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SOLUTION
(a) Parallelogram law:
PROBLEM 2.1
Two forces are applied at point B of beam AB. Determine graphically the magnitude and direction of their resultant using (a) the parallelogram law, (b) the triangle rule.
(b) Triangle rule:
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3
We measure: 3.30kN,66.6 R α ==° 3.30kN = R 66.6°
PROBLEM 2.2
The cable stays AB and AD help support pole AC. Knowing that the tension is 120 lb in AB and 40 lb in AD, determine graphically the magnitude and direction of the resultant of the forces exerted by the stays at A using (a) the parallelogram law, (b) the triangle rule.
SOLUTION
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4
measure: 51.3 59.0 α β =° =° (a) Parallelogram law: (b) Triangle
We measure: 139.1lb, R = 67.0 γ =° 139.1lb R = 67.0°
We
rule:
SOLUTION (a) Parallelogram law:
PROBLEM 2.3
Two structural members B and C are bolted to bracket A. Knowing that both members are in tension and that P = 10 kN and Q = 15 kN, determine graphically the magnitude and direction of the resultant force exerted on the bracket using (a) the parallelogram law, (b) the triangle rule.
(b) Triangle rule:
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5
We measure: 20.1kN, R = 21.2 α =° 20.1kN = R 21.2°
SOLUTION (a) Parallelogram law:
PROBLEM 2.4
Two structural members B and C are bolted to bracket A. Knowing that both members are in tension and that P = 6 kips and Q = 4 kips, determine graphically the magnitude and direction of the resultant force exerted on the bracket using (a) the parallelogram law, (b) the triangle rule.
(b) Triangle rule:
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6
We measure: 8.03kips,3.8 R α ==° 8.03kips = R 3.8°
SOLUTION
PROBLEM 2.5
A stake is being pulled out of the ground by means of two ropes as shown. Knowing that α = 30°, determine by trigonometry (a) the magnitude of the force P so that the resultant force exerted on the stake is vertical, (b) the corresponding magnitude of the resultant.
Using the triangle rule and the law of sines:
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7
(a) 120N sin30sin25 P = °° 101.4N P = (b) 3025180 1802530 125 β β °++°=° =°−°−° =° 120N sin30sin125 = °° R 196.6N = R
PROBLEM 2.6
A trolley that moves along a horizontal beam is acted upon by two forces as shown. (a) Knowing that α = 25°, determine by trigonometry the magnitude of the force P so that the resultant force exerted on the trolley is vertical. (b) What is the corresponding magnitude of the resultant?
SOLUTION
Using the triangle rule and the law of sines:
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8
(a) 1600 N sin25°sin75 P = ° 3660 N P = (b) 2575180 1802575 80 β β °++°=° =°−°−° =° 1600 N sin25°sin80 R = ° 3730 N R =
SOLUTION
PROBLEM 2.7
A trolley that moves along a horizontal beam is acted upon by two forces as shown. Determine by trigonometry the magnitude and direction of the force P so that the resultant is a vertical force of 2500 N.
Using the law of cosines:
Using the law of sines:
222 (1600 N)(2500 N)2(1600 N)(2500 N)cos 75° 2596 N P P =+−
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9
=
α
=° P is directed 9036.5 or 53.5° °−° below the horizontal. 2600 N = P 53.5°
sinsin75 1600 N2596 N 36.5 α
° =
SOLUTION
PROBLEM 2.8
A telephone cable is clamped at A to the pole AB. Knowing that the tension in the left-hand portion of the cable is T1 = 800 lb, determine by trigonometry (a) the required tension T2 in the right-hand portion if the resultant R of the forces exerted by the cable at A is to be vertical, (b) the corresponding magnitude of R.
Using the triangle rule and the law of sines:
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10
(a) 7540180 1807540 65 α α °+°+=° =°−°−° =° 2 800lb sin65sin75 T = °° 2 853 lb T = (b) 800lb sin65sin40 R = °° 567 lb R =
SOLUTION
PROBLEM 2.9
A telephone cable is clamped at A to the pole AB. Knowing that the tension in the right-hand portion of the cable is T2 = 1000 lb, determine by trigonometry (a) the required tension T1 in the left-hand portion if the resultant R of the forces exerted by the cable at A is to be vertical, (b) the corresponding magnitude of R.
