Statics and mechanics of materials 2nd edition beer solutions manual PDF

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Statics and Mechanics of Materials 2nd Edition Beer Solutions Manual Download: http://testbanklive.com/download/statics-and-mechanics-ofmaterials-2nd-edition-beer-solutions-manual/ PROBLEM 2.1 SOLUTION (a)

Parallelogram law:

(b)

Triangle rule:

We measure:

.

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.

R = 1391 kN, α = 47.8°

R = 1391 N

47.8°

PROBLEM 2.2 Two forces are applied as shown to a bracket support. Determine graphically the magnitude and direction of their resultant using (a) the parallelogram law, (b) the triangle rule.

SOLUTION

(a)

Parallelogram law:

(b)

Triangle rule:

We measure:

R = 906 lb,

α = 26.6°

R = 906 lb

26.6°

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 dir ection 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 = 20.1 kN, α = 21.2°

R = 20.1 kN

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

3.8°

PROBLEM 2.5 The 300-lb force is to be resolved into components along lines a-a′ and b-b′. (a) Determine the angle α by trigonometry knowing that the component along line a-a′ is to be 240 lb. (b) What is the corresponding value of the component along b-b′?

SOLUTION (a)

Using the triangle rule and law of sines: sin b

=

sin 60°

240 lb 300 lb sin b = 0.69282

b = 43.854° α + b + 60° = 180° α = 180° − 60° − 43.854° = 76.146° (b)

Law of sines:

Fbb′ sin 76.146°

=

300 lb sin 60°

α = 76.1°  F bb′

= 336 lb

Copyright © McGraw-Hill Education. All rights reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education.

PROBLEM 2.6 The 300-lb force is to be resolved into components along lines a-a′ and b-b′ . (a) Determine the angle α by trigonometry knowing that the component along line b-b′ is to be 120 lb. (b) What is the corresponding value of the component along a-a′?

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

sin α

=

120 lb

sin 60° 300 lb

sin α = 0.34641

α = 20.268° (b)

α = 20.3°

α + β + 60° = 180° β = 180° − 60° − 20.268° = 99.732° Faa′ sin 99.732°

=

300 lb sin 60°

F aa′

= 341 lb

PROBLEM 2.7 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: (a) (b)

1600 N P = sin 25° sin 75°

P = 3660 N 

25° + β + 75°= 180° β = 180° − 25° − 75° = 80° R 1600 N = sin 25° sin 80°

R = 3730 N

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: TAC R 2.2 kN = = sin 30° si n125° sin 25A TAC = 2.603 kN R = 4.264 kN (a) (b)

TAC = 2.60 kN R = 4.26 kN

PROBLEM 2.9 Two forces are applied as shown to a hook support. Knowing that the magnitude of P is 35 N, determine by trigonometry (a) the requir ed 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: (a)

sin α sin 25° = 50 N 35 N sin α = 0.60374 α = 37.138°

(b)

α = 37.1°

α + β + 25° = 180° β = 180° − 25° − 37.138° = 117.862° R sin117.862°

=

35 N sin 25°

R = 73.2 N

PROBLEM 2.10 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 4.8kN force directed along the axis of the automobile.

SOLUTION

Using the law of cosines:

TAC 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°

PROBLEM 2.11 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.

SOLUTION

Using the law of cosines:

P 2 = (1600 N) 2 + (2500 N) 2 − 2(1600 N)(2500 N) cos 75° P = 2596 N

Using the law of sines:

sin α sin 75° = 1600 N 2596 N α = 36.5°

P is directed 90° − 36.5° or 53.5° below the horizontal.

P = 2600 N

53.5°

PROBLEM 2.12 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: R2 = (200 lb)2 + (300 lb)2 − 2(200 lb)(300 lb) cos (45 + 65°) R = 413.57 lb

Using the law of sines:

sinα

=

sin (45 +65°)

300 lb 413.57 lb α = 42.972°

β = 90 + 25 − 42.972°

R = 414 lb

72.0°

PROBLEM 2.13 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

8 10 a = 38.66° 6 tan β = 10 β = 30.96° tana =

Using the triangle rule:

Using the law of cosines:

a + β +ψ = 180° 38.66° + 30.96° + ψ = 180° ψ = 110.38° R 2 = (120 lb) 2 + (40 lb) 2 − 2(120 lb)(40 lb)cos110.38° R = 139.08 lb

Using the law of sines:

sin γ 40 lb

=

sin 110.38° 139.08 lb

γ = 15.64° φ = (90° − a ) + γ φ = (90° − 38.66°) + 15.64° φ = 66.98°

R = 139.1 lb

67.0°

PROBLEM 2.14 Solve Problem 2.4 by trigonometry. 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 Using the force triangle and the laws of cosines and sines: We have:

Then

γ = 180° − (50° + 25°) = 105°

R2 = (4 kips) 2 + (6 kips)2 − 2(4 kips)(6 ki ps) cos105° = 64.423 kips 2 R = 8.0264 kips

And

4 kips 8.0264 kips = sin105° sin(25° + α ) sin(25° + α ) = 0.48137 25° + α = 28.775°

α = 3.775° R = 8.03 kips

3.8°

PROBLEM 2.15 For the hook support of Prob. 2.9, 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 ho rizontal, (b) the corresponding magnitude of R.

