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SOLUTIONS MANUAL FOR Vector Mechanics for Engineers Statics and Dynamics

11th Edition By Beer ISBN13-9780073398242

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CHAPTER 2  

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

 SOLUT TION (a)

Parallelogram law: l

(b)

Triangle rule:

W measure: We

R = 139 1 kN, α = 477.8°

R = 1391 N

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47.8° 

PR OBLEM 2.2 Tw o forces are applied as shown s to a bracket b support. Determiine r usiing grapphically the magnitude and directioon of their resultant (a) the paralleloogram law, ((b) the trianggle rule.

 TION SOLUT

(a)

Parallelogram law: l

(b)

Triangle rule:

R = 9006 lb,

W measure: We

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α = 2 6.6°

R = 9066 lb

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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

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21.2° 

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

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3.8° 

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°

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R = 196.6 N 

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)

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R = 4.26 kN 

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°

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R = 73.2 N 

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 th he 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 °

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α = 42.6°  R = 551 lb 

2.13 PROBLEM P A steel tank is to be positioned p in an excavation. D etermine byy trigonometry (a) the magnitude aand 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 perpendicular n to R. (a)

P = (425 lb)co s30°

(b)

R = (425 lb)sin n30°

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P = 368 lb

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R = 213 lb 

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

 TION SOLUT

The sma a llest force P will w be perpendicular n 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

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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

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47.8° 

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 la aw, (b ) the triangle rule.

 TION SOLUT 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 ) = 0.481 37 25° + α = 28.7755° α = 3.775° R = 8.03 kips

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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

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44.7° 

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 and a sines: Using th he force trianggle 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

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86.7°° 

P PROBLEM M 2.20 Two forces P and Q are appplied to the lid of a storag T ge bin as show 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 and a 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° = 83.2 95°

R = 104.44 N

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83.3° 

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BLEM 2.21 PROB Determiine the x and y components of each of the 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:

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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:

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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 

Fy = +(120 N)sin 70°

F y = 112.8 N 

Fx = −(150 N) cos35°

Fx = −122. 9 N 

F y = + (150 N)sin 35°

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Fy = 86.0 N 

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PROBLEM 2.24 Determine the x and y components of each of the forces shown.

 SOLUTION 40-lb Force:

50-lb Force:

60-lb Force:

Fx =+ (40 lb) cos60°

Fx = 20.0 lb 

F y = −(40 lb)sin 60°

Fy = −34.6 lb 

Fx =− (50 lb)sin 50°

Fx = −38.3 lb 

F y = −(50 lb)cos50°

Fy = − 32.1 lb 

Fx = +(60 lb) cos25°

Fx = 54.4 lb 

F y = +(60 lb)sin 25°

F y = 25.4 lb 

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PROBLEM 2.25 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.

 SOLUTION BC = (650 mm) 2 + (720 mm)2 = 970 mm

⎛ 650 ⎞ Px = P ⎜ ⎟ ⎝ 970 ⎠

(a)

⎛ 970 ⎞ P = Px ⎜ ⎟ ⎝ 650 ⎠ ⎛ 970 ⎞ = 325 N ⎜ ⎟ ⎝ 650 ⎠ = 485 N

or

P = 485 N  ⎛ 720 ⎞ Py = P ⎜ ⎟ ⎝ 970 ⎠ ⎛ 720 ⎞ = 485 N ⎜ ⎟ ⎝ 970 ⎠ = 360 N

(b)

Py = 970 N 

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PRO OBLEM 2.26 6 B Memb ber BD exerts on memberr ABC a forcce P directed along line BD. Know wing that P mu ust have a 300--lb horizontal component, determine (a) the t magniitude of the force P , (b) its vertical v compoonent.

 TION SOLUT

P sin 35° = 300 lb

(a)

P= (b)

300 lb sin 35°

Pv = P cos35°

V Vertical compoonent

= (523 lb)cos355°

 

P = 523 lb 

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Pv = 428 lb 

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PR ROBLEM 2.27 2 Thhe hydraulic cylinder BC exxerts on membber AB a forcee P dirrected along lii ne BC. Know wing that P muust have a 6000-N co mponent perppendicular to member m AB, ddetermine (a) the B. maagnitude of thee force P , (b) its componentt along line AB

 SOLUT TION 180° = 45° + α + 90° + 300 ° − 30 ° α = 180 ° − 45° − 90°− = 15 °

Px P P P= x coss α 600 N = coss15 ° = 621.17 N

cos α =

(a)

P = 621 N  tan α =

(b)

Py Px

P y = Px tan t α = (60 0 N) tan15 ° = 160.770 N

P y = 160.8 N 

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PRO OBLEM 2.2 28 Cablee AC exerts oon beam AB a force P directed along linne AC. Know ing that P must have a 350-lb verticcal componentt, determine (aa) th...