Solutions Manual for Engineering Mechanics: Statics (14th Edition, 2019) by Hibbeler

ALL 11 CHAPTERS COVERED




SOLUTION MANUAL

,
,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.

,1–2.

Wood has a density of 4.70 slug>ft3. What is its density
expressed in SI units?




SOLUTION

(4.70 slug>ft3) b r = 2.42 Mg>m3
(1 ft3)(14.59 kg)
Ans.
(0.3048 m)3(1 slug)

,1–3.

Represent each of the following combinations of units in
the correct SI form using an appropriate prefix: (a) kN>ms,
(b) Mg>mN, and (c) MN>(kg # ms).




SOLUTION
(103) N (109) N
a) kN>ms = = = GN>s Ans.
(10-6) s s

(106) g (109) g
b) Mg>mN = = = Gg>N Ans.
(10 ) N
-3 N

(106) N (109) N
c) MN>(kg # ms) = = = GN>(kg # s) Ans.
kg # (10-3) s kg # s

,*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
11023 m
a) m>ms = ¢ ≤ ¢ ≤ = km>s
m
1102-3 s
= Ans.
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
3
d) km # mN = 10 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.

,1–6.

If a car is traveling at 55 mi>h, determine its speed in
kilometers per hour and meters per second.




SOLUTION

55 mi>h = a ba ba ba b
55 mi 5280 ft 0.3048 m 1 km
1h 1 mi 1 ft 1000 m
= 88.5 km>h Ans.

88.5 km>h = a ba ba b = 24.6 m>s
88.5 km 1000 m 1h
Ans.
1h 1 km 3600 s

,1–7.

The pascal (Pa) is actually a very small unit of pressure. To
show this, convert 1 Pa = 1 N>m2 to lb>ft2. Atmospheric
pressure at sea level is 14.7 lb> in2. How many pascals is this?




SOLUTION
Using Table 1–2, we have

b = 20.9 A 10 - 3 B lb>ft2
0.30482 m2
a b a
1N 1 lb
1 Pa = Ans.
m2 4.4482 N 1 ft2

144 in2 1 ft2
a b a b a b
14.7 lb 4.448 N
1 ATM =
in2 1 lb 1 ft2 0.30482 m2

= 101.3 A 103 B N>m2

= 101 kPa Ans.

, *1–8.

The specific weight (wt.> vol.) of brass is 520 lb>ft3.
Determine its density (mass> vol.) in SI units. Use an
appropriate prefix.




SOLUTION
3
520 lb>ft3 = a ba b a ba b
520 lb 1 ft 4.448 N 1 kg
3 0.3048 m 1 lb 9.81 N
ft

= 8.33 Mg>m3 Ans.