Engineering Mechanics Statics 14th Edition Hibbeler Solutions Manual Full clear download( no error farmatting) at: https://testbanklive.com/download/engineering-mechanics-statics-14th-editionhibbeler-solutions-manual/ 2–1. If
60° and
450 N, determine the magnitude of the y
resultant force and its direction, measured counterclockwise from the positive x axis.
F 15 700 N
SOLUTION The parallelogram law of addition and the triangular rule are shown in Figs. a and b, respectively. Applying the law of consines to Fig. b, 7002 497.01 N
4502
2(700)(450) cos 45°
497 N
Ans.
This yields sin 700
sin 45° 497.01
Thus, the direction of angle positive axis, is
60°
of F
95.19°
95.19°
measured counterclockwise from the
60°
155°
Ans.
x
Ans: FR = 497 N f = 155 22
2–2. y
If the magnitude of the resultant force is to be 500 N, directed along the positive y axis, determine the magnitude of force F and its direction u.
F u 15
x
700 N
SOLUTION The parallelogram law of addition and the triangular rule are shown in Figs. a and b, respectively. Applying the law of cosines to Fig. b, F = 25002 + 7002 - 2(500)(700) cos 105° = 959.78 N = 960 N
Ans.
Applying the law of sines to Fig. b, and using this result, yields sin (90° + u) sin 105° = 700 959.78 u = 45.2°
Ans.
Ans: F = 960 N u = 45.2 23
2–3. Determine the magnitude of the resultant force FR = F1 + F2 and its direction, measured counterclockwise from the positive x axis.
y F1
250 lb
30
SOLUTION
x
FR = 2(250)2 + (375)2 - 2(250)(375) cos 75° = 393.2 = 393 lb 393.2
Ans.
45
250 =
sin 75° sin u u = 37.89° f = 360° - 45° + 37.89° = 353°
F2
375 lb
Ans.
Ans: FR = 393 lb f = 353 24
*2–4. The vertical force F acts downward at on the two-membered frame. Determine the magnitudes of the two components of 500 N. F directed along the axes of and . Set
B
SOLUTION
A
Parallelogram Law: The parallelogram law of addition is shown in Fig. a. Trigonometry: Using the law of sines (Fig. b), we have
sin 60°
448 N
sin 45°
F
500 sin 75°
C
Ans.
500 sin 75°
366 N
Ans.
Ans: FAB = 448 N FAC = 366 N
25
2–5. Solve Prob. 2-4 with F = 350 lb. B
45
SOLUTION
A
Parallelogram Law: The parallelogram law of addition is shown in Fig. a. Trigonometry: Using the law of sines (Fig. b), we have FAB
F
350
30 C
=
sin 60° sin 75° FAB = 314 lb FAC
Ans.
350
= sin 45° sin 75° FAC = 256 lb
Ans.
Ans: FAB = 314 lb FAC = 256 lb 26
2–6. v
Determine the magnitude of the resultant force FR = F1 + F2 and its direction, measured clockwise from the positive u axis.
30 75
F1
4 kN
30 u F2
6 kN
Solution Parallelogram Law. The parallelogram law of addition is shown in Fig. a, Trigonometry. Applying Law of cosines by referring to Fig. b, FR = 242 + 6 2 - 2(4)(6) cos 105 = 8.026 kN = 8.03 kN
Ans.
Using this result to apply Law of sines, Fig. b, sin 105 sin u ; = 8.026 6
u = 46.22
Thus, the direction f of FR measured clockwise from the positive u axis is f = 46.22 - 45 = 1.22
Ans.
Ans: f = 1.22 27
2–7. v
Resolve the force F1 into components acting along the u and v axes and determine the magnitudes of the components. 30 75
F1
4 kN
30 u F2
6 kN
Solution Parallelogram Law. The parallelogram law of addition is shown in Fig. a, Trigonometry. Applying the sines law by referring to Fig. b. (F1)v 4 = ; sin 45 sin 105
(F1)v = 2.928 kN = 2.93 kN
Ans.
(F1)u 4 = ; sin 30 sin 105
(F1)u = 2.071 kN = 2.07 kN
Ans.
