Engineering
Mechanics: Statics (Solutions Manual Chapters 1 - 9) 14th
Edition Russell C. Hibbeler
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1–1.
What is the weight in newtons of an object that has a mass of (a) 8 kg, (b) 0.04 kg, and (c) 760 Mg?
Solution
(a) W = 9 81(8) = 78 5 N
(b) W = 9 81(0 04) ( 10 - 3 ) = 3 92 ( 10 - 4 ) N = 0 392 mN
Ans.
Ans.
(c) W = 9.81(760) ( 103 ) = 7.46 ( 106 ) N = 7.46 MN Ans. Ans: W = 78 5 N W = 0 392 mN W = 7 46 MN
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1–2.
Represent each of the following combinations of units in the correct SI form: (a) > KN > ms, (b) Mg > mN, and (c) MN > (kg # ms).
Solution
(a) kN > ms = 103N > ( 10 - 6 ) s = GN > s
(b) Mg > mN = 106g > 10 - 3 N = Gg > N
(c) MN > (kg # ms) = 106 N > kg(10 - 3 s) = GN > (kg # s)
Ans.
Ans.
Ans.
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1–3.
Represent each of the following combinations of unit in the correct SI form: (a) Mg > ms, (b) N > mm, (c) mN > (kg # ms)
Solution
(a)
(b)
Mg ms = 103 kg 10 - 3 s = 106 kg > s = Gg > s
N mm = 1 N 10 - 3 m = 103 N > m = kN > m
(c) mN (kg ms) = 10 - 3 N 10 - 6 kg # s = kN > (kg s)
Gg > s kN > m kN > (kg # s)
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*1–4.
Convert: (a) to (b) to ( c) Express the result to three significant figures Use an appropriate prefix. 8ft> h
SOLUTION
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1–5.
Represent each of the following as a number between 0.1 and 1000 using an appropriate prefix: (a) 45 320 kN, (b) 568(105) mm, and (c) 0.00563 mg.
Solution
(a) 45 320 kN = 45 3 MN
Ans. (b) 568 ( 105 ) mm = 56 8 km Ans. (c) 0.00563 mg = 5.63 mg Ans.
2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1–6.
Round off the following numbers to three significant figures: (a) 58 342 m, (b) 68.534 s, (c) 2553 N, and (d) 7555 kg.
SOLUTION
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1–7.
Represent each of the following quantities in the correct SI form using an appropriate prefix: (a) 0.000 431 kg , (b) ,a nd (c) 0.005 32 km 35.3(103)N
SOLUTION
a)
b)
c)
0.000 431 kg = 0.000 431 A 103 B g = 0.431g
35.3 A 103 B N = 35.3kN
Ans.
Ans.
Ans. 0.005 32 km = 0.005 32 A 103 B m = 5.32 m
2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*1–8.
Represent each of the following combinations of units in the correct SI form using an appropriate prefix: (a) Mg/mm, (b) (c) mm # Mg. mN> ms,
SOLUTION
2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1–9.
Represent each of the following combinations of units in the correct SI form using an appropriate prefix: (a) (b) (c)
SOLUTION
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1–10.
Represent each of the following combinations of units in the correct SI form: (a) GN # mm, (b) kg > mm, (c) N > ks2, and (d) KN > ms.
Solution
(a) GN # mm = 109 ( 10 - 6 ) N # m = kN # m
(b) kg > mm = 103 g > 10 - 6 m = Gg > m
(c) N > ks2 = N > 106 s2 = 10 - 6 N > s2 = mN > s2
(d) kN > ms = 103 N > 10 - 6 s = 109 N > s = GN > s
Ans.
Ans.
Ans.
Ans.
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
1–11.
Represent each of the following with SI units having an appropriate prefix: (a) 8653 ms, (b) 8368 N, (c) 0.893 kg
SOLUTION
a)
b)
c)
8368 N = 8.368 kN
0.893 kg = 893(10 - 3)(103)g = 893 g
8653 ms = 8.653(10)3(10 - 3)s = 8.653 s Ans: 8 653 s
kN 893 g
2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
*1–12.
Evaluate each of the following to three significant figures and express each answer in SI units using an appropriate prefix: (a) (b) (c) (2.68 mm)(426 Mg).
(28 ms)(0.0458 Mm)> (348 mg) =
348(10 - 3)(10 - 3)kg = 15.9 mm> s (684 mm)> 43 ms = 684(10 - 6)m 43(10 - 3)s = 15.9(10 - 3)m s
28(10 - 3)s DC
- 3)(10)6 m
mm)(426
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1–13.
The density (mass volume) of aluminum is Determine its density in SI units Use an appropriate prefix. 5.26 slug> ft3
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1–14.
Evaluate each of the following to three significant figures and express each answer in SI units using an appropriate prefix: (a) (212 mN)2, (b) (52800 ms)2, and (c) [548(106)]1 > 2 ms.
