lb Py = 120 sin 50° = 91.9 lb ------------------------------------------------------------
2-13
Prob. 2.9 Fx = +300 sin 42° = +201 lb Fy = −300 cos 42° = −223 lb -----------------------------------------------------------Prob. 2.10 (a) Fy = 650 cos 20° = −611 lb Fx = 650 sin 20° = +222 lb (b)
Fx Fy 300 = = 12 5 13 Fx = (12/13)(300) = +277 lb Fy = (95/13)(300) = −115 lb
-----------------------------------------------------------Prob. 2.11 (a) Px = 120 cos 30° = +103.9 kN Py = 120 sin 30° = −60.0 kN
(b) Px = 120 cos 75° = +31.1 kN Py = 120 sin 75° = −115.9 kN
(c) Px = 120 sin 5° = +10.46 kN Py = 120 cos 5° = +119.5 kN ----------------------------------------------------------Prob. 2.12 (a) Fy = 800cos30° = −693 lb Fx = 800sin 30° = −400 lb
2-14
(b)
Fy 4
=
Fx 15 50 = 3 5
4 Fy = (150) = −120 1 lb 5 3 Fx = (150) = −90 9 lb 5 -------------------------------------------------------------Prob.. 2.13
FH = 1000 sin 40° = −643 lb 0° = −766 lb FV = 1000 cos 40 -------------------------------------------------------------Prob.. 2.14 (a) Py = 80cos20°° = −75.2 k Px = 80sin20° = −27.4 k (b) Px = 60 sin 30 0° = −30.0 kN N Py = 60 cos 30 0° = +52.0 kN -------------------------------------------------------------Prob.. 2.15 FV = 500sin42° = +335 lb FH = 500cos42° = +372 lb -------------------------------------------------------------Prob.. 2.16 FH = 443 lb = 500 0 cos θ
443 = 27.6° 500
θ = cos c −1
--------------------------------------------------------------
2-15
Prob.. 2.17
Fx Fy 120 = = 2 1 5 2 120 = 10 Fx = 07.3kN 5 120 = 53.7 N Fy = 5 -------------------------------------------------------------Prob.. 2.18 W = weight w of thee skiers W = 38(175) 3 = 66 650 lb T = Wx = 6650 sin n 17° = 1944 1 lb -------------------------------------------------------------Prob.. 2.19
6N F1x = F2x = 10 cos 40° = 7.66
Hypo otenuse of slope triangle =
s2 +1
By siimilar triangles: 1 = 7.66
s2 +1 → s = 1.68 84 15
Resultant: F1 y
F1x 1.684 4 1 F1 y = 1.684(7.66) = 12.90 N =
F2y = 10 sin 40° = 6.43 N R = 6.43 6 + 12.90 = 19.33 N ↑ 2-16
-------------------------------------------------------------Prob.. 2.20
Fy
Fx 9500 = 4 8.06 7 7 Fy = (9500) = −8250 lb 8.06 4 Fx = (9500) = −4710 lb 8.06 =
---------------------------------------------------------------Prob.. 2.21 (a) P = 300 2 + 200 2 = 361 lb θx = tan–1(200 0/300) = 33.7 7° (b) P = 500 2 + 300 2 = 583 lb θx = tan–1(300 0/500) = 31.0 0° (c) P = 2402 + 3602 = 433 lb θx = tan–1(360 0/240) = 56.3 3° (d) P = 2502 + 4602 = 524 lb θx = tan–1(46 60/250) = 61.5° ---------------------------------------------------------------Prob.. 2.22 P1x = 7.78 cos 80 0° = +1.351 k P1y = 7.78 sin 80°° = +7.66 k P2x = 10 sin 40° = + 6.43 k P2y = 10 cos 40° = −7.66 k --------------------------------------------------------------
2-17
Prob.. 2.23 R = 1502 + 44.12 = 156.3 kN N 150 = 73.6° 44.1