Ans: (F1)v = 2.93 kN (F1)u = 2.07 kN 28
*2–8. v
Resolve the force F2 into components acting along the u and v axes and determine the magnitudes of the components. 30 75
F1
4 kN
30 u F2
6 kN
Solution Parallelogram Law. The parallelogram law of addition is shown in Fig. a, Trigonometry. Applying the sines law of referring to Fig. b, (F2)u 6 = ; sin 75 sin 75
(F2)u = 6.00 kN
Ans.
(F2)v 6 = ; sin 30 sin 75
(F2)v = 3.106 kN = 3.11 kN
Ans.
Ans: (F2)u = 6.00 kN (F2)v = 3.11 kN 29
2–9. If the resultant force acting on the support is to be 1200 lb, directed horizontally to the right, determine the force F in rope A and the corresponding angle u.
F A u B 60
900 lb
Solution Parallelogram Law. The parallelogram law of addition is shown in Fig. a, Trigonometry. Applying the law of cosines by referring to Fig. b, F = 29002 + 12002 - 2(900)(1200) cos 30 = 615.94 lb = 616 lb
Ans.
Using this result to apply the sines law, Fig. b, sin u 900
sin 30 =
615.94
;
u = 46.94 = 46.9
Ans.
Ans: F = 616 lb u = 46.9 30
2–10. y
Determine the magnitude of the resultant force and its direction, measured counterclockwise from the positive x axis.
800 lb 40
x 35
Solution 500 lb
Parallelogram Law. The parallelogram law of addition is shown in Fig. a, Trigonometry. Applying the law of cosines by referring to Fig. b, FR = 28002 + 5002 - 2(800)(500) cos 95 = 979.66 lb = 980 lb
Ans.
Using this result to apply the sines law, Fig. b, sin u 500
sin 95 =
979.66
;
u = 30.56
Thus, the direction f of FR measured counterclockwise from the positive x axis is f = 50 - 30.56 = 19.44 = 19.4
Ans.
Ans: FR = 980 lb f = 19.4 31
2–11. The plate is subjected to the two forces at A and B as shown. If u = 60°, determine the magnitude of the resultant of these two forces and its direction measured clockwise from the horizontal.
FA u
8 kN
A
SOLUTION Parallelogram Law: The parallelogram law of addition is shown in Fig. a. Trigonometry: Using law of cosines (Fig. b), we have FR = 282 + 62 - 2(8)(6) cos 100° 40
= 10.80 kN = 10.8 kN
Ans. B
The angle u can be determined using law of sines (Fig. b).
FB
6 kN
sin u sin 100° = 6 10.80 sin u = 0.5470 u = 33.16° Thus, the direction f of FR measured from the x axis is f = 33.16° - 30° = 3.16°
Ans.
Ans: FR = 10.8 kN f = 3.16 32
*2–12. Determine the angle of u for connecting member A to the plate so that the resultant force of FA and FB is directed horizontally to the right. Also, what is the magnitude of the resultant force?
FA u
8 kN
A
SOLUTION Parallelogram Law: The parallelogram law of addition is shown in Fig. a. Trigonometry: Using law of sines (Fig .b), we have sin (90° - u) sin 50° = 6 8
40 B
sin (90° - u) = 0.5745
FB
u = 54.93° = 54.9°
6 kN
Ans.
From the triangle, f = 180° - (90° - 54.93°) - 50° = 94.93°. Thus, using law of cosines, the magnitude of FR is FR = 282 + 62 - 2(8)(6) cos 94.93° = 10.4 kN
Ans.
Ans: u = 54.9 FR = 10.4 kN 33
2–13.
Engineering Mechanics Statics 14th Edition Hibbeler Solutions Manual Full clear download( no error farmatting) at: https://testbanklive.com/download/engineering-mechanics-statics-14th-editionhibbeler-solutions-manual/ engineering mechanics statics 14th edition free pdf statics hibbeler 14th edition solutions engineering mechanics statics 14th edition ebook engineering mechanics statics and dynamics 14th edition engineering mechanics statics hibbeler 14th edition pdf download engineering mechanics statics rc hibbeler 14th edition solution manual pdf engineering mechanics statics 13th edition pdf engineering mechanics statics pdf
60° and
450 N, determine the magnitude of the y
resultant force and its direction, measured counterclockwise from the positive x axis.
F 15 700 N
SOLUTION The parallelogram law of addition and the triangular rule are shown in Figs. a and b, respectively. Applying the law of consines to Fig. b, 7002 497.01 N
4502
2(700)(450) cos 45°
497 N
Ans.
