Engineering Mechanics - Statics Chapter 1 Problem 1-1 Represent each of the following combinations of units in the correct SI form using an appropriate prefix: (a) m/ms (b) μ km (c) ks/mg (d) km μN⋅ Units Used: μN10 6− N= μkm 10 6− km= Gs 10 9 s= ks 10 3 s= mN 10 3− N= ms 10 3− s= Solution: a() m ms 110 3 × m s = m ms 1 km s = b() μkm 1 10 3− × m= μkm 1mm= c() ks mg 110 9 × s kg = ks mg 1 Gs kg = d() km μN⋅ 110 3− × mN= km μN⋅ 1mmN⋅= 1 © 2007 R. C. Hibbeler. Published by 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. Engineering Mechanics - Statics Chapter 1 Problem 1-2 Wood has a density d. What is its density expressed in SI units? Units Used: Mg 1000 kg= Given: d 4.70 slug ft 3 = Solution: 1slug 14.594 kg= d 2.42 Mg m 3 = Problem 1-3 Represent each of the following combinations of units in the correct SI form using an appropriate prefix: (a) Mg/mm (b) mN/ μ s (c) μmMg⋅ Solution: a() Mg mm 10 3 kg 10 3− m = 10 6 kg m = Gg m = Mg mm Gg m = b() mN μs 10 3− N 10 6− s = 10 3 N s = kN s = mN μs kN s = c() μmMg⋅ 10 6− m () 10 3 kg () = 10 3− mkg⋅= μmMg⋅ mm kg⋅= 2 © 2007 R. C. Hibbeler. Published by 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. Engineering Mechanics - Statics Chapter 1 Problem 1-4 Represent each of the following combinations of units in the correct SI form: (a) Mg/ms, (b) N/mm, (c) mN/( kg μs⋅ ). Solution: a() Mg ms 10 3 kg 10 3− s = 10 6 kg s = Gg s = Mg ms Gg s = b() N mm 1N 10 3− m = 10 3 N m = kN m = N mm kN m = c() mN kg μs⋅ 10 3− N 10 6− kg s⋅ = kN kg s⋅ = mN kg μs⋅ kN kg s⋅ = Problem 1-5 Represent each of the following with SI units having an appropriate prefix: (a) S 1 , (b) S 2 , (c) S 3 . Units Used: kg 1000 g= ms 10 3− s= kN 10 3 N= Given: S 1 8653 ms= S 2 8368 N= S 3 0.893 kg= Solution: a() S 1 8.653 s= 3 © 2007 R. C. Hibbeler. Published by 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. Engineering Mechanics - Statics Chapter 1 b() S 2 8.368 kN= c() S 3 893g= Problem 1-6 Represent each of the following to three significant figures and express each answer in SI units using an appropriate prefix: (a) x, (b) y, and (c) z. Units Used: MN 10 6 N= μg110 6− × gm= kN 10 3 N= Given: x 45320 kN= y 568 10 5 × () mm= z 0.00563 mg= Solution: a() x 45.3MN= b() y 56.8km= c() z 5.63μg= Problem 1-7 Evaluate ( ab⋅ )/c to three significant figures and express the answer in SI units using an appropriate prefix. Units Used: μm10 6− m= Given: a 204 mm()= 4 © 2007 R. C. Hibbeler. Published by 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. Engineering Mechanics - Statics Chapter 1 b 0.00457 kg()= c 34.6 N()= Solution: l ab c = l 26.945 μmkg⋅ N = Problem 1-8 If a car is traveling at speed v, determine its speed in kilometers per hour and meters per second. Given: v 55 mi hr = Solution: v 88.514 km hr = v 24.6 m s = Problem 1-9 Convert: (a) S 1 to Nm⋅ , (b) S 2 to kN/m 3 , (c) S 3 to mm/s. Express the result to three significant figures. Use an appropriate prefix. Units Used: kN 10 3 N= Given: S 1 200g lb ft⋅= S 2 350g lb ft 3 = S 3 8 ft hr = 5 © 2007 R. C. Hibbeler. Published by 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. Engineering Mechanics - Statics Chapter 1 Solution: a() S 1 271N m⋅= b() S 2 55.0 kN m 3 = c() S 3 0.677 mm s = Problem 1-10 What is the weight in newtons of an object that has a mass of: (a) m 1 , (b) m 2 , (c) m 3 ? Express the result to three significant figures. Use an appropriate prefix. Units Used: Mg 10 3 kg= mN 10 3− N= kN 10 3 N= Given: m 1 10 kg= m 2 0.5 gm= m 3 4.50 Mg= Solution: a() Wm 1 g= W 98.1 N= b() Wm 2 g= W 4.90 mN= c() Wm 3 g= W 44.1 kN= 6 © 2007 R. C. Hibbeler. Published by 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. Engineering Mechanics - Statics Chapter 1 Problem 1-11 If an object has mass m, determine its mass in kilograms. Given: m 40 slug= Solution: m 584 kg= Problem 1-12 The specific weight (wt./vol.) of brass is ρ . Determine its density (mass/vol.) in SI units. Use an appropriate prefix. Units Used: Mg 10 3 kg= Given: ρ 520 lb ft 3 = Solution: ρ 8.33 Mg m 3 = Problem 1-13 A concrete column has diameter d and length L. If the density (mass/volume) of concrete is ρ , determine the weight of the column in pounds. Units Used: Mg 10 3 kg= kip 10 3 lb= Given: d 350 mm= L 2m= 7 © 2007 R. C. Hibbeler. Published by 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. Engineering Mechanics - Statics Chapter 1 ρ 2.45 Mg m 3 = Solution: V π d 2 ⎛ ⎜ ⎝ ⎞ ⎟ ⎠ 2 L= V 192.423 L= W ρ V= W 1.04 kip= Problem 1-14 The density (mass/volume) of aluminum is ρ . Determine its density in SI units. Use an appropriate prefix. Units Used: Mg 1000 kg= Given: ρ 5.26 slug ft 3 = Solution: ρ 2.17 Mg m 3 = Problem 1-15 Determine your own mass in kilograms, your weight in newtons, and your height in meters. Solution: Example W 150 lb= mW= m 68.039 kg= Wg 667.233 N= h 72 in= h 1.829 m= 8 © 2007 R. C. Hibbeler. Published by 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. Engineering Mechanics - Statics Chapter 1 Problem 1-16 Two particles have masses m 1 and m 2 , respectively. If they are a distance d apart, determine the force of gravity acting between them. Compare this result with the weight of each particle. Units Used: G 66.73 10 12− × m 3 kg s 2 ⋅ = nN 10 9− N= Given: m 1 8kg= m 2 12 kg= d 800 mm= Solution: F Gm 1 m 2 d 2 = F 10.0 nN= W 1 m 1 g= W 1 78.5 N= W 1 F 7.85 10 9 ×= W 2 m 2 g= W 2 118 N= W 2 F 1.18 10 10 ×= Problem 1-17 Using the base units of the SI system, show that F = G(m 1 m 2 )/r 2 is a dimensionally homogeneous equation which gives F in newtons. Compute the gravitational force acting between two