θ = tan −1
1 = 0.294 m 6° tan 73.6 x1 = 1.5 + 0.2 294 = 1.794 m x=
-------------------------------------------------------------Prob.. 2.24
TV = 2000 sin 70° = 1879 lb TH = 2000 cos 70 0° = 684 lb ----------------------------------------------------------------Prob.. 2.25 For th he 2-kip load ds:
FV FH 2 = = 2 1 5 2(2) = 1.789 k FV = 5 1(2) = 0.894 4k FH = 5
For the 4-kip load: FV = 1.789(2) = 3.58 k FH = 0.894(2) = 1.789 k
----------------------------------------------------------------
2-18
Prob. 2.26
θ = tan–1(4/5) = 38.66° 6-lb force: Fy = 6 cos 19.66° = 5.65 lb Fx = 6 sin 19.66° = 2.02 lb 9-lb force: Py = 9 sin 63.66° = 8.07 lb Px = 9 cos 63.66° = 3.99 lb
2-19
Prob. 3.1
Using triangle OAC: R2 = 252 + 452 − 2(25)(45) cos 130° R = 64.0 lb 64.0 45 = sin 130° sin θ
45 θ = sin −1 sin130° = 32.59° 64.0 θx = 60° − 32.59° = 27.4°
-----------------------------------------------------------Prob. 3.2
θ1= tan−1(4/3) = 53.13° θ2 = tan−1(5/12) = 22.62° φ = 180° − θ1 − θ2 = 104.25° α = 180° − φ = 75.75° R2 = 122 + 102 − 2(12)(10) cos 75.75° R = 13.60 k φ1 = sin−1(sin75.75°(10/13.60)) = 45.45° 3-20
θx = θ2 +φ1 = 22.62° + 45.45° = 68.07°
-----------------------------------------------------------Prob. 3.3
θ1 = tan−1(1/2) = 26.57° θ = 180° − 26.57°−30° = 123.43° φ = 180° − θ = 56.57° R2 = 752 + 602 −2(75)(60)cos 56.57° R = 65.32 lb
R 60 = sin 56.57° sin θ2 60 θ2 = sin −1 sin 56.57° = 50.05° 65.32 θ x = θ1 + θ2 = 76.6° -----------------------------------------------------------Prob. 3.4
Rx = 45 cos 10° + 25 cos 60° = +56.82 lb Ry = 45 sin 10° + 25 sin 60° = +29.46 lb
3-21
R = 56.82 2 + 29.46 2 R = 64.0 lb 29.46 56.82 = 27.4°
θ x = tan −1
-----------------------------------------------------------Prob. 3.5
Rx = +12(12/13)−10(3/5) = +5.08 k Ry = +12(5/13) + 10(4/5) = +12.62 k
R = 5.082 + 12.622 = 13.60 k 12.62 θ x = tan −1 5.08 = 68.1° -----------------------------------------------------------Prob. 3.6
2 R x = −75 + 60 cos 30° = −15.12 lb 5 1 R y = −75 − 60 sin 30° = −63.54 lb 5
R = (−15.12) 2 + ( −63.5) 2 R = 65.3 lb 63.54 θ x = tan −1 15.12 = 76.6°
------------------------------------------------------------
3-22
Prob. 3.7
Px = −75 cos 40° = −57.45 lb Py = +75 sin 40° = + 48.21 lb Rx = −150 cos 65° = −63.39 lb Ry = +150 sin 65° = + 135.95 lb Fy + Py = Ry Fy + 48.21 lb = 135.95 lb Fy = + 87.74 lb Fx + Px = Rx Fx − 57.45 lb = −63.39 lb
F = 5.942 + 87.742 = 87.9 lb 5.94 θ = tan −1 87.74 = 3.87°
Fx = −5.94 lb -----------------------------------------------------------Prob. 3.8
Force
Horiz. (N)
Vert. (N)
100 N
90.6
42.3
120 N
75.0
93.7
Σ
165.6
136.0
3-23
R = 165.62 + 136.02 = 214 N 136.0 θ = tan −1 165.6 = 39.4° -----------------------------------------------------------Prob. 3.9 Force (N) Horiz. (N)
Vert. (N)
160
+120.4
−105.4
200
−120.4
−159.7
Σ
0
−265