SOLUTION

The smallest force P will be perpendicular to R. (a)

P = (50 N) sin 25°

(b)

R = (50 N) cos 25°

P = 21.1 N R = 45.3 N

PROBLEM 2.16 Determine the x and y components of each of the forces shown.

SOLUTION Compute the following 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 = +(800 N) x

800

424-N Force:

F = −(424 N) x

408-N Force:

F = +(408 N) x

600

F = +480 N y

1000 560

F = −224 N x

1060

F = −(424 N) y

x

1000

F = +(800 N) y

F = +640 N

900

F = −360 N y

1060 480 1020

F = − (408 N)

900

F = +192.0 N x

F = −360 N

y

1020

y

PROBLEM 2.17 Determine the x and y components of each of the forces shown.

SOLUTION Compute the following distances: OA = (84) 2 + (80)2 = 116 in. OB = (28) 2 + (96)2 = 100 in. OC = (48) 2 + (90)2 = 102 in. 29-lb Force:

84

F = +(29 lb) x

50-lb Force:

28 100

F = +(50 lb) y

51-lb Force:

F = +(51 lb) x

80

F = +20.0 lb y

116

F = −(50 lb) x

x

116

F = + (29 lb) y

F = +21.0 lb

96

F = −14.00 lb x

F = +48.0 lb y

100 48 102

F = +24.0 lb x

F = −(51 lb)

90

F = −45.0 lb

y

102

y

PROBLEM 2.18 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) cos 60°

Fx = 20.0 lb 

Fy = −(40 lb) sin 60°

Fy = −34.6 lb 

Fx = −(50 lb) sin 50°

Fx = −38.3 lb 

Fy = −(50 lb) cos50°

Fy = −32.1 lb 

Fx = +(60 lb) cos 25°

Fx = 54.4 lb 

Fy = +(60 lb) sin 25°

Fy = 25.4 lb 

PROBLEM 2.19 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) cos 70°

Fx = 41.0 N 

Fy = +(120 N) sin 70°

Fy = 112.8 N 

Fx = −(150 N) cos35°

Fx = −122. 9 N 

Fy = +(150 N) sin 35°

Fy = 86.0 N 

PROBLEM 2.20 Member BD exerts on member ABC a force P directed along line BD. Knowing that P must have a 300-lb horizontal component, determine (a) the magnitude of the force P, (b) its vertical component.

SOLUTION

(a)

P sin 35° = 300 lb

P= (b)

Vertical component

300 lb sin 35°

P = 523 lb

Pv = P cos35°

= (523 lb) cos 35°

Pv = 428 lb

PROBLEM 2.21 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 Px P

(a)

or

650  970 

P Px

970  650 

325 N

970  650 

485 N P 485 N

(b)

720  970  720 485 N  970  

Py P

360 N Py 970 N

PROBLEM 2.22 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.

SOLUTION

(a)

P=

Py cos 55°

=

350 lb cos 55°

= 610.21 l b (b)

P = 610 lb

Px = P sin 55°

= (610.21 lb) sin 55° = 499.85 lb

Px = 500 lb

PROBLEM 2.23 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.

SOLUTION

(a)

750 N = P sin20° P = 2192.9 N

(b)

P = 2190 N

PABC = P cos 20°

= (2192.9 N) cos 20°

PABC = 2060 N

PROBLEM 2.24 Determine the resultant of the three forces of Problem 2.16. PROBLEM 2.16 Determine the x and y components of each of the forces shown.

SOLUTION Components of the forces were determined in Problem 2.16: Force

x Comp. (N)

y Comp. (N)

800 lb

+640

+480

424 lb

–224

–360

408 lb

+192

–360

Rx = +608

Ry = −240

R = Rx i + Ry j = (608 lb)i + (−240 lb) j Ry Rx 240 = 608 a = 21.541°

tana =

240 N sin(21.541°) = 653.65 N

R=

R = 654 N

21.5°

PROBLEM 2.25 Determine the resultant of the three forces of Problem 2.17. PROBLEM 2.17 Determine the x and y components of each of the forces shown.

SOLUTION Components of the forces were determined in Problem 2.17: Force

x Comp. (lb)

y Comp. (lb)

29 lb

+21.0

+20.0

50 lb

–14.00

+48.0

51 lb

+24.0

–45.0

Rx = + 31.0

Ry = +23.0

R = Rx i + Ry j = (31.0 lb)i + (23.0 lb) j Ry Rx 23.0 = 31.0 a = 36.573°

tana =

R=

23.0 lb sin (36.573°)

= 38.601 lb

R = 38.6 lb

36.6°

PROBLEM 2.26 Determine the resultant of the three forces of Problem 2.18. PROBLEM 2.18 Determine the x and y components of each of the forces shown.