identical spheres that are touching each other. The mass of each sphere is m 1 , and the radius is r. Units Used: μN10 6− N= G 66.73 10 12− m 3 kg s 2 ⋅ = 9 © 2007 R. C. Hibbeler. Published by 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. Engineering Mechanics - Statics Chapter 1 Given: m 1 150 kg= r 275 mm= Solution: F Gm 1 2 2r() 2 = F 4.96 μN= Since the force F is measured in Newtons, then the equation is dimensionally homogeneous. Problem 1-18 Evaluate each of the following to three significant figures and express each answer in SI units using an appropriate prefix: (a) x, (b) y, (c) z. Units Used: MN 10 6 N= kN 10 3 N= μm10 6− m= Given: x 200 kN() 2 = y 0.005 mm() 2 = z 400 m() 3 = Solution: a() x 0.040MN 2 = b() y 25.0μm 2 = c() z 0.0640km 3 = 10 © 2007 R. C. Hibbeler. Published by 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. [...]... publisher Engineering Mechanics - Statics Chapter 2 Problem 2 -1 5 Resolve the force F 1 into components acting along the u and v axes and determine the magnitudes of the components Given: F 1 = 250 N F 2 = 15 0 N θ 1 = 30 deg θ 2 = 30 deg θ 3 = 10 5 deg Solution: F 1v sin ( θ 1 ) = F1 sin ( θ 3 ) ⎛ sin ( θ 1 ) ⎞ ⎜ ⎟ ⎝ sin ( θ 3 ) ⎠ F 1v = F 1 F 1v = 12 9 N F 1u sin ( 18 0 deg − θ 1 − θ 3 ) F 1u = F1 = F1 sin... writing from the publisher Engineering Mechanics - Statics Chapter 2 F 2 = 15 0 N θ 1 = 30 deg θ 2 = 30 deg θ 3 = 10 5 deg Solution: F 1v sin ( θ 1 ) F2 = F 1v = F 2 sin ( 18 0 deg − θ 3 ) sin ( θ 1 ) ⎛ ⎞ ⎜ ⎟ ⎝ sin ( 18 0 deg − θ 3 ) ⎠ F 1v = 77.6 N F2u sin ( 18 0 deg − θ 3 ) F 2u = F2 = F2 sin ( 18 0 deg − θ 3 ) ⎛ sin ( 18 0 deg − θ 3) ⎞ ⎜ ⎟ ⎝ sin ( 18 0 deg − θ 3) ⎠ F 2u = 15 0 N Problem 2 -1 7 Determine the magnitude... means, without permission in writing from the publisher Engineering Mechanics - Statics F 2 = 500 N Chapter 2 β = 45 deg γ = 70 deg Solution: F1u = sin ( γ − α ) F1 sin ( 18 0 deg − γ ) sin ( γ − α ) F 1u = F1 sin ( 18 0 deg − γ ) F 1u = 205 N F 1v sin ( α ) = F 1v = F 1 F1 sin ( 18 0 deg − γ ) sin ( α ) sin ( 18 0 deg − γ ) F 1v = 16 0 N Problem 2-6 Resolve the force F2 into components acting along the.. .Engineering Mechanics - Statics Chapter 1 Problem 1- 1 9 Evaluate each of the following to three significant figures and express each answer in SI units using an appropriate prefix: (a) a 1/ b1, (b) a2b2/c2, (c) a3b3 Units Used: μm = 10 −6 6 m Mm = 10 m Mg = 10 gm kg = 10 gm 6 −3 ms = 10 3 s Given: a1 = 684 μm b1 = 43 ms a2 = 28 ms b2 = 0.0458 Mm c2 = 348 mg a3 = 2.68 mm b3 = 426 Mg Solution: a1 (... without permission in writing from the publisher Engineering Mechanics - Statics Chapter 2 Given: F a = 30 lb θ 1 = 80 deg θ 2 = 60 deg Solution: Fa = sin ( θ 1 ) sin 18 0 deg − ( θ 1 + θ 2 )⎤ ⎣ ⎦ ⎛ sin ( 18 0 deg − θ 1 − θ 2) ⎞ ⎜ ⎟ sin ( θ 1 ) ⎝ ⎠ F = Fa Fa = sin ( θ 