This yields sin 700
sin 45° 497.01
Thus, the direction of angle positive axis, is
60°
of F
95.19°
95.19°
measured counterclockwise from the
60°
155°
Ans.
x
Ans: FR = 497 N f = 155 22
2–2. y
If the magnitude of the resultant force is to be 500 N, directed along the positive y axis, determine the magnitude of force F and its direction u.
F u 15
x
700 N
SOLUTION The parallelogram law of addition and the triangular rule are shown in Figs. a and b, respectively. Applying the law of cosines to Fig. b, F = 25002 + 7002 - 2(500)(700) cos 105° = 959.78 N = 960 N
Ans.
Applying the law of sines to Fig. b, and using this result, yields sin (90° + u) sin 105° = 700 959.78 u = 45.2°
Ans.
Ans: F = 960 N u = 45.2 23
2–3. Determine the magnitude of the resultant force FR = F1 + F2 and its direction, measured counterclockwise from the positive x axis.
y F1
250 lb
30
SOLUTION
x
FR = 2(250)2 + (375)2 - 2(250)(375) cos 75° = 393.2 = 393 lb 393.2
Ans.
45
250 =
sin 75° sin u u = 37.89° f = 360° - 45° + 37.89° = 353°
F2
375 lb
Ans.
Ans: FR = 393 lb f = 353 24
*2–4. The vertical force F acts downward at on the two-membered frame. Determine the magnitudes of the two components of 500 N. F directed along the axes of and . Set
B
SOLUTION
A
Parallelogram Law: The parallelogram law of addition is shown in Fig. a. Trigonometry: Using the law of sines (Fig. b), we have
sin 60°
448 N
sin 45°
F
500 sin 75°
C
Ans.
500 sin 75°
366 N
Ans.
Ans: FAB = 448 N FAC = 366 N
25
2–5. Solve Prob. 2-4 with F = 350 lb. B
45
SOLUTION
A
Parallelogram Law: The parallelogram law of addition is shown in Fig. a. Trigonometry: Using the law of sines (Fig. b), we have FAB
F
350
30 C
=
sin 60° sin 75° FAB = 314 lb FAC
Ans.
350
= sin 45° sin 75° FAC = 256 lb
Ans.
Ans: FAB = 314 lb FAC = 256 lb 26
2–6. v
Determine the magnitude of the resultant force FR = F1 + F2 and its direction, measured clockwise from the positive u axis.
30 75
F1
4 kN
30 u F2
6 kN
Solution Parallelogram Law. The parallelogram law of addition is shown in Fig. a, Trigonometry. Applying Law of cosines by referring to Fig. b, FR = 242 + 6 2 - 2(4)(6) cos 105 = 8.026 kN = 8.03 kN
Ans.
Using this result to apply Law of sines, Fig. b, sin 105 sin u ; = 8.026 6
u = 46.22
Thus, the direction f of FR measured clockwise from the positive u axis is f = 46.22 - 45 = 1.22
Ans.
Ans: f = 1.22 27
2–7. v
Resolve the force F1 into components acting along the u and v axes and determine the magnitudes of the components. 30 75
F1
4 kN
30 u F2
6 kN
Solution Parallelogram Law. The parallelogram law of addition is shown in Fig. a, Trigonometry. Applying the sines law by referring to Fig. b. (F1)v 4 = ; sin 45 sin 105
(F1)v = 2.928 kN = 2.93 kN
Ans.
(F1)u 4 = ; sin 30 sin 105
(F1)u = 2.071 kN = 2.07 kN
Ans.
Ans: (F1)v = 2.93 kN (F1)u = 2.07 kN 28
*2–8. v
Resolve the force F2 into components acting along the u and v axes and determine the magnitudes of the components. 30 75
F1
4 kN
30 u F2
6 kN
Solution Parallelogram Law. The parallelogram law of addition is shown in Fig. a, Trigonometry. Applying the sines law of referring to Fig. b, (F2)u 6 = ; sin 75 sin 75
(F2)u = 6.00 kN
Ans.
(F2)v 6 = ; sin 30 sin 75
(F2)v = 3.106 kN = 3.11 kN
Ans.
Ans: (F2)u = 6.00 kN (F2)v = 3.11 kN 29
2–9. If the resultant force acting on the support is to be 1200 lb, directed horizontally to the right, determine the force F in rope A and the corresponding angle u.
