nd
STATICS AND MECHANICS OF MATERIALS, 2 Edition
RILEY, STURGES AND MORRIS
Chapter 1 1-1
Calculate the mass m of a body that weighs 600 lb at the surface of the earth. SOLUTION
m 1-2
W g
600 32.2
......................................... ......................... ...........Ans. 18.63 slug ........................... Ans.
Calculate the weight W of a body at the surface of the earth if it has a mass m of 675 kg. SOLUTION
W mg 6759.81 6.62 10 3 N 6.62 kN ..................................Ans. Ans. 1-3
If a man weighs 180 lb at sea level, determine the weight W of the man (a) At the top of Mt. McKinley (20,320 ft above sea level). (b)
At the top of Mt. Everest (29,028 ft above sea level).
SOLUTION
W
r 0 2.090 10 7 ft
where
r h r 0 h 2.090 10 7 2.0320 10 4 2.092032 10 7 ft
(a)
W h
1802.090 10 7
2
W 0 r 02
r h2
2.092032 10
7 2
...................................... ...........Ans. 179.7 lb ...........................
r h r 0 h 2.090 10 7 2.9028 10 4 2.0929028 10 7 ft
(b)
W h
W 0 r 02 r h2
2.0929028 10 2
180 2.090 107
7 2
179.5 lb ....................................Ans. Ans.
Calculate the weight W of a navigation satellite at a distance of 20,200 km above the earth’s surface if the satellite weighs 9750 N at the earth’s surface. SOLUTION
W
Gme m r e2
W 0 r 02 W h r h2 Gme m
Therefore:
r 0 6.370 106 m
where (a)
r e2
W 0 r 02 W h r h2 Gme m
Therefore:
1-4
Gme m
rh r0 h 6370 20, 20 200 26, 57 570 km W h
W0r 02 r h2
9750 6370
26,570
1
2
2
......................................... ................ ..Ans. 560 N ...........................
nd
STATICS AND MECHANICS OF MATERIALS, 2 Edition 1-5
Compute the gravitational force acting between two spheres that are touching each other if each sphere weighs 1125 lb and has a diameter of 20 in. SOLUTION
F
1-6
Gm1m2
Gm s m s
r 2
r s r s 2
3.439 108 1125
2012
Gm s2 d s2
32.2
2
2
15.11 106 lb ..................................Ans. Ans.
Two spherical bodies have masses of 60 kg and 80 kg, respectively. Determine the gravitational force of attraction between the spheres if the distance from center to center is 600 mm. SOLUTION
F 1-7
RILEY, STURGES AND MORRIS
Gm1 m2 2
r
6.673 10 60 80 11
0.600
0.890 106 N
2
........................Ans.
Determine the weight W of a satellite when it is in orbit 8500 mi above the surface of the earth if the satellite weighs 7600 lb at the earth’s surface. SOLUTION
W
Gm1m2 r 2
W 0 r 02 W h r h2 Gme mb
Therefore:
r 0 2.090 10 7 ft
where
r h r o h 2.090 10 7 8500(5280) 6.578 10 7 ft W h 1-8
2 o o 2 h
W r r
7600 2.090 10
7
6.578 10 7
2
2
768 lb .....................................Ans. Ans. 6
Determine the weight W of a satellite when it is in orbit 20.2(10 ) m above the surface of the earth if the satellite weighs 8450 N at the earth’s surface. SOLUTION
W
Gm1m2 r 2
W 0 r 02 W h r h2 Gme mb
Therefore:
r 0 6.370 106 m
where
rh ro h 6.370 106 20.2 106 26.570 106 m W h
2 o o 2 h
W r r
8450 6.370 106
26.570 10 6
2
2
2
486 N .....................................Ans. Ans.
nd
STATICS AND MECHANICS OF MATERIALS, 2 Edition 1-9
RILEY, STURGES AND MORRIS
If a woman weighs 135 lb when standing on the surface of the earth, how much would she weigh when standing on the surface of the moon? SOLUTION
W Therefore on the surface of the earth where
Gm1m2 r 2
me 4.095 1023 slugs and r 3960 mi
135
G 4.095 1023 m
3960 5280
2
Gm 1.4413 107 lb ft 2 /slug Then on the surface of the moon where
mm 5.037 1021 slugs and r 1080 mi
1.4413 10 5.037 10 7
W
21
1080 5280
2
22.33 lb ...................................Ans. Ans.
