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Engineering Mechanics Statics, by Plesha Solution manual Material Type: Notes; Professor: Baladi; Class: Statics; Subject: Civil Engineering; University: Michigan State University; Term: Spring 2011;
Typology: Study notes
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Engineering Mechanics: Statics
Version: August 24, 2009
Copyright © 2002– Michael E. Plesha, Gary L. Gray, and Francesco Costanzo
4 Solutions Manual
Accuracy of Numbers in Calculations
Throughout this solutions manual, we will generally assume that the data given for problems is accurate to 3 significant digits. When calculations are performed, all intermediate numerical results are reported to 4 significant digits. Final answers are usually reported with 3 significant digits. If you verify the calculations in this solutions manual using the rounded intermediate numerical results that are reported, you should obtain the final answers that are reported to 3 significant digits.
This solutions manual, in any print or electronic form, remains the property of McGraw-Hill, Inc.of McGraw-Hill, and must be surrendered upon request of McGraw-Hill. Any duplication or distribution, either in print or electronic form, without the It may be used and/or possessed only by permission August 24, 2009
Statics 1e 5
Problem 1.
(a) Consider a situation in which the force F applied to a particle of mass m is zero. Multiply the scalar form of Eq. (1.2) on page 7 (i.e., a D dv=dt) by dt, and integrate both sides to show that the velocity v (also a scalar) is constant. Then use the scalar form of Eq. (1.1) to show that the (scalar) position r is a linear function of time.
(b) Repeat part (a) when the force applied to the particle is a non zero constant, to show that the velocity and position are linear and quadratic functions of time, respectively.
Solution
Part (a) Consider the scalar form of Eq. (1.3) on page 7 for the case with F D 0 ,
F D ma ) 0 D ma ) a D 0: (1)
Next, consider the scalar form of Eq. (1.2) on page 7,
dv dt D^ a^ )^ dv^ D^ adt^ )
dv D v D
a dt: (2)
Substituting a D 0 into Eq. (2) and evaluating the integral provides
v D constant D v 0 ; (3)
demonstrating that the velocity v is constant when the acceleration is zero. Next, consider the scalar form of Eq. (1.1), dr dt
D v ) dr D vdt )
dr D r D
v dt: (4)
For the case with constant velocity given by Eq. (3), it follows that
r D
v 0 dt D v 0
dt D v 0 t C c 1 ; (5)
where c 1 is a constant of integration. Thus, the position r is a linear function of time when the acceleration is zero. Note that in the special case that v 0 D 0 , then the position r does not change with time.
Part (b) When the force F is constant, then Newton’s second law provides
F D constant D ma ) a D F=m D constant: (6)
This solutions manual, in any print or electronic form, remains the property of McGraw-Hill, Inc.of McGraw-Hill, and must be surrendered upon request of McGraw-Hill. Any duplication or distribution, either in print or electronic form, without the It may be used and/or possessed only by permission August 24, 2009
Statics 1e 7
Problem 1.
Using the length and force conversion factors in Table 1.2 on p. 10, verify that 1 slug D 14:59 kg.
Solution
1 slug D 1 slug
lbs^2 /ft slug
lb
ft 0:3048 m
kgm/s^2 N
D 14:59 kg (1)
This solutions manual, in any print or electronic form, remains the property of McGraw-Hill, Inc.of McGraw-Hill, and must be surrendered upon request of McGraw-Hill. Any duplication or distribution, either in print or electronic form, without the It may be used and/or possessed only by permission August 24, 2009
8 Solutions Manual
Problems 1.3 through 1.
Convert the numbers given in U.S. Customary units to the corresponding SI units indicated.
Problem 1.
(a) Length: Convert l D 2:35 in. to m.
(b) Mass: Convert m D 0:156 slug to kg. (c) Force (weight): Convert F D 100 lb to N.
(d) Moment (torque): Convert M D 32:9 ftlb to Nm.
Problem 1. (a) Length: Convert l D 0:001 in. to m.
(b) Mass: Convert m D 0:305 lbs^2 =in. to kg. (c) Force (weight): Convert F D 2:56 kip to kN. (Recall: 1 kip = 1000 lb.)
