Solution Manual for Vector Mechanics for 11th ed PDF

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Solution Manual for Vector Mechanics for 11th ed...

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CHAPTER 2  

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OBLEM 2.1 1 PRO Two forces are applied as shoown to a hoook. Determine e graphically the mag nitude and dirrection of theii r resultant using (a) the paarallelogram laaw, (b) thhe triangle rulle.

 TION SOLUT (a)

Parallelogram law: l

(b)

Triangle rule:

W measure: We

R = 139 1 kN, α = 477.8°

R = 1391 N

47.8° 



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PR OBLEM 2.2 b support. Determiine Tw o forces are applied as sshown to a bracket on of their rresultant usiing grapphically the magnitude and directio (a) the paralleloogram law, ((b) the trianggle rule.

 SOLUT TION

(a)

Parallelogram law: l

(b)

Triangle rule:

R = 9006 lb,

W measure: We



α = 2 6.6°

R = 9066 lb

26.6° 

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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.

 SOLUTION (a)

Parallelogram law:

(b)

Triangle rule:

R = 20.1 kN,

We measure:

α = 21.2°

R = 20.1 kN

21.2° 

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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.

 SOLUTION (a)

Parallelogram law:

(b)

Triangle rule:

We measure:

R = 8.03 kips, α = 3.8°

R = 8.03 kips

3.8° 



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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.

 SOLUTION

Using the triangle rule and the law of sines: (a) (b)

120 N P = sin 30° sin 25°

P = 101.4 N 

30° + β + 25° = 180°

β =180° − 25 ° − 30 ° =125 °

120 N R = sin 30° sin125°

R = 196.6 N 

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PROBLEM 2.6 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.

 SOLUTION

Using the triangle rule and the law of sines: 75° + 40° + α = 180° α = 180° − 75° − 40° = 65 °

(a)

(b)

T2 800 lb = sin 65° sin 75°

T2 = 853 lb 

800 lb R = sin 65° sin 40°

R = 567 lb 

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PROBLEM 2.7 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.

 SOLUTION

Using the triangle rule and the law of sines: (a)

75° + 40° + β = 180 ° β = 180° − 75° − 40° = 65°

(b)

T1 1000 lb = sin 75° sin 65°

T1 = 938 lb 

R 1000 lb = sin 75° sin 40°

R = 665 lb 

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PROBLEM 2.8 A disabled automobile is pulled by means of two ropes as shown. The tension in rope AB is 2.2 kN, and the angle α is 25°. Knowing that the resultant of the two forces applied at A is directed along the axis of the automobile, determine by trigonometry (a) the tension in rope AC, (b) the magnitude of the resultant of the two forces applied at A.  SOLUTION

Using the law of sines:

T AC sin 30°

=

R 2.2 kN = ° sin125 sin 25D

TAC = 2.603 kN R = 4.264 kN

(a)

TAC = 2.60 kN

(b)

R = 4.26 kN 

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PROBLEM 2.9 A disabled automobile is pulled by means of two ropes as shown. Knowing that the tension in rope AB is 3 kN, determine by trigonometry the tension in rope AC and the value of α so that the resultant force exerted at A is a 4.8-kN force directed along the axis of the automobile.  SOLUTION

Using the law of cosines:

T AC 2 = (3 kN) 2 + (4.8 kN) 2 − 2(3 kN)(4.8 kN)cos 30° TAC = 2.6643 kN

Using the law of sines:

sin α sin 30° = 3 kN 2.6643 kN α = 34.3°

TAC =

2.66 kN 34.3° 

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PROBLEM M 2.10 Two forces aare applied as shown to a hoook support. K Knowing that the magnitude of P is 35 N, determine d by trigonometry (a) the requirred angle α if thee resultant R of o the two forcces applied to the support iss to be horizontal, (b) the corree sponding maggnitude of R.