R = 265 N ↓ -----------------------------------------------------------Prob. 3.10
↑→ (+)
Use components: For sloping force Slope triang. hypt = 5 2 + 3 2 = 5.83 Fx Fy 200 = = 3 5 5.83 Fx = 102.9; Fy = 171.5 Total horiz. = −200 −102.9 = − 302.9 lb
R = 302.9 2 + 171.52 = 348 lb 171.5 θ = tan −1 = 29.5° 302.9
------------------------------------------------------------
3-24
Prob. 3.11 Force (lb)
Fx (lb)
Fy (lb)
400
−400
0
300
−212.1
−212.1
200
+68.4
−187.9
Σ
−543.7
−400
R = 543.7 2 + 4002 = 675 lb 400 θ x = tan −1 = 36.3° 543.7
-----------------------------------------------------------Prob. 3.12 (a)
R1, 2 = 50 2 + 70 2 − 2(50)(70) cos100° = 92.82 N sinθ 1 sin 100° = 50 92.82 50 sin 100° θ 1 = sin −1 = 32.04° 92.82 θ x = θ 1 − 10° = 22.04°
3-25
(b)
R3,4 = 602 + 202 − 2(60)(20) cos 76.33° = 58.59 N sin θ1 sin 76.33° = → θ1 = 84.30° 60 58.59 θ x = 84.30° − 36.87° = 47.43° (more) (c)
R = R3,4 2 + R1,2 2 − 2 R3,4 R1,2 cos 25.39° = 47.14 N sinθ1 sin 24.39° = → θ1 = 57.59°(122.41°) 92.82 47.14 θ x = 10.16°
-----------------------------------------------------------Prob. 3.13
→↑ +
(a) Method of components Force (N)
Horiz. (N)
Vert. (N)
50
+17.10
+46.98
70
+68.94
−12.16
20
−16.00
+12.00
60
−23.63
−55.15
Σ +46.41
−8.33
3-26
R = 46.412 + (−8.33) 2 = 47.2 N 8.33 θ x = tan −1 = 10.18° 46.41 -----------------------------------------------------------→↑ +
Prob. 3.14
Rx = ΣFx = +200 cos 30° +100 cos 45° −400 −300 cos 60° = −306.1 lb Ry = ΣFy = +200 sin 30° −100 sin 45° −50 +300 sin 60° = +239.1 lb
R = (−306.1) 2 + 239.12 = 388 lb 239.1 θ x = tan −1 = 38.0° 306.1 --------------------------------------------------------→↑ +
Prob. 3.15 R = Ry = +300 lb
Rx = ΣFx = 0 = −500 + F1x + 240 cos 30°
∴ F1x = +292.2 lb Ry = ΣFy = +300 = +F1y −240 sin 30°
∴ F1y = +420 lb F1 = 4202 + 292.22
= 512 lb 420 θ x = tan −1 = 55.2° 292.2 --------------------------------------------------------Prob. 3.16
→↑ +
Rx = ΣFx = +1000 + 600 sin30° = +1300 lb
3-27
Ry = ΣFy = +900 −600 cos30° = +380.4 lb R = 380.4 2 + 1300 2 = 1355 lb 380.4 = 16.3° 1300
θ x = tan −1
-----------------------------------------------------------↑→ +
Prob. 3.17
Rx = ΣFx = +100 cos 20° − 95 cos 80°
− 110 cos 70° = + 39.9 N → Ry = ΣFy = +95 sin 80° − 110 sin 70°
− 100 sin 20° = − 44.0 N ↓ R = 39.92 + (−44.0) 2 = 59.5 N 44.0 θ x = tan −1 = 47.8° 39.9 -----------------------------------------------------------Prob. 3.18
↑→ +
For the 200-lb resultant force: Rx = 200 cos 60° = +100 lb Ry = 200 sin 60° = +173.2 lb Rx = ΣFx = +100 = − 160 + F2x ∴ F2x = 260 lb Ry = ΣFy = +173.2 = F2y F2 = 173.22 + 26.02 = 312 lb 173.2 θ x = tan −1 = 33.7° 260 ------------------------------------------------------------
3-28
Prob. 3.19
↑→ +
For the resultant: Rx = 300 cos45° = − 212.1 lb Ry = 300 sin45° = 212.1 lb Also: Rx = ΣFx = −212.1 = +F1 cos60° − F2 cos 30° ∴
−212.1 = 0.5F1 − 0.866F2 F1 = −424.2 + 1.732F2
(I.)
