Engineering Mechanics: Statics – Instructor’s Solutions Manual (13th Edition, R.C. Hibbeler) | Complete Worked Solutions

1–1.

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
a) 58.3 km b) 68.5 s c) 2.55 kN d) 7.56 Mg Ans.

,*1–4.

Represent each of the following combinations of units in
the correct SI form using an appropriate prefix: (a) m>ms,
(b) mkm, (c) ks>mg, and (d) km # mN.




SOLUTION
m 11023 m
a) m>ms = ¢ ≤ = ¢ ≤ = km>s
1102-3 s
Ans.
s

b) mkm = 1102-611023 m = 1102-3 m = mm Ans.

11023 s 11029 s
c) ks>mg = = = Gs>kg
1102-6 kg
Ans.
kg

d) km # mN = 10 3
m 10 -6
N = 10 -3
m # N = mm # N Ans.

,1–5.

Represent each of the following quantities in the correct
SI form using an appropriate prefix: (a) 0.000 431 kg,
(b) 35.3(103) N, and (c) 0.005 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.

, 1–9.

A rocket has a mass of 250(103) slugs on earth. Specify
(a) its mass in SI units and (b) its weight in SI units. If the
rocket is on the moon, where the acceleration due to gravity
is gm = 5.30 ft>s2, determine to three significant figures
(c) its weight in SI units and (d) its mass in SI units.




SOLUTION
Using Table 1–2 and applying Eq. 1–3, we have

a) 250 A 103 B slugs = C 250 A 103 B slugs D a b
14.59 kg
1 slugs

= 3.6475 A 106 B kg

= 3.65 Gg Ans.

b) We = mg = C 3.6475 A 106 B kg D A 9.81 m>s2 B

= 35.792 A 106 B kg # m>s2

= 35.8 MN Ans.

c) Wm = mgm = C 250 A 103 B slugs D A 5.30 ft>s2 B

= C 1.325 A 106 B lb D a b
4.448 N
1 lb

= 5.894 A 106 B N = 5.89 MN Ans.
Or
gm 5.30 ft>s2
Wm = We a b = (35.792 MN) a b = 5.89 MN
g 32.2 ft>s2

d) Since the mass is independent of its location, then

mm = me = 3.65 A 106 B kg = 3.65 Gg Ans.