 SOLUT TION Using th he triangle rulee and law of sines: siin α sin 25 ° = 50 N 35 N sin α = 0.60374

(a)

α = 37.138 ° (b)

α = 37.1° 

α + β + 25 ° = 180 ° β = 180° − 25 ° − 37.138° = 117.86 2°

35 N R = sin1177 .862° sin25°

R = 73.2 N 

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PROBLEM P 2.11 A steel tank is to be positionned in an excavation. Knowiing th hat α = 20°, determine d by trigonometry (a) the requirred magnitude m of the force P if i the resultannt R of the t wo fo orces applied at A is to be vertical, v (b) thhe correspondiing magnitude m of R. R

 TION SOLUT

Using th he triangle rulee and the law of sines: (a)

β + 50° + 60° = 180° β = 180° − 50 ° − 60° = 70 °

(b)

P 425 4 lb = s 70° sin 6 0° sin

P = 392 lb 

R 425 4 lb = s 70° sin 50° sin

R = 346 lb 

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PROBLEM P 2.12 A steel tank is to be positionned in an excavation. Knowiing th hat the maggnitude of P is 500 lb,, determine by trrigonometry (a) the required angle α if thhe resultant R of th he two forces applied at A is to be vertical, (b) the corresponding magnitude of R.

 TION SOLUT

Using th he triangle rulee and the law of sines: (a)

(α +330 °) + 60 ° + β = 180° β = 180 ° − (α + 30°) − 60° β = 90 ° − α sin (90° −α ) sin 60° = 500 lb 425 lb

90° − α = 47.402° (b)

R 500 lb = sin (42.598° + 30°) sin 60 °

α = 42.6°  R = 551 lb 

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2.13 PROBLEM P A steel tank is to be positioned p in an excavation. D etermine by trigonometry (a) the magnitude and a direction d of thee smallest forcce P for which h the resultantt R of the two forces f appliedd at A is v ertical, (b) the corresponding magnitude of R.

 TION SOLUT

The sma a llest force P will w be perpen ndicular to R. (a)

P = (425 lb)co s30°

(b)

R = (425 lb)sin n30°

P = 368 lb



R = 213 lb 

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P PROBLEM 2.14 FFor the hook ssupport of Proob. 2.10, deterr mine by trigoonometry (a) the m magnitude and d direction of the smallest foorce P for wh h ich the resultant R of the tw o forces appl ied to the suupport is horrizontal, (b) the c orresponding magnitude of R.

 SOLUT TION

The sma a llest force P will w be perpen ndicular to R. (a)

P = (50 N)sin 25°

P = 21.1 N 

(b)

R = (50 N)cos 25°

R = 45.3 N 

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PROBLEM 2.15 For the hook support shown, determine by trigonometry the magnitude and direction of the resultant of the two forces applied to the support.

SOLUTION

Using the law of cosines: R 2 = (200 lb) 2 + (300 lb) 2 − 2(200 lb)(300 lb) cos(45D + 65° ) R = 413.57 lb

sin α sin (45D + 65° ) = 300 lb 413.57 lb α = 42.972°

Using the law of sines:

D D β = 90 + 25 − 42.972°

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R = 414 lb

72.0° 

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PROBLEM 2.16 Solve Prob. 2.1 by trigonometry.

PROBLEM 2.1 Two forces are applied as shown to a hook. Determine graphically the magnitude and direction of their resultant using (a) the parallelogram law, (b) the triangle rule.

 SOLUTION

Using the law of cosines: R 2 = (900 N) 2 + (600 N ) 2 − 2(900 N )(600 N) cos(135 °) R = 1390.57N

sin(α − 30D ) sin (135° ) = 600N 1390.57N α − 30 D = 17.7642 °

Using the law of sines:

α = 47.764D R = 1391N

47.8° 

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PROBLEM 2.17 2 Soolve Problem 22.4 by trigonoometry. PR ROBLEM 2.44 Two structuural members B and C are bbolted to bracket A. Knowing thaat both membbers are in tennsion and thatt P = 6 kips and a Q = 4 kips, de termine graphhically the magnitude and direction of the resultant force exerted on thhe bracket usi ng (a) the parallelogram laaw, (b ) the triangle rule.