Ry = ΣFy = +212.1 = F1 sin 60° − F2 sin 30° ∴
+212.1 = +0.866F1 − 0.5F2 F1 = +244.9 + 0.577F2
(II.)
Solve simultaneous eqns. (I.) and (II.): F2 = 579 lb F1 = 579 lb -----------------------------------------------------------Prob. 3.20
P1 P2 30 = = sin 84.09° sin14.04° sin 81.87° from which: P1 = 30.1 kN P2 =
7.35 kN
-----------------------------------------------------
3-29
↑→ +
Prob. 3.21 For the resultant:
Rx = +150 cos 60° = +75 lb Ry = −150 sin 60° = −129.9 lb Rx = ΣFx = +75 = +F1 cos 30° − F2 cos 67° +75 = +0.866 F1 − 0.39 F2
∴
F1 = +86.6 + 0.45F2
(I.)
Ry = ΣFy = −129.9 = +F1 sin 30° − F2 sin 67° −129.9= +0.5F1 − 0.92F2
∴
F1 = −259.8 + 1.84F2
(II.)
Solve simultaneous eqns. (I.) and (II.): F2 = 249.2 lb F1 = 198.7 lb -----------------------------------------------------------↑→ +
Prob. 3.22 Force
Horiz.
Vert.
(lb)
(lb)
(lb)
20
−5.176
−19.32
30
−9.271
+28.53
Σ
−14.45
+9.21
R = (−14.45) 2 + 9.212 = 17.14 lb 9.21 θ = tan −1 = 32.5° 14.45 ----------------------------------------------------------
3-30
↑→ +
Prob. 3.23
Rx= +20 cos 30° + 50 cos 70° − 40 cos 15° = − 4.216 lb Ry = +20 sin 30° − 50 sin 70° − 40 sin 15° = − 47.34 lb
R = ( −4.216) 2 + (−47.34) 2 = 47.5 lb 47.34 θ x = tan −1 = 84.9° 4.216 -----------------------------------------------------------↑→ +
Prob. 3.24
Rx = +75 cos 45° − 250 cos 70° −200 − 90 cos 30° = −310.4 lb Ry = +75 sin45° − 250 sin 70° + 150 + 90 sin 30° = +13.11 lb R = (−310.4) 2 + 13.112 = 311 lb 13.11 θ x = tan −1 = 2.42° 310.4
-----------------------------------------------------------Prob. 3.25
↑→ +
For the resultant: Rx = 100 cos 20° = +93.97 lb Ry = 100 sin 20° = +34.20 lb Also: Rx = ΣFx = +93.97 = +F2 cos 30° − F1 cos 45° 3-31
∴
+93.97 = +0.866F2 − 0.707F1 F1 = +1.225F2 −132.91
(I.)
Ry = ΣFy = +34.20 = +F2 sin30° + F1 sin45°−200 +34.20 = + 0.5F2 +0.707F1 − 200
∴
F1 = − 0.707F2 + 331.26
(II.)