1 ) Fb = F F = 19 .6 lb Fb sin ( θ 2 ) Fa sin ( θ 2 ) sin ( θ 1 ) F b = 26.4 lb Problem 2 -1 3 A resultant force F is necessary to hold the... writing from the publisher Engineering Mechanics - Statics Chapter 2 Given: F R = 500 lb F 1 = 200 lb F 2 = 300 lb φ = 30 deg Solution: Cosine Law: F R1 = F 1 + F2 − 2 F 1 F2 cos ( 90 deg − φ ) 2 2 F R1 = 264.6 lb Sine Law: Make F parallel to FR1 sin ( φ + θ ) sin ( 90 deg − φ ) = F1 F R1 ⎛ θ = −φ + asin ⎜ sin ( 90 deg − φ ) ⎝ ⎞ ⎟ FR1 ⎠ F1 θ = 10 .9 deg When F is directed along F R1, F will be minimum to... publisher Engineering Mechanics - Statics Chapter 2 Problem 2 -1 9 The riveted bracket supports two forces Determine the angle θ so that the resultant force is directed along the negative x axis What is the magnitude of this resultant force? Given: F 1 = 60 lb F 2 = 70 lb θ 1 = 30 deg Solution: sin ( θ 1 ) sin ( θ ) = F1 F2 ⎛ F1 ⎞ ⎝ F2 ⎠ θ = asin ⎜ sin ( θ 1 ) ⎟ θ = 25.4 deg φ = 18 0 deg − θ − θ 1 φ = 12 4.6... Engineering Mechanics - Statics Chapter 2 sin ( θ 1 ) ⎡ ⎤ ⎢ ⎥ ⎣ sin 18 0 deg − ( θ 1 + θ 2)⎤⎦ ⎣ ⎦ F AB = F F AB = 18 6 lb F AC sin ( θ 2 ) F = sin 18 0 deg − ( θ 1 + θ 2 )⎤ ⎣ ⎦ sin ( θ 2 ) ⎡ ⎤ ⎢ ⎥ ⎣sin 18 0 deg − ( θ 1 + θ 2 )⎤ ⎦ ⎣ ⎦ F AC = F F AC = 239 lb Problem 2 -1 4 The post is to be pulled out of the ground using two ropes A and B Rope A is subjected to force F 1 and is directed at angle 1 from the... in writing from the publisher Engineering Mechanics - Statics Chapter 2 θ = 12 0 deg Solution: FR = F1 + F 2 − 2 F1 F 2 cos ( 18 0 deg − θ ) 2 2 F R = 72 .1 lb ⎛ ⎝ β = asin ⎜ F1 sin ( 18 0 deg − θ ) ⎞ ⎟ FR ⎠ β = 73.9 deg Problem 2-3 Determine the magnitude of the resultant force F R = F1 + F 2 and its direction, measured counterclockwise from the positive x axis Given: F 1 = 250 lb F 2 = 375 lb θ = 30... permission in writing from the publisher Engineering Mechanics - Statics Chapter 2 Given: F 1 = 25 lb F 2 = 50 lb θ 1 = 30 deg θ 2 = 30 deg θ 3 = 45 deg Solution: −F u = sin ( θ 3 − θ 2 ) Fu = F1 sin ( θ 2 ) −F1 sin ( θ 3 − θ 2 ) sin ( θ 2 ) F u = 12 .9 lb Fv sin ( 18 0 deg − θ 3 ) Fv = = F1 sin ( θ 2 ) F 1 sin ( 18 0 deg − θ 3 ) sin ( θ 2 ) F v = 35.4 lb Problem 2-9 Resolve the force F2 into components . Used: μN10 6− N= μkm 10 6− km= Gs 10 9 s= ks 10 3 s= mN 10 3− N= ms 10 3− s= Solution: a() m ms 11 0 3 × m s = m ms 1 km s = b() μkm 1 10 3− × m= μkm 1mm= c() ks mg 11 0 9 × s kg = ks mg 1 Gs kg = d(). of each particle. Units Used: G 66.73 10 12 − × m 3 kg s 2 ⋅ = nN 10 9− N= Given: m 1 8kg= m 2 12 kg= d 800 mm= Solution: F Gm 1 m 2 d 2 = F 10 .0 nN= W 1 m 1 g= W 1 78.5 N= W 1 F 7.85 10 9 ×= W 2 m 2 g=. the publisher. Engineering Mechanics - Statics Chapter 1 Problem 1- 1 1 If an object has mass m, determine its mass in kilograms. Given: m 40 slug= Solution: m 584 kg= Problem 1- 1 2 The specific