1-10 Determine the weight W of a body that has a mass of 1000 kg (a.) At the surface of the earth.
(b.)
At the top of Mt. McKinley (6193 m above sea level).
(c.)
In a satellite at an altitude of 250 km.
SOLUTION
W
Gm1m2 r 2
6.673 10 5.976 10 1000 W 9830 N .............................Ans. Ans. 6370 10 6.673 10 5.976 10 1000 W 9810 N .............................Ans. Ans. 6370 10 6193 6.673 10 5.976 10 1000 W 9100 N .............................Ans. Ans. 6370 10 250 10 11
(a)
24
3
2
11
(b)
24
2
3
11
(c)
24
3
3
2
1-11 If a man weighs 210 lb at sea level, determine the weight W of the man (a.) At the top of Mt. Everest Everest (29,028 ft above sea level).
(b.)
In a satellite at an altitude of 200 mi.
SOLUTION
W Therefore
210
Gm1m2 r 2 Gmem
3960 5280
2
Gmem 9.181 1016 lb ft 2
3
nd
STATICS AND MECHANICS OF MATERIALS, 2 Edition
W
(a)
RILEY, STURGES AND MORRIS
9.181 1016 2
....................................... ...........Ans. 209.4 lb ............................
2
........................................ ..............Ans. 190.3 lb ..........................
028 3960 5280 29, 02 9.181 10
16
W
(b)
3960 200 5280
25 6 1-12 A space traveler weighs 800 N on earth. A planet having a mass of 5(10 ) kg and a diameter of 30(10 ) m orbits a distant star. Determine the weight W of the traveler on the surface of this planet. SOLUTION
W Therefore on the surface of the earth where
Gm1m2 r 2
me 5.976 1024 kg and r 6370 km
800
G 5.976 1024 m
6370 10 3
2
Gm 5.432 109 N m2 /kg Then on the surface of the planet where
m 5 1025 kg and r 15 106 m
5.432 10 5 10 W ......................................... ..............Ans. 1207 N ........................... 15 10 9
25
6
2
26 1-13 The planet Jupiter has a mass of 1.302(10 ) slug and a visible diameter (top of the cloud layer) of 88,700 mi. Determine the gravitational acceleration g (a.) At a point 100,000 miles miles above the top of the clouds.
(b.)
At the top of the cloud layers.
SOLUTION
W
Gm1m2 r 2
3.439 10 1.302 10 m 8
W mg
(a)
26
350 100, 00 000 5280 44, 35
2
g 7.71 ft/s 2 ............................ .......................................... ............................ ....................... .........Ans. Ans.
3.439 10 1.302 10 m 8
W mg
(b)
26
,350 5280 44,35
2
g 81.7 ft/s 2 ........................... ......................................... ............................ ....................... .........Ans. Ans. 26 1-14 The planet Saturn has a mass of 5.67( 10 ) kg and a visible diameter (top of the cloud layer) of 120,000 km. The weight W of a planetary probe on earth is 4.50 kN. Determine (a.) The weight weight of the probe when it is 600,000 km above the top of the clouds.
(b.)
The weight weight of the probe as it begins its penetration of the cloud layers. layers.
SOLUTION
W
Gm1m2 r 2
4
nd
STATICS AND MECHANICS OF MATERIALS, 2 Edition Therefore on the surface of the earth where
RILEY, STURGES AND MORRIS
me 5.976 1024 kg and r 6370 km
4500
G 5.976 1024 m
6370 10 3
2
Gm 3.055 108 N m2 /kg
3.055 10 5.67 10 8
W
(a)
26
60, 00 000 600, 00 000 103
2
39.8 N .....................................Ans. Ans.
3.055 10 5.67 10 W ...................................... ...........Ans. 4810 N ........................... 000 10 60, 00 8
(b)
26
3
2
1-15 The first U.S. satellite, Explorer 1 , had a mass of approximately 1 slug. Determine the force exerted on the satellite by the earth at the low and high points of it s orbit, which were 175 mi and 2200 mi, respectively, above the surface of the earth. SOLUTION
Gm1 m2
F
r 2
3.439 10 4.095 10 1 8
F
23
3960 175 5280
3.439 10 4.095 10 1 8
F
2
29.5 lb ...................................Ans. Ans.