Solve simultaneous eqns. (I.) and (II.): F2 = 240.3 lb F1 = 161.4 lb -----------------------------------------------------------Prob. 3.26 FH = Fsin18° FH = AH AH = 5000 sin 25° = 2113 lb
F=
FH 2113 = = 6840 lb sin18° 0.309
-----------------------------------------------------------Prob. 3.27 Solve the force triangle: F2 = 60002 + 50002 −2(6000)(5000) cos 25° F = 2570 lb
sin θ sin 25° = 5000 2570 θ = 55.3° -----------------------------------------------------------Prob. 3.28
Refer to Prob. 2.8. CC moment: +
W = weight of barrel and contents 100 W = 60 + (62.4) = 894 lb 7.481
3-32
ΣMo = +(894 − 120 sin 50°)(1.25) − (120 cos 50°)(2.75) = +791 lb-ft --------------------------------------------------------Prob. 3.29 (a) MA : (counterclockwise: +) Force
Moment
Moment
(lb)
arm (ft)
(lb-ft)
20
4
−80
40
9
−360
50
15
+750
30
23
−690
−−
−380
Σ
(b) ΣMA = −380 lb-ft (clockwise) ------------------------------------------------------------Prob. 3.30 [CC moment: +]; [↑→ +] (a) [F1] Mo = 0 [F2] Mo = 50(6) = +300 lb-ft [F3] Mo = 75 sin 15°(6) = +116.5 lb-ft (b) ΣMo = +300 + 116.5 = + 417 lb-ft (c) Rx = ΣFx = +25 − 75 cos 15° = −47.4 lb Ry = ΣFy = +50 + 75 sin 15° = +69.4 lb (more) R = 69.42 + (−47.4) 2 = 84.0 lb 69.4 θ x = tan −1 = 55.7° 47.4
3-33
(d) Mo = 84.0(6 sin55.7°) = + 416 lb-ft (Say OK) -----------------------------------------------------------Prob. 3.31
[CC moment is +]
(a) [F1] Mo = 30 sin60°(4) = +103.9 lb-ft [F2] Mo = 40 sin45°(4) = −113.1 lb-ft [F3] Mo = 10(4) = −40 lb-ft [F4] Mo = 0 (b) ΣMo = +103.9 − 113.1 − 40 = − 49.2 lb-ft -----------------------------------------------------------↑→ +
Prob. 3.32
Rx = ΣFx = 30 cos 60° + 40 cos 45° − 50 = −6.716 lb Ry = ΣFy = 30 sin 60° −40 sin 45° −10 = −12.304 lb R = (−6.716) 2 + (−12.304)2 = 14.02 lb 12.304 θ x = tan −1 = 61.4° 6.716
Moment of resultant: ΣMo= 14.02 sin 61.4°(4) = 49.2 lb-ft (checks) -----------------------------------------------------------Prob. 3.33
CC moment: +
For 30 kN force: Fx = 26.49 kN, Fy = 14.08 kN (a) ΣMA Force (kN) Arm (m)
Moment (kN⋅m)
70
1.50
−105.0
26.49
2.00
+52.98
14.08
1.70
−23.94 3-34
60 Σ
2.35
−141.0
−−
−217
(b) ΣMB Force (kN) Arm (m)
Moment (kN⋅m)
70
1.50
+105.0
26.49
2.00
+52.98
14.08
1.30
+18.30
60
0.65
+39.0
−−
+215
Σ
-----------------------------------------------------------Prob. 3.34
CC moment: +
For the 900-lb forces: Px Py 900 lb = = 4 3 5
Px = 720 lb Py = 540 lb 3-35
ΣMA = −540(2)(20) + 720(78) + 720(68) + 400(48) = +102,700 lb-ft -------------------------------------------------------Prob. 3.35
CC moment: +
ΣMB = −12(3) − 14(10) + 25(4) = −76.0 k-ft -------------------------------------------------------Prob. 3.36 Resultant of dist. force = 10(6) = 60 N R = ΣFy = −50 − 60 + 30 − 60 = −140 N ΣMA = −50(2) − 60(6) + 30(11) − 60(14) = −970 N⋅m x is measured to the right from point A:
x=
ΣM A 970 = = 6.93 m R 140
------------------------------------------------------------Prob. 3.37