(d) Mass moment of inertia: Convert Imass D 23:0 in.lbs^2 to Nms^2.
Problem 1.
(a) Pressure: Convert p D 25 lb=ft^2 to N=m^2. (b) Elastic modulus: Convert E D 30 106 lb=in.^2 to GN=m^2.
(c) Area moment of inertia: Convert Iarea D 63:2 in.^4 to mm^4. (d) Mass moment of inertia: Convert Imass D 15:4 in.lbs^2 to kgm^2.
Solution to 1.
Part (a)
l D 2:35 in:
25:4 mm in.
m 103 mm
D 0:0597 m: (1)
Part (b)
m D 0:156 slug
14:59 kg slug
D 2:28 kg: (2)
Part (c)
F D 100 lb
lb
This solutions manual, in any print or electronic form, remains the property of McGraw-Hill, Inc.of McGraw-Hill, and must be surrendered upon request of McGraw-Hill. Any duplication or distribution, either in print or electronic form, without the It may be used and/or possessed only by permission August 24, 2009
10 Solutions Manual
Problems 1.6 through 1.
Convert the numbers given in SI units to the corresponding U.S. Customary units indicated.
Problem 1.
(a) Length: Convert l D 1:53 m to in.
(b) Mass: Convert m D 65 kg to slug. (c) Force (weight): Convert F D 89:2 N to lb.
(d) Moment (torque): Convert M D 32:9 Nm to in.lb.
Problem 1. (a) Length: Convert l D 122 nm to in.
(b) Mass: Convert m D 3:21 kg to lbs^2 =in: (c) Force (weight): Convert F D 13:2 kN to lb.
(d) Mass moment of inertia: Convert Imass D 93:2 kgm^2 to slugin:^2.
Problem 1.
(a) Pressure: Convert p D 25 kN=m^2 to lb=in:^2. (b) Elastic modulus: Convert E D 200 GN=m^2 to lb=in:^2.
(c) Area moment of inertia: Convert Iarea D 23:5 105 mm^4 to in:^4. (d) Mass moment of inertia: Convert Imass D 12:3 kgm^2 to in.lbs^2.
Solution to 1. Part (a)
l D 1:53 m
ft 0:3048 m
12 in: ft
D 60:2 in: (1)
Part (b)
m D 65 kg
slug 14:59 kg
D 4:46 slug: (2)
Part (c)
F D 89:2 N
lb 4:448 N
D 20:1 lb: (3)
This solutions manual, in any print or electronic form, remains the property of McGraw-Hill, Inc.of McGraw-Hill, and must be surrendered upon request of McGraw-Hill. Any duplication or distribution, either in print or electronic form, without the It may be used and/or possessed only by permission August 24, 2009
Statics 1e 11
Part (d)
M D 32:9 N m
lb 4:448 N
in. 0:0254 m
D 291 in: lb: (4)
Solution to 1.
Part (a)
l D 122 nm
10 ^9 m nm
in. 0:0254 m
D 4:80 10 ^6 in: (1)
Part (b)
m D 3:21 kg
(^) slug
14:59 kg
(^) lbs (^2) /ft
slug
(^) ft
12 in:
D 0:0183 lb s^2 =in: (2)
Part (c)
F D 13:2 kN
kN
(^) lb
4:448 N
D 2970 lb: (3)
Part (d)
Imass D 93:2 kg m^2
(^) slug
14:59 kg
(^) in.
0:0254 m
D 9:90 103 slug in:^2 : (4)
Solution to 1.
Part (a)
p D 25 kN=m^2
kN
lb 4:448 N
0:0254 m in.
D 3:63 lb=in:^2 : (1)
Part (b)
E D 200 GN=m^2
lb 4:448 N
0:0254 m in.
D 29:0 106 lb=in:^2 : (2)
Part (c)
Iarea D 23:5 105 mm^4
in. 25:4 mm
D 5:65 in:^4 : (3)
Part (d)
Imass D 12:3 kg m^2
lbs^2 /ft 14:59 kg
ft 12 in:
in. 0:0254 m
D 109 lb s^2 in: (4)
This solutions manual, in any print or electronic form, remains the property of McGraw-Hill, Inc.of McGraw-Hill, and must be surrendered upon request of McGraw-Hill. Any duplication or distribution, either in print or electronic form, without the It may be used and/or possessed only by permission August 24, 2009
Statics 1e 13
Problem 1.
