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 A
(b) The current must be from west to east to produce
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 = 40RV and the number of electrons is
q 4 0 R V
( 1.0 106
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 C ) ( 3.2 10 3 m ) cos145 = 1.5 10 2 N m 2 C.
z
2. We u
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 ) 40 = (6.4 10 18 N)
i
i
C
which means the magnitude of t
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 understood to mean the magnitude of acceleration, Newton
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 in the
loop is clockwise. Thus, the current is to left
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 )
14
= 6.2 10 N k.
)
r
r
Thus, the magnitude of FB is 6.2
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 / min) = 67 J.
U = i Rt = G J Rt = G
H+ R K H + 5.0 J
r
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 V .
C 3.5 pF
2. Charge flows until the potential differ
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 is +0.70 mWb as given in the problem statement. Since
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 number of electrons N is given by q = Ne, where e is t
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 in which the cross sections of
the shaft remain undefo
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(2) xy sin(2)
2
2
xx yy
xy =
sin(2) + xy cos(2)
2
xx =
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 ) lbs in.2
(2)
The free-body diagram of the system is giv
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 needs to leave
Leave, pilot, leave
The canopy is gone, get
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 both shafts have the same mass and are made of the same ma
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
0.5 m
0.5 m
Figure 1: Question 6.53
Solution
5 kN / m
y
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 a beam cross section is given as Mz = 10 kN m.
Determi
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
Gir
=
70 109 N/m2
Gcu
=
40 109 N/m2
T
d2
y
t
Ti
A
x
1
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 wide and 1/2 inch thick plate is bonded onto the tank to
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 zero. To nd the displacement u of a point at x = L/2, we m
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 are 50 mm and 70 mm and the wall thickness of each
pipe
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 elasticity, divide the axial stress by the axial strain:
xx =
1
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 A is 36 inches, determine the movement of point B .
Fi
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, use the crosssectional area of the column Ac and do a
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)