Using the triangle rule and the law of sines:
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11
(a) 7540180 1807540 65 β β °+°+=° =°−°−° =° 1 1000 lb sin75°sin65 T = ° 1 938 lb T = (b) 1000 lb sin75°sin40 R = ° 665 lb R =
PROBLEM 2.10
Two forces are applied as shown to a hook support. Knowing that the magnitude of P is 35 N, determine by trigonometry (a) the required angle
α if the resultant R of the two forces applied to the support is to be horizontal, (b) the corresponding magnitude of R.
SOLUTION
Using the triangle rule and law of sines:
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12
(a) sinsin25 50 N35 N sin0.60374 α α ° = = 37.138 α =° 37.1 α =° (b) 25180 1802537.138 117.862 α β β ++°=° =°−°−° =° 35 N sin117.862sin25 R = °° 73.2 N R =
PROBLEM 2.11
A steel tank is to be positioned in an excavation. Knowing that α = 20°, determine by trigonometry (a) the required magnitude of the force P if the resultant R of the two forces applied at A is to be vertical, (b) the corresponding magnitude of R.
SOLUTION
Using the triangle rule and the law of sines:
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13
(a) 5060180 1805060 70 β β +°+°=° =°−°−° =° 425 lb sin70sin60 P = °° 392 lb P = (b) 425 lb sin70sin50 R = °° 346 lb R =
PROBLEM 2.12
A steel tank is to be positioned in an excavation. Knowing that the magnitude of P is 500 lb, determine by trigonometry (a) the required angle α if the resultant R of the two forces applied at A is to be vertical, (b) the corresponding magnitude of R.
SOLUTION
Using the triangle rule and the law of sines:
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14
(a) (30)60180 180(30)60 90 sin(90)sin60 425 lb500 lb α β βα βα α +°+°+=° =°−+°−° =°− °−° = 9047.402 α °−=° 42.6 α =° (b) 500 lb sin(42.59830)sin60 R = °+°° 551 lb R =
SOLUTION
PROBLEM 2.13
A steel tank is to be positioned in an excavation. Determine by trigonometry (a) the magnitude and direction of the smallest force P for which the resultant R of the two forces applied at A is vertical, (b) the corresponding magnitude of R
The smallest force P will be perpendicular to R
(
) (425 lb)cos30 P =°
(
) (425 lb)sin30 R =°
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15
a
368
P
213
R
lb =
b
lb
=
PROBLEM 2.14
For the hook support of Prob. 2.10, determine by trigonometry (a) the magnitude and direction of the smallest force P for which the resultant R of the two forces applied to the support is horizontal, (b) the corresponding magnitude of R.
SOLUTION
The smallest force P will be perpendicular to R.
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16
(a) (50 N)sin25 P =° 21.1 N = P (b) (50 N)cos25 R =° 45.3 N R =
PROBLEM 2.15
Solve Problem 2.2 by trigonometry.
PROBLEM2.2 The cable stays AB and AD help support pole AC. Knowing that the tension is 120 lb in AB and 40 lb in AD, determine graphically the magnitude and direction of the resultant of the forces exerted by the stays at A using (a) the parallelogram law, (b) the triangle rule.
SOLUTION
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17
8 tan 10 38.66 6 tan 10 30.96 α α β β = =° = =° Using the triangle rule: 180 38.6630.96180 110.38 α βψ ψ ψ ++=° °+°+=° =° Using the law of cosines: 22 2 (120lb)(40 lb)2(120 lb)(40 lb)cos110.38 139.08 lb R R =+− ° = Using the law of sines: sinsin110.38 40lb139.08lb γ ° = 15.64 (90) (9038.66)15.64 66.98 γ φαγ φ φ =° =°−+ =°−°+° =° 139.1lb = R 67.0°
PROBLEM 2.16
Solve Problem 2.4 by trigonometry.
PROBLEM2.4 Two structural members B and C are bolted to bracket A Knowing that both members are in tension and that P = 6 kips and Q = 4 kips, determine graphically the magnitude and direction of the resultant force exerted on the bracket using (a) the parallelogram law, (b) the triangle rule.