If the weight of a certain object on the surface of the Earth is 0.254 lb, determine its mass in kilograms.
Solution
We first determine the mass of the object in slugs using
w D mg ) m D w=g D .0:254 lb/=.32:2 ft=s^2 / D 7:888 10 ^3 lb s^2 =ft D 7:888 10 ^3 slug: (1)
Next, we convert the mass into SI units, such that
m D 7:888 10 ^3 slug
14:59 kg slug
D 0:115 kg: (2)
This solutions manual, in any print or electronic form, remains the property of McGraw-Hill, Inc.of McGraw-Hill, and must be surrendered upon request of McGraw-Hill. Any duplication or distribution, either in print or electronic form, without the It may be used and/or possessed only by permission August 24, 2009
14 Solutions Manual
Problem 1.
If the mass of a certain object is 69.1 kg, determine its weight on the surface of the Earth in pounds.
Solution
We first determine the weight of the object in newtons using
w D mg D .69:1 kg/
9:81 m=s^2
D 677:9 kg m=s^2 D 677:9 N: (1)
Next, we convert the weight into pounds using
w D 677:9 N
lb 4:448 N
D 152 lb: (2)
This solutions manual, in any print or electronic form, remains the property of McGraw-Hill, Inc.of McGraw-Hill, and must be surrendered upon request of McGraw-Hill. Any duplication or distribution, either in print or electronic form, without the It may be used and/or possessed only by permission August 24, 2009
16 Solutions Manual
Problem 1.
Two identical asteroids travel side by side while touching one another. If the asteroids are composed of homogeneous pure iron and are spherical, what diameter in feet must they have for their mutual gravitational attraction to be 1 lb?
Solution Begin by considering Eq. (1.6) on p. 15, F D G
m 1 m 2 r^2
where F D 1 lb according to the problem statement. The mass of each asteroid is given by m D .4 r^3 =3/, where is the density and r is the radius of the spherical asteroids. Thus, Eq. (1) becomes
.4 r^3 =3/
.2r/^2
(^2) G (^2) r 4 9
Solving for r, we obtain
r D
4^2 .66:74 10 ^12 m^3 =.kg s^2 //.7860 kg=m^3 /^2
D 3:960 m; (3)
where F D 1 lb D 4:448 N. Since the diameter d of the asteroid is twice its radius r (i.e., d D 2r), it follows that
d D 2.3:960 m/
ft 0:3048 m
D 26:0 ft: (4)
This solutions manual, in any print or electronic form, remains the property of McGraw-Hill, Inc.of McGraw-Hill, and must be surrendered upon request of McGraw-Hill. Any duplication or distribution, either in print or electronic form, without the It may be used and/or possessed only by permission August 24, 2009
Statics 1e 17
Problem 1.
The mass of the Moon is approximately 7:35 1022 kg, and its mean distance from the Earth is about 3:80 108 km. Determine the force of mutual gravitational attraction in newtons between the Earth and Moon. In view of your answer, discuss why the Moon does not crash into the Earth.
Solution
Use Eq. (1.6) to solve for the force F :
m 1 m 2 r^2
D 66:74 10 ^12 m^3 =.kg s^2 /
.5:9736 1024 kg/.7:35 1022 kg/ 3:80 108 km.10^3 m=km/
2 D^ 2:02^1014 kg^ ^ m=s^2 ;
(1)
so that the force may be written as
F D 2:02 1014 N: (2)
Although this force is large, the force due to centripetal acceleration equilibrates this, so that the Moon maintains its orbit.
This solutions manual, in any print or electronic form, remains the property of McGraw-Hill, Inc.of McGraw-Hill, and must be surrendered upon request of McGraw-Hill. Any duplication or distribution, either in print or electronic form, without the It may be used and/or possessed only by permission August 24, 2009
Statics 1e 19
Problem 1.
The specific weights of several materials are given in U.S. Customary units. Convert these to specific weights in SI units (kN/m^3 ), and also compute the densities of these materials in SI units (kg/m^3 ).
(a) Zinc die casting alloy, D 0:242 lb/in.^3. (b) Oil shale (30 gal/ton rock), D 133 lb/ft^3.
(c) Styrofoam (medium density), D 2:0 lb/ft^3. (d) Silica glass, D 0:079 lb/in.^3.
Solution
The equation to be used to convert the specific weight to density is
D g ) D =g: (1)
Part (a)
D 0:242 lb=in^3
lb
in. 0:0254 m
D 65:69 103 N=m^3 D 65:7 kN=m^3 ;
65:69 103 N=m^3 9:81 m=s^2 D^
65:69 103 kg=.m^2 s^2 / 9:81 m=s^2 D^ 6:70^10
(^3) kg=m (^3) :
Part (b)
D 133 lb=ft^3
lb
ft 0:3048 m
D 20:89 103 N=m^3 D 20:9 kN=m^3 ;
(^3) N=m 3 9:81 m=s^2
(^3) kg=.m (^2) s (^2) / 9:81 m=s^2
D 2:13 103 kg=m^3 :
Part (c)
D 2:0 lb=ft^3
lb
ft 0:3048 m
D 314:2 N=m^3 D 0:314 kN=m^3 ;
314:2 N=m^3 9:81 m=s^2 D^
314:2 kg=.m^2 s^2 / 9:81 m=s^2 D^ 32:0^ kg=m
Part (d)
D 0:079 lb=in^3
lb
in. 0:0254 m
D 21:44 103 N=m^3 D 21:4 kN=m^3 ;
21:44 103 N=m^3 9:81 m=s^2
21:44 103 kg=.m^2 s^2 / 9:81 m=s^2
D 2:19 103 kg=m^3 :
This solutions manual, in any print or electronic form, remains the property of McGraw-Hill, Inc.of McGraw-Hill, and must be surrendered upon request of McGraw-Hill. Any duplication or distribution, either in print or electronic form, without the It may be used and/or possessed only by permission August 24, 2009
20 Solutions Manual
Problem 1.
The densities of several materials are given in SI units. Convert these to densities in U.S. Customary units (slug/ft^3 ), and also compute the specific weights of these materials in U.S. Customary units (lb/ft^3 ).
(a) Lead (pure), D 11:34 g/cm^3. (b) Ceramic (alumina Al 2 O 3 ), D 3:90 Mg/m^3.
(c) Polyethylene (high density), D 960 kg/m^3. (d) Balsa wood, D 0:2 Mg/m^3.
Solution
The equation to be used to determine the specific weight is
D g: (1)
Part (a)
D 11:34 g=cm^3
kg 103 g
slug 14:59 kg
100 cm m
0:3048 m ft
D 22:0 slug=ft^3 ;
D 22:01 slug=ft^3 .32:2 ft=s^2 /
lbs^2 /ft slug
D 709 lb=ft^3 :
Part (b)
D 3:90 Mg=m^3
(^3) kg Mg
slug 14:59 kg
0:3048 m ft
D 7:569 slug=ft^3 D 7:57 slug=ft^3 ;
D 7:569 slug=ft^3 .32:2 ft=s^2 /
lbs^2 /ft slug
D 244 lb=ft^3 :
Part (c)
D 960 kg=m^3
slug 14:59 kg
0:3048 m ft
D 1:863 slug=ft^3 D 1:86 slug=ft^3 ;
D 1:863 slug=ft^3 .32:2 ft=s^2 / lbs
(^2) /ft slug
D 60:0 lb=ft^3 :
Part (d)
D 0:2 Mg=m^3
103 kg Mg
slug 14:59 kg
0:3048 m ft
D 0:3882 slug=ft^3 D 0:388 slug=ft^3 ;
D 0:3882 slug=ft^3 .32:2 ft=s^2 / lbs
(^2) /ft slug
D 12:5 lb=ft^3 :
This solutions manual, in any print or electronic form, remains the property of McGraw-Hill, Inc.of McGraw-Hill, and must be surrendered upon request of McGraw-Hill. Any duplication or distribution, either in print or electronic form, without the It may be used and/or possessed only by permission August 24, 2009