Solution
(a) (212 mN)2 = 3 212(10) - 3 N 4 2 = 0 0449 N 2 = 44 9(10) - 3 N 2
(b) (52 800 ms)2 = 3 52 800(10) - 3 4 2 s2 = 2788 s2 = 2.79 ( 103 ) s2
(c) 3 548(10)6 4 1 2 ms = (23 409)(10) - 3 s = 23 4(10)3(10) - 3 s = 23 4 s
1–15.
Using the SI system of units, show that Eq. 1–2 is a dimensionally homogeneous equation which gives F in newtons. Determine to three significant figures the gravitational force acting between two spheres that are touching each other. The mass of each sphere is 200 kg and the radius is 300 mm.
SOLUTION
Using Eq. 1–2,
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*1–16.
The pascal (Pa) is actually a very small unit of pressure.To show this, convert to Atmospheric pressure at sea level is in2 How many pascals is this? 14.7 lb> lb> ft2 1Pa = 1N> m2
SOLUTION
Using Table 1–2, we have
1–17.
Water has a density of 1.94 slug > ft 3 What is the density expressed in SI units? Express the answer to three signi cant gures.
SOLUTION
Using Table 1–2, we have
999.8 kg > m3 = 1.00 Mg > m3
2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.
Evaluate each of the following to three significant figures and express each answer in Sl units using an appropriate prefix:
(a) 354 mg (45 km) (0.0356 kN), (b) (0.004 53 Mg) (201 ms), >
1–18. and (c) 435 MN 23.2 mm. >
1–19.
SOLUTION
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*1–20.
If a man weighs 155 lb on earth, specify (a) his mass in slugs,( b) his mass in kilograms,a nd (c) his weight in newton s. If the man is on the moon, w here th e acceleration due to gravity is determine (d) his weight in pounds,a nd (e) his mass in kilogram s. gm = 5.30 ft> s2,
SOLUTION
d = 25.5 lb W = 15514.44822 = 689 N
m = 155 c 14.59 kg 32.2 d = 70.2 kg W = 155 c 5.30
m = 25.5 14.59 kg 5.30 = 70.2 kg
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1–21.
Two particles have a mass of 8 kg and 12 kg,respectively. If they are 800 mm apart, determine the force of gravity acting between them. Compare this result with the weight of each particle
SOLUTION
F = G m1 m2 r2
Where
G = 66.73 A 10 - 12 B m3 > (kg # s2)
F = 66.73 A 10 - 12 B B 8(12) (0.8)2 R = 10.0 A 10 - 9 B N = 10.0 nN
W1 = 8(9.81) = 78.5 N
W2 = 12(9.81) = 118 N
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2–1.
If and , determine the magnitude of the resultant force and its direction, measured counterclockwise from the positive x axis 450 N 60°
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 4502 2(700)(450) cos 45°
This yields Th us, the direction of an g le of measured counterclockwi se from the positive a xis,i s
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2–2.
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
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 = 2 5002 + 7002 - 2(500)(700) cos 105°
= 959.78 N = 960 N
Applying the law of sines to Fig. b, and using this result, yields
sin (90° + u)
700 = sin 105° 959.78
u = 45.2°
Determine the magnitude of the resultant force and its direction, measured counterclockwise from the positive x axis
SOLUTION
The vertical force acts downward at on the two-membered frame Determine the magnitudes of the two components of directed along the axes of and Set 500 N
SOLUTION
Parallelogram Law: The parallelogram law of addition is shown in Fig. a
Trigonometr y: Using the law of sines (Fig. b), we have
2–5.
Solve Prob 2-4 with F = 350 lb
SOLUTION
Parallelogram Law: The parallelogram law of addition is shown in Fig. a
Trigonometr y: Using the law of sines (Fig. b), we have
2–6.
Determine the magnitude of the resultant force
FR = F1 + F2 and its direction, measured clockwise from the positive u axis.
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 u 6 = sin 105° 8 026 ; u = 46 22°
Thus, the direction f of FR measured clockwise from the positive u axis is f = 46.22° - 45° = 1.22° Ans.
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Ans: f = 1.22°
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2–7.
Resolve the force F1 into components acting along the u and v axes and determine the magnitudes of the components.
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 sin 45° = 4 sin 105° ; (F1)v = 2 928 kN = 2 93 kN
(F1)u sin 30° = 4 sin 105° ; (F1)u = 2 071 kN = 2 07 kN
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*2–8.
Resolve the force F2 into components acting along the u and v axes and determine the magnitudes of the components.
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 sin 75° = 6 sin 75° ; (F2)u = 6 00 kN
(F2)v sin 30° = 6 sin 75° ; (F2)v = 3 106 kN = 3 11 kN