SOLUTION
Using the force triangle and the laws of cosines and sines: We have:
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18
105 γ =°−°+° =° Then 222 2 (4kips)(6kips)2(4kips)(6kips)cos105 64.423kips 8.0264kips R R =+− ° = = And 4kips8.0264kips sin(25)sin105 sin(25)0.48137 2528.775 3.775 α α α α = °+° °+= °+=° =° 8.03kips = R 3.8°
180(5025)
SOLUTION
PROBLEM 2.17
For the stake of Prob. 2.5, knowing that the tension in one rope is 120 N, determine by trigonometry the magnitude and direction of the force P so that the resultant is a vertical force of 160 N.
PROBLEM2.5 A stake is being pulled out of the ground by means of two ropes as shown. Knowing that α = 30°, determine by trigonometry (a) the magnitude of the force P so that the resultant force exerted on the stake is vertical, (b) the corresponding magnitude of the resultant.
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19
Using the laws of cosines and sines: 222(120N)(160N)2(120N)(160N)cos25 72.096N P P =+− ° = And sinsin25 120N72.096N sin0.70343 44.703 α α α ° = = =° 72.1N = P 44.7°
PROBLEM 2.18
For the hook support of Prob. 2.10, knowing that P = 75 N and α = 50°, determine by trigonometry the magnitude and direction of the resultant of the two forces applied to the support.
PROBLEM2.10 Two forces are applied as shown to a hook support. Knowing that the magnitude of P is 35 N, determine by trigonometry (a) the required angle α if the resultant R of the two forces applied to the support is to be horizontal, (b) the corresponding magnitude of R
Using the force triangle and the laws of cosines and
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20
SOLUTION
sines: We have 180(5025) 105 β =°−°+° =° Then 222 22 (75N)(50N) 2(75N)(50N)cos105 10,066.1 N 100.330 N R R R =+ −° = = and sinsin105 75 N100.330 N sin0.72206 46.225 γ γ γ ° = = =° Hence: 2546.2252521.225 γ −°=°−°=° 100.3 N = R 21.2°
PROBLEM 2.19
Two forces P and Q are applied to the lid of a storage bin as shown. Knowing that P = 48 N and Q = 60 N, determine by trigonometry the magnitude and direction of the resultant of the two forces.
SOLUTION
Using the force triangle and the laws of cosines and sines: We have
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21
180(2010) 150 γ =°−°+° =° Then 222 (48 N)(60 N) 2(48 N)(60 N)cos150 104.366 N R R =+ −° = and 48 N104.366N sinsin150 sin0.22996 13.2947 α α α = ° = =° Hence: 18080 18013.294780 86.705 φ α =°−−° =°−°−° =° 104.4 N = R 86.7°
PROBLEM 2.20
Two forces P and Q are applied to the lid of a storage bin as shown. Knowing that P = 60 N and Q = 48 N, determine by trigonometry the magnitude and direction of the resultant of the two forces.
SOLUTION
Using the force triangle and the laws of cosines and sines:
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22
We have 180(2010) 150 γ =°−°+° =° Then 222 (60N)(48N) 2(60N)(48N)cos150 104.366 N R R =+ −° = and 60 N104.366 N sinsin150 sin0.28745 16.7054 α α α = ° = =° Hence: 180180 18016.705480 83.295 φ α =°−−° =°−°−° =° 104.4 N = R 83.3°
SOLUTION
PROBLEM 2.21
Determine the x and y components of each of the forces shown.
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23
80-N Force: (80 N)cos40 Fx =+° 61.3 N Fx = (80 N)sin40 Fy =+° 51.4 N Fy = 120-N Force: (120 N)cos70 Fx =+° 41.0 N Fx = (120 N)sin70 Fy =+° 112.8 N Fy = 150-N Force: (150 N)cos35 Fx =−° 122. 9 N Fx =− (150 N)sin35 Fy =+° 86.0 N Fy =
SOLUTION
PROBLEM 2.22
Determine the x and y components of each of the forces shown.
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24
40-lb Force: (40 lb)cos60 Fx =+° 20.0 lb Fx = (40 lb)sin60 Fy =−° 34.6 lb Fy =− 50-lb Force: (50lb)sin50 Fx =−° 38.3lb Fx =− (50 lb)cos50 Fy =−° 32.1 lb Fy =− 60-lb Force: (60 lb)cos25 Fx =+° 54.4 lb Fx = (60 lb)sin25 Fy =+° 25.4lb Fy =
SOLUTION
Compute the following distances:
PROBLEM 2.23
Determine the x and y components of each of the forces shown.
