Chapter 29
1. (a) The field due to the wire, at a point 8.0 cm from the wire, must be 39 T and must
be directed due south. Since B = 0i 2 r ,
b
g
c
h
6
2 rB 2 0.080 m 39 10 T
i=
=
= 16 A.
0
4 10 7 T m
Chapter 24
1. If the electric potential is zero at infinity then at the surface of a uniformly charged
sphere it is V = q/40R, where q is the charge on the sphere and R is the sphere radius.
Thus q =
Chapter 23
1. The vector area A and the electric field E are shown on the diagram below. The angle
between them is 180 35 = 145, so the electric flux through the area is
2
= E A = EA cos = ( 1800 N
Chapter 22
1. (a) We note that the electric field points leftward at both points. Using F = q0 E , and
orienting our x axis rightward (so points right in the figure), we find
i
N
F = ( +1.6 1019 C ) 4
Chapter 21
1. Eq. 21-1 gives Coulombs Law, F = k
k | q1 | q2 |
r=
=
F
( 8.99 10 N m
9
2
q1 q2
r2
, which we solve for the distance:
C 2 ) ( 26.0 10 6 C ) ( 47.0 10 6 C )
5.70N
= 1.39m.
2. (a) With a u
Chapter 30
1. (a) The magnitude of the emf is
=
c
bg
h
d B
d
=
6.0t 2 + 7.0t = 12t + 7.0 = 12 2.0 + 7.0 = 31 mV.
dt
dt
(b) Appealing to Lenzs law (especially Fig. 30-5(a) we see that the current flow
Chapter 28
1. (a) The force on the electron is
(
)(
)
r
r
rr
FB = qv B = q vx + v y Bx + By j = q ( v x B y v y Bx ) k
i
j
i
(
(
)(
)
(
)
= 1.6 1019 C 2.0 106 m s ( 0.15 T ) 3.0 106 m s ( 0.030 T )
1
Chapter 27
1. (a) The energy transferred is
2t
(2.0 V) 2 (2.0 min) (60 s / min)
U = Pt =
=
= 80 J .
r+R
1.0 + 5.0
(b) The amount of thermal energy generated is
F I F 2.0 V I (5.0 ) (2.0 min) (60 s /
Chapter 25
1. (a) The capacitance of the system is
C=
q
70 pC
=
= 3.5 pF .
V
20 V
(b) The capacitance is independent of q; it is still 3.5 pF.
(c) The potential difference becomes
V =
q 200 pC
=
= 57
Chapter 32
1. We use
6
n =1
Bn = 0 to obtain
5
B 6 = Bn = ( 1Wb + 2 Wb 3 Wb + 4 Wb 5 Wb ) = +3 Wb .
n =1
2. (a) The flux through the top is +(0.30 T)r2 where r = 0.020 m. The flux through the
bottom
Chapter 26
1. (a) The charge that passes through any cross section is the product of the current and
time. Since t = 4.0 min = (4.0 min)(60 s/min) = 240 s,
q = it = (5.0 A)(240 s) = 1.2 103 C.
(b) The
SD 281
Febuary 2003
Torsion
Assumptions
1. Plane sections remain plane.
2. Diameters remain diameters.
3. No normal strains are present.
Under assumption 3 the only possible mode of deformation is one
Potentially Useful Information
1
1 u v
xy = xy =
+
2
2 y x
1
1 v w
yz = yz =
+
2
2 z
y
1
1 w u
zx = zx =
+
2
2 x
z
u
x
v
yy =
y
w
zz =
z
xx =
xx + yy xx yy
+
cos(2) + xy sin(2)
2
2
xx + yy xx yy
=
cos
SYDE 281 Supplemental Solutions
Tim Lahey
April 5, 2009
Question 7.20
The given displacement can be written as:
v (x) = 20x3 40x2 (106 )
(1)
and the bending rigidity (EI ) is given as,
EI = 135(106 )
Structural Failure
US National Guard F-15
Everything looks okay
But maybe its not
Whats happening here?
Yes, there is a pilot
The plane is in two
parts, part number 1
And part number 2
The pilot need
SD 281
February 2009
Torsion of Solid Shafts versus Thin Walled Tubes
Compare the maximum stress and angle of twist per unit length for a solid shaft to those for a
thin walled tubular shaft where bot
SYDE 281 Assignment 9 Solutions
Tim Lahey and Ian Mackenzie
March 25, 2008
Question 6.53
Write the equations for shear force and bending moments as a function of x for the entire
beam.
5 kN / m
y
x
z
SYDE 281 Assignment 8 Solutions
Tim Lahey
March 19, 2007
Question 6.24
The cross section of a beam with a coordinate system at its origin at the centroid C is shown
in Figure 1. The internal moment at
SYDE 281 Assignment 7 Solutions
Ian Mackenzie
March 15, 2007
Question 5.73
d2
y
T
T
t
d1
L AB
x
Dt
BC
A
L BC
d1
=
0.05m
d2
t
=
=
LC D
0.07m
0.01 m
LAB
=
0.5 m
LBC
LCD
=
=
0.15 m
0.4 m
allow
=
0.02 rad
SYDE 281 Assignment 6 Solutions
Tim Lahey
March 8, 2007
Question 4.105
A pressure tank 15 feet long and 40 inches in diameter is to be fabricated from a 1/2 inch
thick sheet. A 15 feet long, 8 inch wi
SYDE 281 Assignment 5 Solutions
Ian Mackenzie
February 9, 2007
Question 4.91
y
x
L
T = TL
x
L
2
Since the bar can expand freely and there are no applied forces, the axial stress in the
bar will be zer
SYDE 281 Assignment 4 Solutions
Tim Lahey
February 6, 2007
Question 4.17
Two cast iron pipes (E = 100 GPa) are adhesively bonded together as shown (see Figure 1).
The outer diameters of the two pipes
SYDE 281 Assignment 3 Solutions
Ian Mackenzie
January 30, 2007
Question 3.19
P
P
y
h0
h1
L0
L1
x
L0
L1
h0
h1
A
P
=
=
=
=
=
=
5 in
5.005 in
2 in
1.9996 in
2 in2
50 000 lbs
To nd the modulus of elastici
SYDE 281 Assignment 2 Solutions
Tim Lahey
January 24, 2007
Question 2.12
Due to the application of force P , the average normal strain in bar A in Figure 1, was found
to be 6000 . If the length of bar
SYDE 281 Assignment 1 Solutions
Ian Mackenzie
2007-01-09
Question 1.12
P
r
R
z
w
y
x
R
r
w
P
=
=
=
=
2 in
1.75 in
8 in
150 103 lbs
1
a)
P
z
y
c Ac
To nd the average compressive stress c in the column
SYDE 281: Mechanics of Deformable Solids
February 25, 2009
Problem Set #7
Questions from the text by Vable:
5.73,
5.77,
5.104 (include 5.101 as one of the design cases)