CC moment: +
b = 5/(tan40°) = 5.959 ft a = 10 − 5.959 = 4.041 ft d = 4.041 sin 40° = 2.598 ft Mo = 550(2.598) = −1429 lb-ft Using Varig. theorem w/ components at A: ΣMo = −550 sin 40°(10) + 550 cos 40°(5) = −1429 lb-ft
(checks)
-----------------------------------------------------------3-36
Prob. 3.38
CC moment: +
Components at B: ΣMo = 550 sin40° = −1429 lb-ft (checks) Components at C: ΣMo = 550 cos 40°(4.041 tan40°) = −1429 lb-ft (checks) --------------------------------------------------------------Prob. 3.39 Fx = 2 cos20° = 1.879 k Fy = 2 sin20° = 0.684 k ΣMA = −1.879(3 sin 50°) − 0.684(3 cos 50°) = −5.64 k-ft --------------------------------------------------------------Prob. 3.40 CC moment: + Fx = 150 cos 60°= 75.0 lb Fy = 150 sin 60°=129.9 lb ΣMA= +129.9(5) + 75.0(20) = +2150 lb-ft ------------------------------------------------------------Prob. 3.41
CC moment: +
ΣMA = −3(5) − (4/5)(5)(10) = −55.0 k-ft ------------------------------------------------------------Prob. 3.42
CC moment: +; ↑→ +
R = −300 − 450 − 200 = −950 lb (↓) ΣMA = −300(3) − 450(8) − 200(6) = −7700 lb-ft x = 7700 / 950 = 8.11 ft
-----------------------------------------------------------3-37
Prob. 3.43
CC moment: +; ↑→ +
R = +5 + 3 − 8 + 12 − 2 = +10 k (↑) ΣMA = +3(3) − 8(6) + 12(11) − 2(13) = +67 k-ft
x = 67 / 10 = 6.7 ft -----------------------------------------------------------Prob. 3.44
CC moment: +; ↑→ +
R = + 40 − 20 − 20 + 80 = +80 lb (↑) ΣMA = − 20(2) − 20(10) + 80(12) = +720 lb-ft x = 720 / 80 = 9.0 ft
-----------------------------------------------------------Prob. 3.45
CC moment: +; ↑→ +
R = + 4 − 5 − 22 = −3 k (↓) ΣMA = −4(7) + 5(3) − 2(2) = −17 k-ft ∴ R is located to the right of A x = 17 / 3 = 5.67 ft
-----------------------------------------------------------Prob. 3.46
CC moment: +; ↑→ +
R = −8 − 32 − 32 = − 72 k(↓) ΣMA = −32(14 − 32(28) = − 1344 k-ft x = 1344 / 72 = 18 .67 ft
-----------------------------------------------------------Prob. 3.47
CC moment: +; ↑→ +
Note: F1 (assumed ↑) is located x1 to the right of A. + F1 − 10 − 20 = − 60 k F1 = −30 k (↓) ΣMA = −10(4) − 20(9) − 30x1 = −60(10) = −600 k-ft ∴ x1 = 12.67 ft -----------------------------------------------------------3-38
Prob. 3.48
CC moment: +; ↑→ +
Note: F1 (assumed ↑)is located x1 to the right of A. +F1 + 9 + 16 = +10 kN F1 = − 15 kN (↓) ΣMA: +10(5) = +9(1) +16(8) −15x1 ∴ x1 = 5.80 m -------------------------------------------------------------Prob. 3.49
R = −10 − 28 = − 38 k (↓) ΣMA = −10(5) − 28(17) = − 526 k-ft
x=
526 = 13.84 ft 38
--------------------------------------------------------Prob. 3.50
R = −1000 − 7000 − 1600 = −9600 lb (↓) ΣMA = −7000(8) − 1600(12) = −75,200 lb-ft
x=
75,200 = 7.83 ft 9600
-------------------------------------------------------------
3-39
