PA R T I : E L E C T R I C I T Y
UNIT
1
COULOMBS LAW
1.1 Overview
Electricity and magnetism are the central topics to be developed in this course. The one
new big idea we will develop is that we will need to modify our description of the world
to include
Professor Patrick O' Connor
10/12/11
Khashayar Moez
Mohammad Amin Alimardani
Ohms Law
Equipment
o
1-35V DC power supply
o
Ammeter
o
Digital multimeter (DMM) measures voltages (volts) and resistances (mega ohms)
o
Assortment of patch cords, clip leads and
Part 1: Conceptual Questions (3 points each) Circle and justify your answers.
Question 1 :
The long, straight wire AB carries a 14.0-A current as shown. The rectangular loop has long edges parallel to AB and carries a
clockwise 5.00-A current.
. . . . I
15—111
The ill-leg erate is pulled by the ennstant farce P. Ifthe crate
starts [mm rest and achieve-s a speed elf III] :11}: [I1 5 5. dete r-
mine the magnitude eII FLThe cuefﬁeient nf kinetic frietiun
between the crate and the ground is m = Ell
SOLUTION
=Mil—£5.
If the SD—kg crate is subjected to a force of P = EDD N. p
determine its speed when it has traveled 15 In starting from
rest. The coefficient of kinetic friction between the crate
and the ground is ,ut = 0.3.
SOLUTION
RITE-Body Diagram: Referring
13—3.
If the coefficient of kinetic friction between the 50-11 g crate P
and the ground is M = 0.3. determine the distance the
crate travels and its 1.relocitgtI when t = 3 s. The crate starts
from rest. and P = 200 N. 31}
SOLUTION
Free-Body Diagram: Th
12—11}.
The position of a particle along a straight line is given by
5 = {15:3 — 13.5:1 + 22.50 0. where t is in seconds.
Determine the position of the particle when t = 6 5 and the
total distance it travels during the ﬁ—s lime interval. Him:
Plot the pat
Mechanics of Materials Exam 1 Practice
Hibbeler 8ed.
Chapter 1 & 2
1
Mechanics of Materials Exam 1 Practice
Hibbeler 8ed.
CHAPTER 3
2
Mechanics of Materials Exam 1 Practice
Hibbeler 8ed.
3
Mechanics of Materials Exam 1 Practice
Hibbeler 8ed.
CHAPTER 4
4
M
STATICS FORMULAS
Sine Law
Moments
Moment 2D
A
B
C
=
=
sina
sinb
sinc
MO = F d
Moment 3D
Cosine Law
A2 + B 2 2ABcos
Vectors
i
j
k
r F = rx ry rz
Fx Fy Fz
Simplication of a System
Magnitude
FR =
F =
2
2
Fx + Fy + Fz
F
2
(MR )O =
M+
MO
Direction (2D)
= arct
Electric Forcesand Fields
Problem1
Threepoint charges refixed in placein a right triangle. What is the electricforce on the
a
-cfw_.60-pC chargedueto the othertwo charges?
p
+0.80 C
(t^r1.,*-^ 9 d*
9'.
ca
'110.0
cm
8.0c m
o
d\
.60tr
K= t "cq* l oq N , '/c
PH 262 Discussion
Simple Harmonic Motion
Session 1
Concepts this Week
Relevant Lectures for Discussion Session:
Lecture 1: Simple Harmonic Motion
1. Restoring force leads to oscillations
2. Initial conditions determine the amplitude and phase of the motio
Last Time
Resistors in series:
Current through is same.
General Physics II
Lecture 12
Reffective
Voltage drop across is IRi
R1
R2
R3 .
Resistors in parallel:
1
Reffective
Voltage drop across is same.
Current through is V/Ri
Todays Concept:
1
R1
1
R2
1
.
R
Key Concepts:
1) Understanding the behavior of capacitors in
circuits with resistors
2) Understanding the RC time constant
General Physics II
Lecture 13
Todays Plan:
1) Examples with switches
closing and opening
- What changes?
- What is constant?
2) Exam
Biot-Savart Law:
What is it?
dB
Fundamental law for determining the
direction and magnitude of the magnetic
field due to an element of current
General Physics II
Lecture 16
0
4
I ds r
r2
We can use this law to calculate the magnetic field produced by ANY
Last Time:
Force on a moving charge
General Physics II
Lecture 15
F
qv B
This Time:
z
Force on a current-carrying wire
F q
Todays Concept:
y
vi B
F
i
Torques
F qNvavg B
N nAL
I qnAvavg
I
F
Electricity & Magnetism Lecture 13, Slide 1
x
IL B
Electricity
Magnetic Observations
Bar Magnets
N
S
N
S
S
General Physics II
Lecture 14
N
N
S
Compass Needles
N
S
Todays Concept:
Magnetic Force on Moving Charges
F
qv B
Magnetic Charge?
N
S
cut in half
N
S
Electricity & Magnetism Lecture 12, Slide 1
Magnetic Observa
Simple Capacitor Circuit
Q
General Physics II
Lecture 10
V
V
C
Q
C
VC
Todays Concept:
Q
Capacitors
(Capacitors in a circuits, Dielectrics, Energy in capacitors)
This Q really means that the battery has
moved charge Q from one plate to the other,
so that o
Part 1: Conceptual Questions (3 points each) Circle and justify your answers.
Question 1:
(“hinge =2
—q
(,‘luu'ge =l _ ~
The electric potential due to a point charge approaches Zero as you move farther away from the charge. If the three point charges
s