23
3960 2200 5280
2
13.31 lb ..................................Ans. Ans.
30
1-16 A neutron star has a mass of 2(10 ) kg and a diameter of 10 km. Determine the gravitational force of attraction on a 10-kg space probe (a.) When it is 1000 km from the center of the star.
(b.)
At the instant of impact with the surface of the star. star.
SOLUTION
Gm1 m2
F
r 2
6.673 10 2 10 10 F 1.335 10 1000 10 6.673 10 2 10 10 F 5.34 10 5 10 11
30
3
11
9
N ...............................Ans. Ans.
13
N ...............................Ans. Ans.
2
30
3
2
1-17 Determine the weight W , in U.S. customary units, of a 75-kg steel bar under standard conditions (sea level at a latitude of 45 degrees). SOLUTION
W mg 75 9.81 735.75 N 0.2248 lb/N 165.4 lb ......................Ans.
5
nd
STATICS AND MECHANICS OF MATERIALS, 2 Edition
RILEY, STURGES AND MORRIS
1-18 Determine the mass m, in SI units, for a 500-lb steel beam under standard conditions (sea level at a latitude of 45 degrees). SOLUTION
W mg m
W g
500 32.2
slug 14.59 kg kg/slug 227 kg kg ........................Ans. 15.528 sl
displacement. Determine the engine displacement in liters. 1-19 An automobile has a 440 cubic inch engine displacement. SOLUTION 3
V 440 in.3 16.39 103 mm3 /in.3 101 cm/mm 103 L/cm3 .......................................... ............................ ......................... ...........Ans. V 7.21 L ............................ Ans. 1-20 How many barrels of oil are contained in 100 kL of oil? One barrel (petroleum) equals 42.0 gal. SOLUTION
V 100 103 L 0.2642 gal/L 1 barrel 42 gal 629 barrel ..................Ans. 1-21* Express the density, in SI units, of a specimen of material that has a specific weight of 0.025 lb/in. SOLUTION
3
g
1000 mm 3 in. 0.025 slug 14.59 kg g 32.2 in.3 slug 16.39 103 mm3 m 3
691 kg/m ............................ .......................................... ............................ ..................... .......Ans. Ans. 3
-3
2
1-22 The viscosity of crude oil under conditions of standard temperature and pressure is 7.13(10 ) N-s/m . Determine the viscosity of crude oil in U.S. Customary units. SOLUTION
0.2248 lb 0.0929 m 3N s 7.13 10 2 2 m N ft
2
3 lb s 0 . 1 4 8 9 1 0 2 ft
...........Ans.
2 3 1-23 One acre equals 43,560 ft . One gallon equals 231 in. Determine the number of liters of water required to cover 2000 acres to a depth of 1 foot.
SOLUTION 3
43, 560 ft 2 12 in gal 3.785 L V 2000 2000 acr acree ft 3 acre ft 231 in. gal V 2.47 109 L ........................... ......................................... ............................ ..................... .......Ans. Ans. 1-24 The stress in a steel bar is 150 MPa. Express the stress in appropriate U.S. Customary units (ksi) by using the values listed in Table 1-6 for length and force as defined values. SOLUTION
ft stress 150 10 N/m 0.2248 lb/N 0.0929 m /ft 12 in. 6
2
2
2
2
stress 21.75 10 lb/in. 21.75 ksi .......................... ........................................ ................ Ans. 3
2
6
nd
STATICS AND MECHANICS OF MATERIALS, 2 Edition
RILEY, STURGES AND MORRIS
ft-lb/min and 1 W = 1 N-m/s. N-m/s. Verify the conversion factors factors listed in Table 1-6 1-25 By definition, 1 hp = 33,000 ft-lb/min for converting power from U.S. Customary units to SI units by using the values listed for length and force as defined values. SOLUTION
1 hp 33, 000
ft lb 4.448 N 0.3048 m min
min
lb
60 s 745.7
ft
N m s
.............Ans.