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25
22 22 22 (600)(800) 1000 mm (560)(900) 1060 mm (480)(900) 1020 mm OA OB OC =+ = =+ = =+ = 800-N Force: 800 (800 N) 1000 Fx =+ 640 N Fx =+ 600 (800 N) 1000 y F =+ 480 N y F =+ 424-N Force: 560 (424 N) 1060 Fx =− 224 N Fx =− 900 (424 N) 1060 y F =− 360 N y F =− 408-N Force: 480 (408 N) 1020 Fx =+ 192.0 N Fx =+ 900 (408 N) 1020 y F =− 360 N y F =−
SOLUTION
Compute the following distances:
PROBLEM 2.24
Determine the x and y components of each of the forces shown.
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26
22 22 22 (24in.)(45in.) 51.0 in. (28in.)(45in.) 53.0 in. (40in.)(30in.) 50.0 in. OA OB OC =+ = =+ = =+ = 102-lb Force: 24in. 102 lb 51.0in. Fx =− 48.0 lb Fx =− 45in. 102 lb 51.0in. y F =+ 90.0 lb y F =+ 106-lb Force: 28in. 106 lb 53.0in. =+ Fx 56.0 lb Fx =+ 45in. 106 lb 53.0in. y F =+ 90.0 lb y F =+ 200-lb Force: 40in. 200 lb 50.0in. Fx =− 160.0 lb Fx =− 30in. 200 lb 50.0in. y F =− 120.0 lb y F =−
PROBLEM 2.25
The hydraulic cylinder BD exerts on member ABC a force P directed along line BD. Knowing that P must have a 750-N component perpendicular to member ABC, determine (a) the magnitude of the force P, (b) its component parallel to ABC
distributed in any
the
distribution to teachers and
27
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form or by any means, without the prior written permission of the publisher, or used beyond
limited
educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
SOLUTION (a) 750 Nsin20 P =° 2192.9 N P = 2190 N P = (b) cos20 ABC PP=° (2192.9 N)cos20 =° 2060 N PABC =
PROBLEM 2.26
Cable AC exerts on beam AB a force P directed along line AC. Knowing that P must have a 350-lb vertical component, determine (a) the magnitude of the force P, (b) its horizontal component.
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28
SOLUTION (a) cos55 y P P = ° 350 lb cos55 610.21 lb = ° = 610 lb P = (b) sin55 x PP=° (610.21 lb)sin55 499.85 lb =° = 500lb Px =
PROBLEM 2.27
Member BC exerts on member AC a force P directed along line BC. Knowing that P must have a 325-N horizontal component, determine (a) the magnitude of the force P, (b) its vertical component.
22 (650mm)(720mm)
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29
SOLUTION
970mm BC =+ = (a) 650 970 x PP = or 970 650 970 325N 650 485N PPx = = = 485 N P = (b) 720 970 720 485N 970 360N y PP = = = 970 N y P =
PROBLEM 2.28
Member BD exerts on member ABC a force P directed along line BD Knowing that P must have a 240-lb vertical component, determine (a) the magnitude of the force P, (b) its horizontal component.
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30
SOLUTION (a) 240 lb sin40sin40° == ° y P P or 373 lb P = (b) 240 lb tan40tan40° y x P P == ° or 286 lb Px =
PROBLEM 2.29
The guy wire BD exerts on the telephone pole AC a force P directed along BD. Knowing that P must have a 720-N component perpendicular to the pole AC, determine (a) the magnitude of the force P, (b) its component along line AC
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31
SOLUTION (a) 37 12 37 (720N) 12 2220N = = = PPx 2.22 kN P = (b) 35 12 35 (720N) 12 2100N yx PP = = = 2.10 kN = y P
PROBLEM 2.30
The hydraulic cylinder BC exerts on member AB a force P directed along line BC. Knowing that P must have a 600-N component perpendicular to member AB, determine (a) the magnitude of the force P, (b) its component along line AB
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32
SOLUTION 180459030 180459030 15 α α °=°++°+° =°−°−°−° =° (a) cos cos 600N cos15 621.17N x x P P P P α α = = = ° = 621 N P = (b) tan tan (600N)tan15 160.770N y x yx P P PP α α = = =° = 160.8N y P =
PROBLEM 2.31
Determine the resultant of the three forces of Problem 2.23.