Prob. 3.51 (a) pmax = 62.4(18) = 1123 psf (b) R = (1123/2)(1)(18) = 10,100 lb (c) y =18 / 3 = 6 ft (up from base) ------------------------------------------------------------Prob. 3.52 Resultant of the triangular load is 36 k (↓) located 3 ft right of A. R = 36 + 5 = 41 k ↓ ΣMA = −36(3) − 5(13) = − 173 k-ft
x=
173 = 4.22 ft 41
------------------------------------------------------------Prob. 3.53
CC moment: +; ↑→ +
R = − 21 − 15 − 10 = −46 kN ΣMA = −21(7) − 15(11.5) − 10(14) = − 459.5 kN⋅m
x=
459.5 = 9.99 m 46
-----------------------------------------------------------Prob. 3.54
CC moment: +; ↑→ +
Hydrostatic force F = (1/2)(62.4)(18)2 =10,100 lb (located 18/3 = 6 ft up from point A) For the resultant: Rx = +10,100 lb 3-40
Ry = − 19,800 lb R = 10,1002 + (−19,800) 2 = 22, 200 lb 119,800 θ x = tan −1 = 63° 10,100
ΣMA = − 19,800(4) − 10,100(6) = −139,800 lb-ft Resultant intersects the base of the dam x to the right of point A:
x=
ΣM A 139,800 = = 7.06 ft Ry 19,800
(Within middle 1/3: O.K.) -------------------------------------------------------Prob. 3.55
CC moment: +; ↑→ +
F1 = 0.5(900 − 350)(15) = 4125 lb F2 = 350(15) = 5250 lb F3 = 0.5(350)(15) = 2275 R = ΣFy = + 11,650 lb ΣMA = 4125(5) + 5250(7.5) + 2275(28.8.67) = 104,000 lb-ft x=
ΣM A 104,000 = = 8.93 ft 11,650 R
------------------------------------------------------------Prob. 3.56
CC moment: +; ↑→ +
R = ΣFy = − 900 − 600 = −1500 lb ΣMA = − 900(2) − 600(15) = −10,800 lb-ft x=
ΣM A 10,800 = = 7.20 ft R 1500
---------------------------------------------------------
3-41
Prob. 3.57
CC moment: +; ↑→ +
Ry = −53 − 70 = −123 kN Rx = −21 kN R = (−123) 2 + (−21) 2 = 124.8 kN
(
)
θ = tan -1 123 21 = 80.3° ΣMA = −53(1) − 70(2) + 21(1) = − 172 kN⋅m x is measured to the right from point A:
ΣM A 172 = = 1.398 m R 123
x=
-----------------------------------------------------------Prob. 3.58 (a) 312 + 499.2 (3)(3) = 3650 lb 2 R1 = 312(3)(3) = 2808 lb Ra =
187.2 (3)(3) = 842.4 lb 2 R (1.5) + R2 (1) y= 1 = 1.385 ft R1 + R2
R2 =
124.8 ( 2)(3) = 374.4 lb 2 Rb = −3650 + 374.4 = 3276 lb R3 =
y=
3650(1.385) − 374.4(0.667) = 1.467 ft 3276
-----------------------------------------------------------Prob. 3.59 1. R = (52/2)(0.833)(5) = 108.3 lb
y=
0.833 = 0.278 ft 3
2. R1 = 52(0.833)(5) = 216.7 lb
3-42
R2 = 108.3 lb R = R1 + R2 = 325 lb 0.833 216 .7 + 108 .3(0.278) 2 = 0.370 ft y= 325
3. R1 = 104(0.833)(5) = 433 lb R2 = 108.3 lb R = R1 + R2 = 541 lb 0.833 433 + 108.3(0.278) 2 = 0.389 ft y= 541
------------------------------------------------------------Prob. 3.60 R = ΣFy = −25(4) −12−40 = −77 k ΣMA = −25(4)−12(14)−40(28) = −1388 k-ft x=
ΣM A 1388 = = 18.03 ft Ry 77
------------------------------------------------------------Prob. 3.61 F1 + F2 = 150+100 = +250 lb F1 = 250 − F2
I.
ΣMA = +220 = + F1(4) + F2(8) −100(12) 4F1 + 8F2 = +1420 F1 = −2F2 + 355
II.
Solve simultaneous equations I. & II.: F2 = 105 lb ↑ F2 = 145 lb ↑ -------------------------------------------------------------
3-43