1-26 The specific heat of air under standard atmospheric pressure, in SI units, is 1003 N-m/kg-K. Determine the specific heat of air under standard atmospheric pr essure in U.S. customary units (ft-lb/slug-R). SOLUTION
1003 N m
0.2248 lb 3.281 ft 14.59 kg 5 K 9 R m slug kgK N
.......................................... ............................ ................ ..Ans. 6000 lb ft slug R ............................ Ans. 1-27 Newton’s law of gravitation can be expressed in equation form as
m1m2
F G
r 2
F is a force, m1 and m2 are masses, and r is a distance, determine the di mensions of G. If F SOLUTION 2
G
m1m2
FMrL T L 2
M M
2
3 L ............................ .......................................... ..............Ans. 2 M T
1-28 The elongation of a bar of uniform cross section subjected to an axial force i s given by the equation
PL EA . What are the dimensions of E E if if and and L are lengths, P is a force, and A is an area? SOLUTION
E
2 PL ML T L
A
L L2
M 2 L T
F ........................... ...................................... ...........Ans. 2 L
1-29 The period of oscill ation of a simple pendulum is given by the equation
T k L g , where T is in seconds,
L is in feet, g is the acceleration due to gravity, and k is a constant. constant. What are the dimensions dimensions of k for dimensional homogeneity? SOLUTION 12
L T 2 k T g L T L
1 (dimensionless) ...............................Ans. Ans.
1-30 An important parameter in fluid flow problems involving thin films i s the Weber number (We) which can be expressed in equation form as
We
v 2 L
where is is the density of the fluid, v is a velocity, L is a length, and is is the surface surface tension of the fluid. If the Weber number is dimensionless, what are the dimensions of the surface tension ? SOLUTION
v 2 L We
M
L3 L T
2
L M F 2 ................................Ans. Ans. T L
1
7
nd
STATICS AND MECHANICS OF MATERIALS, 2 Edition
RILEY, STURGES AND MORRIS
1-31 In the dimensionally homogeneous equation
P Mc A
I
is is a stress, A is an area, M is a moment of a force, and c is a length. Determine the dimensions dimensions of P and I . SOLUTION 2 2 P ML T L M 2 LT 2 I L
Therefore
M 2 ML L 2 F.......................... ............................ ............ ..................... .......Ans. 2 LT T
P
ML I M 2
T 2 L 2
LT
........................................ ............................ ..............Ans. L4 .......................... Ans.
1-32 In the dimensionally homogeneous equation
Pd
1 2
mv2 12 I 2
is an angular velocity. velocity. Determine the dimensions dimensions of P d is a length, m is a mass, v is a linear velocity, and is P and I . SOLUTION 2
L 1 P L M I 2 T T Therefore
ML2 T 2 ML P ......................................... ..................... .......Ans. 2 F ........................... L T ML2 2 ........................................ ............................ ..............Ans. I 2 T M2L .......................... Ans. T 1-33 In the dimensionally homogeneous equation
Tr VQ J
I b
is is a stress, T is a torque (moment of a force), V is a force, r and b are lengths and I is a second moment of an J and Q. area. Determine the dimensions of J SOLUTION 2 2 2 M ML T L ML T Q 2LT J L4 L
Therefore
ML J M 2
T 2 L 2
LT
8
L4
............................ .......................................... ......................... ...........Ans. Ans.
nd
STATICS AND MECHANICS OF MATERIALS, 2 Edition
RILEY, STURGES AND MORRIS
M LT L L Q .......................................... ......................... ...........Ans. ............................ Ans. ML T 2
5
3
2
1-34 In the dimensionally homogeneous equation
P Tr A
J
is is a stress, A is an area, T is a torque (moment of a force), and r is a length. Determine the dimensions dimensions of P and J . SOLUTION 2 2 P ML T L M 2 LT 2 J L
Therefore
M 2 ML L 2 F.......................... ............................ ............ ..................... .......Ans. 2 L T T
P
ML J M 2
1-35 The equation x
Ae t b sin at
T 2 L 2
LT
L4
............................ .......................................... ......................... ...........Ans. Ans.
is dimensionally dimensionally homogeneous. If A A is a length and t is time,
determine the dimensions of x x, a, b, and . SOLUTION T b
x L e
sin a T
Therefore
x L 1 1 L ........................... ......................................... ............................ .................. ....Ans. Ans. b T ............................ Ans. .......................................... ............................ ............................ ..............Ans. a 1 T .......................... ........................................ ............................ ............................ ..............Ans. Ans. 1 (dimensionless) ............................ .......................................... ............................ ..............Ans. Ans. 1-36 In the dimensionally homogeneous equation
w x 3 ax 2 bx a 2b x , if x x is a length, what are the
dimensions of a, b, and w? SOLUTION If x
L , then each term has the dimension L3 .