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

Rx = +36.08

Ry = −41.42

R = Rx i + Ry j = (+36.08 lb)i + (−41.42 lb) j tan a =

Ry Rx

41.42 lb 36.08 lb tan a = 1.14800 a = 48.942° 41.42 lb R= sin 48.942° tan a =

R = 54.9 lb

48.9°

PROBLEM 2.27 Determine the resultant of the three forces of Problem 2.19. PROBLEM 2.19 Determine the x and y components of each of the forces shown.

SOLUTION Components of the forces were determined in Problem 2.19: 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

Rx = − 20.6

Ry = +250.2

R = Rx i + Ry j = (−20.6 N)i + (250.2 N) j tana =

Ry Rx

250.2 N 20.6 N tana = 12.1456 a = 85.293° tana =

R=

250.2 N sin 85.293°

R = 251 N

85.3°

PROBLEM 2.28 For the collar loaded as shown, 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.

SOLUTION Rx = ΣFx = (100 N) cos α + (150 N) cos (α + 30°) − (200 N) cosα Rx = −(100 N) cos α + (150 N)cos(α + 30°)

(1)

Ry = ΣFy = −(100 N) sin α − (150 N) sin (α + 30°) − (200 N) sinα Ry = −(300 N) sin α − (150 N) sin (α + 30°) (a)

(2)

Fo r R to be vertical, we must have Rx = 0. We make Rx = 0 in Eq. (1): − 100 cos α + 150 cos (α + 30°) = 0 −100 cos α + 150 (cos α cos 30° − sin α sin 30°) = 0 29.904 cos α = 75 sin α 29.904 75 = 0.39872 α = 21.738°

tan α =

(b)

α = 21.7°

Substituting for α in Eq. (2): Ry = −300 sin 21.738° − 150 sin 51.738° = −228.89 N

R = | Ry | = 228.89 N

R = 229 N

PROBLEM 2.29 A hoist trolley is subjected to the three forces shown. Knowing that α = 40°, determine (a) the requir ed magnitude of the force P if the resultant of the three forces is to be vertical, (b) the corresponding magnitude of the resultant.

SOLUTION Rx =

ΣFx = P + (200 lb) sin 40° − ( 400 lb) cos 40°

Rx = P − 177.860 lb

Ry =

ΣFy = (200 lb) cos 40° + (400 lb) sin 40°

Ry = 410.32 lb (a)

(1)

(2)

Fo r R to be vertical, we must have Rx =0. Set

Rx = 0 in Eq. (1) 0 = P − 177.860 lb P = 177.860 lb

(b)

P = 177.9 lb 

Since R is to be vertical: R = Ry = 410 lb

R = 410 lb

PROBLEM 2.30 A hoist trolley is subjected to the three forces shown. Knowing that P = 250 lb, determine (a) the required value of α if the resultant of the three forces is to be vertical, (b) the corresponding magnitude of the resultant.

SOLUTION Rx =

ΣFx = 250 lb + (200 lb) sinα − (400 lb) cos α

(1)

Rx = 250 lb + (200 lb) sin α − (400 lb) cosα

Ry = (a)

ΣFy = (200 lb) cos α + (400 lb) sin α

Fo r R to be vertical, we must have Rx =0. Set

Rx = 0 in Eq. (1)

0 = 250 lb + (200 lb) sin α − (400 lb) cos α (400 lb) cos α = (200 l b)s in α + 250 l b 2 cos α = sin α + 1.25 4cos 2 α = sin 2 α + 2.5 sin α + 1.5625 4(1 − sin 2 α ) = sin 2 α + 2.5 sin α + 1.5625 0 = 5 sin 2 α + 2.5 sin α − 2.4375 Using the quadratic formula to solve for the roots gives sin α = 0.49162

or (b)

α = 29.447°

α = 29.4°

Since R is to be vertical: R = Ry = (200 lb) cos 29.447° + (400 lb) sin 29.447°

R = 371 lb

PROBLEM 2.31 For the post loaded as shown, determine (a) the requir ed 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.

SOLUTION 24 4 T + (500 N) + (200 N) AC 25 5 1460 960

R = ΣF = − x

R =− x

x

48

AC

73

R = ΣF = − y

R =− y

(a)

(1)

+ 640 N

T

1100

y

+

T

1460

AC

7

3 (500 N) − (200 N)

25

55

(2)

T + 20 N AC 73

Fo r R to be horizontal, we must have Ry = 0. Set Ry = 0 in Eq. (2):

−

55 73

+ 20 N = 0

T AC

...