PROBLEM2.23 Determine the x and y components of each of the forces shown.
SOLUTION
Components of the forces were determined in Problem 2.23:
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33
Force x Comp. (N) y Comp. (N) 800 lb +640 +480 424 lb –224 –360 408 lb +192 –360 608 Rx =+ 240 y R =− (608 lb)(240 lb) tan 240 608 21.541 240 N sin(21.541°) 653.65 N xy y x RR R R R α α =+ =+− = = =° = = Rij ij 654N = R 21.5°
PROBLEM 2.32
Determine the resultant of the three forces of Problem 2.21.
PROBLEM2.21 Determine the x and y components of each of the forces shown.
SOLUTION
Components of the forces were determined in Problem 2.21:
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Force x Comp. (N) y Comp. (N) 80 N +61.3 +51.4 120 N +41.0 +112.8 150 N –122.9 +86.0 20.6 Rx =− 250.2 y R =+ (20.6 N)(250.2 N) tan 250.2 N tan 20.6 N tan12.1456 85.293 250.2 N sin85.293 xy y x RR R R R α α α α =+ =−+ = = = =° = ° Rij ij 251N = R 85.3°
PROBLEM 2.33
Determine the resultant of the three forces of Problem 2.22.
PROBLEM2.22 Determine the x and y components of each of the forces shown.
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SOLUTION Force x Comp. (lb) y Comp. (lb) 40 lb +20.00 –34.64 50 lb –38.30 –32.14 60 lb +54.38 +25.36 36.08 Rx =+ 41.42 y R =− (36.08 lb)(41.42 lb) tan 41.42 lb tan 36.08 lb tan1.14800 48.942 41.42 lb sin48.942 xy y x RR R R R α α α α =+ =++− = = = =° = ° Rij ij 54.9lb = R 48.9°
PROBLEM 2.34
Determine the resultant of the three forces of Problem 2.24.
PROBLEM2.24 Determine the x and y components of each of the forces shown.
SOLUTION
Components of the forces were determined in Problem 2.24:
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Force x Comp. (lb) y Comp. (lb) 102 lb 48.0 +90.0 106 lb +56.0 +90.0 200 lb 160.0 120.0 152.0 Rx =− 60.0 y R = (152 lb)(60.0 lb) tan 60.0 lb tan 152.0 lb tan0.39474 21.541 α α α α =+ =−+ = = = =° xy y x RR R R Rij ij 60.0 lb sin21.541 R = ° 163.4lb = R 21.5°
PROBLEM 2.35
Knowing that α = 35°, determine the resultant of the three forces shown.
SOLUTION
100-N Force: (100 N)cos3581.915 N (100 N)sin3557.358 N
150-N Force: (150 N)cos6563.393 N (150 N)sin65135.946 N
200-N Force: (200 N)cos35163.830 N (200 N)sin35114.715 N
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x y F F =+°=+ =−°=−
F F =+°=+ =−°=−
x y
y F F =−°=− =−°=− Force x Comp. (N) y Comp. (N) 100 N +81.915 57.358 150 N +63.393 135.946 200 N 163.830 114.715 18.522 Rx =− 308.02 y R =− (18.522 N)(308.02 N) tan 308.02 18.522 86.559 xy y x RR R R α α =+ =−+− = = =° Rij ij 308.02 N sin86.559 R = 309 N = R 86.6°
x
SOLUTION
Determine force components:
PROBLEM 2.36
Knowing that the tension in rope AC is 365 N, determine the resultant of the three forces exerted at point C of post BC.
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Cable force AC: 960 (365N)240N 1460 1100 (365N)275N 1460 =−=− =−=− x y F F 500-N Force: 24 (500N)480N 25 7 (500N)140N 25 x y F F == == 200-N Force: 4 (200N)160N 5 3 (200N)120N 5 x y F F == =−=− and 22 22 240N480N160N400N 275N140N120N255N (400N)(255N) 474.37N =Σ=−++= =Σ=−+−=− =+ =+− = xx yy xy RF RF RRR Further: 255 tan 400 32.5 α α = =° 474N = R 32.5°
SOLUTION
PROBLEM 2.37
Knowing that α = 40°, determine the resultant of the three forces shown.