Therefore
w L3 ........................... ......................................... ............................ ............................ ..............Ans. Ans.
L L a .......................................... ............................ ..................... .......Ans. ............................ Ans. L L b .......................................... ............................ ..................... .......Ans. L ............................ Ans. L 3 2
3
2
Using the last term as a check,
9
nd
STATICS AND MECHANICS OF MATERIALS, 2 Edition
L L 2
a 2b
x
RILEY, STURGES AND MORRIS
2
L
L3
1-37 Determine the dimensions of a, b, c, and y in the dimensionally homogeneous equation
y Ae bt cos 1 a 2 bt c in which A is a length and t is time. SOLUTION bT
y L e
cos 1 a2 b T c
Therefore
y L 1 1 L ............................ .......................................... ............................ ................ ..Ans. Ans. b T 1
b 1 T .......................... ........................................ ............................ ............................ ..............Ans. Ans. a 1 (dimensionless) ............................ .......................................... ............................ ..............Ans. Ans. c 1 (dimensionless) ............................ .......................................... ............................ ..............Ans. Ans.
1-38 Determine the dimensions of c, , k and P in the differential equation
m
d 2 x
c
2
dt
dx dt
kx P cos t
in which m is a mass, x is a length, and t is time. SOLUTION
M L
T 2
c
L k L P cos T T
Therefore
ML T 2
c
M ......................................... ............................ ................ ..Ans. ........................... Ans. L T T
ML T 2
k
L
M F ......................................... ....................... .........Ans. 2 ........................... Ans. T L
ML ........................................ ............................ ..................... .......Ans. Ans. F .......................... 2 T
P
1 T .......................... ........................................ ............................ ............................ ..............Ans. Ans. numbers to two significant figures. figures. Find the percent difference between each 1-39 Round off the following numbers rounded-off number and the original number by using the original number as the reference. (a) 0.015362 (b) 55.33682 (c) 63,746.27 SOLUTION
(a)
(b)
0.015 0.0153 15362 0.015362 55 55.33682 55.33682
......................................... .................. ....Ans. 100 2.36 % ...........................
........................................ ..................... .......Ans. 100 0.609 % ..........................
10
nd
STATICS AND MECHANICS OF MATERIALS, 2 Edition
64,000 ,000 63,746 ,746.27
(c)
63,746.27
RILEY, STURGES AND MORRIS
......................................... ..............Ans. 100 0.398 % ...........................
numbers to two significant figures. figures. Find the percent difference between each 1-40 Round off the following numbers rounded-off number and the original number by using the original number as the reference. (a) 0.837482 (b) 374.9371 (c) 937,284.9 SOLUTION
0.84 0.837482
(a)
0.837482 370 374.9371
(b)
374.9371
......................................... .................. ....Ans. 100 0.301 % ...........................
100 1.317 %
940,000 ,000 937 937,284 ,284.9
(c)
937,284.9
........................... ......................................... .................. ....Ans.
......................................... ..............Ans. 100 0.290% ...........................
numbers to three significant figures. Find the percent difference between each each 1-41 Round off the following numbers rounded-off number and the original number by using the original number as the reference. (a) 0.034739 (b) 26.39473 (c) 55,129.92 SOLUTION
0.0 0.0347 0.0 0.034739 739
(a)
0.034739 26.4 26.39473
(b)
26.39473
.......................................... ..............Ans. 100 0.01997 % ............................
55,1 55,100 00 55,1 55,129 29.9 .92 2
(c)
......................................... ..............Ans. 100 0.1123 % ...........................