 SOLUT TION Using thhe force trianggle and the law ws of cosines and a sines: We have:

γ = 1800° − (50 ° + 25° ) = 1055°

Then

2 2 2 R = (4 kips) + (6 kips) p − 2(4 kip s)(6 kips) cos1 05° 2 = 64 .423 kips R = 8.00264 kips

And

4 kipps 8.02644 kips = sin(25 ° + α ) sin105 ° sin(25° + α ) = 0.4813 7 25° + α = 28.7755° α = 3.775° R = 8.03 kips

3.8° 

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PROBLEM 2.18 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. 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.

 SOLUTION

Using the laws of cosines and sines: P2 = (120 N)2 + (160 N)2 − 2(120 N)(160 N) cos 25° P = 72.096 N

And

sinα sin 25° = 120 N 72.096 N sin α = 0.70343

α = 44.703 ° P = 72.1 N

44.7° 

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PROBLEM 2.19 a to the lid of a storagge bin as show wn. Two forces P and Q are applied N determine by trigonomeetry Knowing thhat P = 48 N and Q = 60 N, w o forces. the magnitude and direction of the resuultant of the tw

SOLUT TION ws of cosines aand sines: Using th he force triang gle and the law

γ = 180° − (20 ° + 10 °)

We have e

= 150° R 2 = (48 N) 2 + (60 N)2 − 2(48 N)(60 N)cos1550 °

Then

R = 104.366 N

and

48 N 104.366 N = sin150° sin α sinn α = 0.22996 α = 13.2947°

φ = 180° − α − 80°

Hence:

= 180° − 13.2947° − 80° = 86.705°

R = 104.4 N



86.7°° 



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P PROBLEM M 2.20 Two forces P and Q are appplied to the lid of a storagge bin as show T wn. K Knowing that P = 60 N and Q = 48 N, ddetermine by trigonometry the m magnitude andd direction of the resultant o f the two forcces.

SOLUT TION Using th he force trianggle and the law ws of cosines aand sines: We have e

γ = 180° − (20° + 10 °) = 150°

Then

and

Hence:

R2 = (60 N) 2 + (448 N)2 −2(60 N)(488 N) cos150 ° R =104.366 N 60 N 104.366 N = sin150 ° sinα sin α = 0.28745 α =16.7054°

φ =180° − α − 180° =180° − 16.7054° − 80° = 83.295°

R = 104.44 N

83.3° 

  

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

BLEM 2.21 PROB Determiine the x and y components of each of thee forces shownn.

 SOLUT TION Computte the followinng distances:

OA = (84)22 + (80) 2 = 116 in. OB = (28)22 + (96) 2 = 100 in. OC = (48)22 + (90) 2 = 102 in. 29-lb Foorce:

50-lb Foorce:

51-lb Foorce:



Fx = +(29 lbb)

84 116

Fx = + 21.0 lb 

F y = + (29 l b)

80 116

Fy = + 20.0 lb 

Fx = − (50 lb b)

28 100

FFx = −14.00 lb 

F y = + (50 lbb)

96 100

Fy = + 48.0 lb 

Fx = + (51 lbb)

48 102

Fx = + 24.0 lb 

F y = − (51 lbb)

90 102

Fy = − 45.0 lb 



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

PROBL LEM 2.22 Determinne the x and y components o f each of theforces shown.

 SOLUT TION Compute the followinng distances:

OA = (600) 2 + (800) 2 = 1000 mm OB = (560) 2 + (900)2 = 1060 mm OC = (480) 2 + (900)2 = 1020 mm 800-N Force: F

424-N Force: F

F 408-N Force:



Fx = +(800 N) N

800 1000

Fx = + 640 N 

Fy = +(800 N) N

600 1000

Fy = + 480 N 

Fx = −(424 N) N

560 1060

Fx = − 224 N 

Fy = −(424 N) N

900 1060

Fy = − 360 N 

Fx = +(408 N) N

480 1020

Fx = +192.0 N 

Fy = −(408 N) N

900 1020

Fy = − 360 N 



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

PROBLEM 2.23 Determine the x and y components of each of the forces shown.

 SOLUTION 80-N Force:

120-N Force:

150-N Force:

Fx = + (80 N)cos 40°

Fx = 61.3 N 

Fy = +(80 N)sin 40°

Fy = 51.4 N 

Fx = +(120 N)cos70°

Fx = 41.0 N 

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