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(60lb)cos2056.382lb (60lb)sin2020.521lb x y F F =°= =°= 80-lb Force: (80lb)cos6040.000lb (80lb)sin6069.282lb x y F F =°= =°= 120-lb Force: (120lb)cos30103.923lb (120lb)sin3060.000lb x y F F =°= =−°=− and 22 200.305lb 29.803lb (200.305lb)(29.803lb) 202.510lb xx yy RF RF R =Σ= =Σ= =+ = Further: 29.803 tan 200.305 α = 1 29.803 tan 200.305 8.46 α = =° 203lb = R 8.46°
60-lb Force:
SOLUTION
PROBLEM 2.38
Knowing that α = 75°, determine the resultant of the three forces shown.
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Force: (60lb)cos2056.382 lb (60lb)sin2020.521 lb x y F F =° = =° = 80-lb Force: (80lb)cos956.9725 lb (80lb)sin9579.696 lb x y F F =° = =° = 120-lb Force: (120lb)cos5119.543 lb (120lb)sin510.459 lb x y F F =° = =° = Then 168.953 lb 110.676 lb xx yy RF RF =Σ= =Σ= and 22 (168.953 lb)(110.676 lb) 201.976 lb R =+ = 110.676 tan 168.953 tan0.65507 33.228 α α α = = =° 202 lb = R 33.2°
60-lb
SOLUTION
PROBLEM 2.39
For the collar of Problem 2.35, determine (a) the required value of α if the resultant of the three forces shown is to be vertical, (b) the corresponding magnitude of the resultant.
(100 N)cos(150 N)cos(30)(200 N)cos
N)cos(30)
(100 N)sin(150 N)sin(30)(200 N)sin
(a) For R to be vertical, we must have 0. Rx = We make 0 Rx = in Eq. (1):
(b) Substituting for α in Eq. (2):
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(100 N)cos(150
xx x RF R ααα αα =Σ =++°− =−++° (1)
(300 N)sin(150
yy y RF R ααα αα =Σ =−−+°− =−−+° (2)
N)sin(30)
100cos150cos(30)0 100cos150(coscos30sinsin30)0 29.904cos75sin αα ααα αα −++°= −+°−°= = 29.904 tan 75 0.39872 21.738 α α = = =° 21.7 α =°
300sin21.738150sin51.738 228.89 N y R =−°−° =− ||228.89 N y RR== 229 N R =
PROBLEM 2.40
For the post of Prob. 2.36, determine (a) the required tension in rope AC if the resultant of the three forces exerted at point C is to be horizontal, (b) the corresponding magnitude of the resultant.
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SOLUTION 960244 (500 N)(200 N) 1460255 48 640 N 73 xxAC xAC RFT RT =Σ=−++ =−+ (1) 110073 (500 N)(200 N) 1460255 55 20 N 73 yyAC yAC RFT RT =Σ=−+− =−+ (2) (a) For R to be horizontal, we must have 0. y R = Set 0 y R = in Eq. (2): 55 20 N0 73 TAC −+= 26.545N TAC = 26.5 N TAC = (b) Substituting for TAC into Eq. (1) gives 48 (26.545 N)640 N 73 622.55 N 623N =− + = == x x x R R RR 623 N R =
PROBLEM 2.41
A hoist trolley is subjected to the three forces shown. Knowing that α = 40°, determine (a) the required magnitude of the force P if the resultant of the three forces is to be vertical, (b) the corresponding magnitude of the resultant.
For R to be vertical, we must have
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the
to
43
displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond
limited distribution
teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
Rx = (200 lb)sin40(400 lb)cos40 x FP Σ=+°−° 177.860 lb x RP=− (1) y R = (200 lb)cos40(400 lb)sin40 y F Σ=°+° 410.32 lb y R = (2) (a)
0. Rx = Set 0 Rx = in Eq. (1) 0177.860lb 177.860lb P P =− = 177.9 lb P = (b)
410 lb == y RR 410 lb R =
SOLUTION
Since R is to be vertical: