1–1.What is the weight in newtons of an object that has a mass of (a) 8 kg, (b) 0.04 kg, (c) 760 Mg? Solution (a) W = 9.81(8) = 78.5 N Ans. (b) W = 9.81(0.04) ( 10 -3 ) = 3.92 ( 10 ) N = 0.392 mN -4 Ans. (c) W = 9.81(760) ( 103 ) = 7.46 ( 106 ) N = 7.46 MN Ans. Instant download and all chapters Solutions Manual Statics and Mechanics of Materials 5th Edition Russell C. Hibbeler https://testbankdata.com/download/solutions-manual-statics-mechanics-materials-5th-ed ition-russell-c-hibbeler/ Ans: 78.5 N 0.392 mN 7.46 MN 1 1–2. Represent each of the following combinations of units in the correct SI form using an appropriate prefix: (a) kN>ms, (b) Mg>mN, (c) MN>(kg # ms). Solution (a) kN>ms = 103N> ( 10 - 6 ) s = GN>s Ans. 6 -3 (b) Mg>mN = 10 g>10 N = Gg>N Ans. (c) MN>(kg # ms) = 106 N>kg(10 - 3 s) = GN>(kg # s) Ans. Ans: GN>s Gg>N GN>(kg # s) 2 1–3. Represent each of the following combinations of units in the correct SI form: (a) Mg>ms, (b) N>mm, (c) mN>(kg # ms). Solution Mg 103 kg (a) = = 106 kg>s = Gg>s Ans. ms 10-3 s N 1N (b) = = 103 N>m = kN>m Ans. mm 10-3 m 10-3 N = kN>(kg # s) mN (c) # (kg ms) = 10-6 kg # s Ans. Ans: Gg>s kN>m kN>(kg # s) 3 *1–4. Convert: 200 lb # ft to N # m, (b) 350 lb>ft (a) 200 Convert: (a) lb>ft33 to kN>m33, , to kN>m (c) 8 ft>h ft>htotomm>s. mm>s.Express Expressthetheresult resulttotothree threesignificant significant figures. Use Use an an appropriate appropriateprefix. prefix. SOLUTION (a) (200 lb # ft) ¢ ≤¢ ≤ = 271 N # m 4.4482 N 0.3048 m Ans. 1 lb 1 ft (b) ¢ 3 ≤ ¢ 0.3048 m ≤ ¢ ≤ = 55.0 kN>m3 3 350 lb 1 ft 4.4482 N Ans. 1 ft 1 lb (c) ¢ ≤¢ ≤¢ ≤ = 0.677 mm>s 8 ft 1h 0.3048 m Ans. 1h 3600 s 1 ft Ans: 271 N # m 55.0 kN>m3 0.677 mm>s 4 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, (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. Ans: 45.3 MN 56.8 km 5.63 mg 5 1–6. Round off the following numbers to three significant figures: (a) (a)5858342 342 m, m, (b) (b) 68.534 68.534 s, (c)s,2553 (c) N, 2553 N, and (d) 7555 kg. (d) 7555 kg. SOLUTION (a) 58.3 km (b) 68.5 s (c) 2.55 kN (d) 7.56 Mg Ans. Ans: 58.3 km 68.5 s 2.55 kN 7.56 Mg 6 1–7. Represent each Represent each ofof thethe following following quantities quantities in the in the correct correct SI form SI form using using an prefix: an appropriate appropriate prefix: (a) 0.000 431 kg,(a) 35.3 (10 (b)0.000 431 ) N, 3 kg, 3 (b) 35.3(10 (c) 0.005 ) N, and (c) 0.005 32 km. 32 km. SOLUTION (a) 0.000 431 kg = 0.000 431 A 103 B g = 0.431 g Ans. (b) 35.3 A 103 B N = 35.3 kN Ans. (c) 0.005 32 km = 0.005 32 A 103 B m = 5.32 m Ans. Ans: 0.431 g 35.3 kN 5.32 m 7 *1–8. Represent each Represent eachofofthe thefollowing following combinations combinations of units of units in in the the correct correct SI form SI form using using ananappropriate appropriateprefix: prefix:(a) (a) Mg>mm, Mg/mm, (b) mN>ms, (c) mm ## Mg. (c) mm Mg. SOLUTION 103 kg 106 kg (a) Mg>mm = -3 = = Gg>m Ans. 10 m m 10 - 3 N 103 N (b) mN>ms = = = kN>s Ans. 10 - 6 s s (c) mm # Mg = C 10 -6 m D # C 103 kg D = (10)-3 m # kg = mm # kg Ans. Ans: Gg>m kN>s mm # kg 8 1–9. Represent each of the following combinations of units in the correct SI form using an an appropriate appropriate prefix: prefix: (a) m>ms, (a) m>ms, (b) mkm, (c) ks>mg, (d) (c) ks>mg, andkm km # mN. (d)# mN. SOLUTION 11023 m (a) m>ms = ¢ -3 ≤ = ¢ ≤ = km>s m Ans. 1102 s s (b) mkm = 1102-611023 m = 1102-3 m = mm Ans. 11023 s 11029 s (c) ks>mg = = = Gs>kg Ans. 1102-6 kg kg (d) km # mN = 10 3 m 10 -6 N = 10 -3 m # N = mm # N Ans. Ans: km>s mm Gs>kg mm # N 9 1–10. Represent each of the following combinations of units in the correct SI form using an appropriate prefix: (a) GN # mm, (b) kg>mm, (c) N>ks2, and (d) kN>ms. Solution (a) GN # mm = 109 ( 10-6 ) N # m = kN # m Ans. 3 -6 (b) kg>mm = 10 g>10 m = Gg>m Ans. N>ks2 = N>106 s2 = 10-6 N>s2 = mN>s2 (c) Ans. 3 -6 9 (d) kN>ms = 10 N>10 s = 10 N>s = GN>s Ans. Ans: kN # m Gg>m mN>s2 GN>s 10
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