55,129.92
......................................... ..............Ans. 100 0.0543 % ...........................
numbers to three significant figures. Find the percent difference between each each 1-42 Round off the following numbers rounded-off number and the original number by using the original number as the reference. (a) 0.472916 (b) 826.4836 (c) 339,872.8 SOLUTION
0.4 0.473 0.47291 2916
(a)
0.472916 826 826.4836
(b)
826.4836
......................................... ..............Ans. 100 0.01776 % ...........................
.......................................... ................ ..Ans. 100 0.0585 % ............................
340 340,000 ,000 339,872. ,872.8 8
(c)
339,872.8
....................................... ...........Ans. 100 0.0374 % ............................
numbers to four significant figures. figures. Find the percent difference between each 1-43 Round off the following numbers rounded-off number and the original number by using the original number as the reference. (a) 0.056623 (b) 74.82917 (c) 27,382.84 SOLUTION
(a)
(b)
0.056 05662 0.0 0.056623 623 0.056623 74.8 4.83 74. 74.82917 917 74.82917
100 5.30 103 % .....................................Ans. Ans.
...................................... ...........Ans. 100 1.109 103 % ...........................
11
nd
STATICS AND MECHANICS OF MATERIALS, 2 Edition
27,38 ,380 27,38 ,382.84
(c)
27,382.84
RILEY, STURGES AND MORRIS
100 10.37 103 % ....................................Ans. Ans.
numbers to four significant figures. figures. Find the percent difference between each 1-44 Round off the following numbers rounded-off number and the original number by using the original number as the reference. (a) 0.664473 (b) 349.3378 (c) 274,918.2 SOLUTION
(a)
(b)
(c)
0.664 6645 0.66447 4473 0.664473 349. 49.3 349.33 .3378 349.3378
...................................... ...........Ans. 100 4.06 103 % ...........................
...................................... ...........Ans. 100 10.82 103 % ...........................
274,900 ,900 274,918. ,918.2 2 274,918.2
100 6.62 103 % ....................................Ans. Ans.
earth. Determine the force 1-45 The weight of the first Russian satellite, Sputnik I , was 184 lb on the surface of the earth. exerted on the satellite by the earth at the l ow and high points of its orbit which were 149 mi and 597 mi, respectively, above the surface of the earth. SOLUTION
F 184
Therefore
Gm1 m2 r 2 Gme m
3960 5280
2
Gmem 8.044 1016 lb ft 2 8.044 10
16
F
F
3960 149 5280
2
......................................... ..............Ans. 170.9 lb ...........................
2
......................................... ..............Ans. 138.9 lb ...........................
8.044 1016
3960 597 5280 27
7
1-46 The planet Jupiter has a mass of 1.90(10 ) kg and a radius of 7.14(10 ) m. Determine the force of attraction 11 between the earth and Jupiter when the minimum distance between the two planets is 6(10 ) m. SOLUTION
F
Gm1 m2 r 2
6.673 10 5.976 10 1.90 10 F 2.10 10 6 10 11
24
27
18
11
2
2
N ....................Ans. 4
2
1-47 Convert 640 acres (1 square mile) to hectares if 1 acre equals 4840 yd and 1 hectare equals 10 m . SOLUTION 2
4840 yd 2 3 ft 0.0929 m2 hect 640 acres 4 259 hectare ............Ans. 2 a c r e y d f t 10 m
12
nd
STATICS AND MECHANICS OF MATERIALS, 2 Edition
RILEY, STURGES AND MORRIS
1-48 Determine the dimension of c in the dimensionally homogeneous equation
v
mg c
1 e ct m
in which v is a velocity, m is a mass, t is time, and g is the gravitational acceleration. SOLUTION 2 L M L T c T M 1 e T c
Therefore
ML T 2
c
As a check,
ct m
M
M ......................................... ............................ ................ ..Ans. ........................... Ans. L T T
T T M
1 (dimensionless)
1-49 Develop an expression for the change in gravitational acceleration g between the surface of the earth and a height h when h << Re. SOLUTION
W
Gmem r 2
mg
Therefore
mg e Gme m re2 mg Gme m
re h
2
1 1 g g ge Gme 2 2 re h r e g
Gme
re2 re2 2re h h2 re h r e 2
2
Gme
2re h h 2 re h r e 2
Gme 4
r e
2
2 er h
2Gme h r e3
13
............................ .......................................... ............................ ................ ..Ans. Ans.