F A u B 60
900 lb
Solution Parallelogram Law. The parallelogram law of addition is shown in Fig. a, Trigonometry. Applying the law of cosines by referring to Fig. b, F = 29002 + 12002 - 2(900)(1200) cos 30 = 615.94 lb = 616 lb
Ans.
Using this result to apply the sines law, Fig. b, sin u 900
sin 30 =
615.94
;
u = 46.94 = 46.9
Ans.
Ans: F = 616 lb u = 46.9 30
2–10. y
Determine the magnitude of the resultant force and its direction, measured counterclockwise from the positive x axis.
800 lb 40
x 35
Solution 500 lb
Parallelogram Law. The parallelogram law of addition is shown in Fig. a, Trigonometry. Applying the law of cosines by referring to Fig. b, FR = 28002 + 5002 - 2(800)(500) cos 95 = 979.66 lb = 980 lb
Ans.
Using this result to apply the sines law, Fig. b, sin u 500
sin 95 =
979.66
;
u = 30.56
Thus, the direction f of FR measured counterclockwise from the positive x axis is f = 50 - 30.56 = 19.44 = 19.4
Ans.
Ans: FR = 980 lb f = 19.4 31
2–11. The plate is subjected to the two forces at A and B as shown. If u = 60°, determine the magnitude of the resultant of these two forces and its direction measured clockwise from the horizontal.
FA u
8 kN
A
SOLUTION Parallelogram Law: The parallelogram law of addition is shown in Fig. a. Trigonometry: Using law of cosines (Fig. b), we have FR = 282 + 62 - 2(8)(6) cos 100° 40
= 10.80 kN = 10.8 kN
Ans. B
The angle u can be determined using law of sines (Fig. b).
FB
6 kN
sin u sin 100° = 6 10.80 sin u = 0.5470 u = 33.16° Thus, the direction f of FR measured from the x axis is f = 33.16° - 30° = 3.16°
Ans.
Ans: FR = 10.8 kN f = 3.16 32
*2–12. Determine the angle of u for connecting member A to the plate so that the resultant force of FA and FB is directed horizontally to the right. Also, what is the magnitude of the resultant force?
FA u
8 kN
A
SOLUTION Parallelogram Law: The parallelogram law of addition is shown in Fig. a. Trigonometry: Using law of sines (Fig .b), we have sin (90° - u) sin 50° = 6 8
40 B
sin (90° - u) = 0.5745
FB
u = 54.93° = 54.9°
6 kN
Ans.
From the triangle, f = 180° - (90° - 54.93°) - 50° = 94.93°. Thus, using law of cosines, the magnitude of FR is FR = 282 + 62 - 2(8)(6) cos 94.93° = 10.4 kN
Ans.
Ans: u = 54.9 FR = 10.4 kN 33
2–13.
Engineering Mechanics Statics 14th Edition Hibbeler Solutions Manual Full clear download( no error farmatting) at: https://testbanklive.com/download/engineering-mechanics-statics-14th-editionhibbeler-solutions-manual/ engineering mechanics statics 14th edition free pdf statics hibbeler 14th edition solutions engineering mechanics statics 14th edition ebook engineering mechanics statics and dynamics 14th edition engineering mechanics statics hibbeler 14th edition pdf download engineering mechanics statics rc hibbeler 14th edition solution manual pdf engineering mechanics statics 13th edition pdf engineering mechanics statics pdf
Engineering Mechanics Statics 13th Edition Pdf Free Download Free
Engineering Mechanics Statics 14th Edition Pdf
Engineering Mechanics: Statics (SI Edition), Third Edition (Volume 1) Apago PDF Enhancer Apago PDF Enhancer Apago PDF Enhancer CONVERSIONS BETWEEN U.S. CUSTOMARY UNITS AND SI UNITS (Co. Engineering Mechanics: Statics (12th Edition). Engineering mechanics statics 13th edition by rc hibbeler text book in pdf format available for free download and visitors now can read solution manual engineering mechanics statics 13th edition by rc hibbeler online for free. Solution Manual Engineering Mechanics Statics 13th edition by R.C. Hibbeler Text Book in pdf format available for free download and visitors now can read Solution Manual Engineering Mechanics Statics 13th edition by R.C. Hibbeler online for free.