Week 6 Assignments: - MasteringPhysics:

Lecture 11: Newton’s 3rd Law
This Week’s Announcements:
* HW #4 due today
* Midterm I will be as scheduled (on tuesday 9/29)
Week 6 Reading:
Chapter 1-5 - Giancoli
Due: 9/29
Week 6 Assignments:
- HW #5 Chp 4: Q11, Q22, P9, P33, P83 || Q17, Q21 P13, P55, P82
- MasteringPhysics: - Assignment #5
* Midterm class period (09/29) is the last day to turn in late HW# 1 - 4
* Midterm I is tuesday 09/29:
- 20 questions (~10 MC-TF-SA / 10 Word problems)
- one 3x5” card (one side) of notes
- Material to be tested:
- I posted a copy of an old midterm (without
solutions) to give an example (just studying it
will be not be enough to do well)
- no problems will directly require calculus,
but vectors will be required
-
Chp 1
Chp 2 --- 2.2-2.7
Chp 3 --- 3.1-3.4; 3.5-3.9
Chp 4 --- all
Chp 5.1-5.4 --- all
-all clicker Qs; and any examples
worked in class
- no smart devices (even for listening to music)
- arrive early so that seating arrangements can be worked and still provide
you most of the time
- If you need OCD accommodations make sure to get them arranged with me
(privately) before the exam
* Midterm class period (09/29) is the last day to turn in late HW# 1 - 4
Clicker Question:
4) You are dragging luggage up a ramp (making an angle α with the horizontal).
There is friction between you and the ramp, with a rather large coefficient of
kinetic friction, µk = 1. The luggage is heavy (as usual you’ve over-packed),
and you wish to exert the least amount of force pulling the luggage (Fp). What
is the optimum direction, θ, in which to pull the luggage such that it moves up
the ramp at a constant speed with the least amount of effort (force)?
Fp
a) Straight up (θ = 90 – α)
N
θ
Fk
b) Along ramp (θ = 0)
c) Somewhere in between
α
mg
α
Vector Forces
T2
T2
T1
T1 sinθ1
θ1
T1 cosθ1
θ2
Fg
T2 sinθ2
T2 cosθ2
Fg
X: -T1cosθ1 + T2cosθ2 = 0
T1
Fnet = 0
- Mass isn’t accelerating
- Equilibrium
Y: T1sinθ1 + T2sinθ2 = mg
Clicker Question:
6) You are in the process of designing a suspension bridge. At one junction six
cables intersect and you measure the tension forces (direction and
magnitude) listed. Assuming that you wish your bridge junction to remain at
rest for a long while, should you be satisfied with the following cable tensions.
50 kN
a) yes
b) no
20 kN
20 kN
30o
10 kN
30o
10 kN
80 kN
Clicker Question:
5) What is each piece of the string’s tension, T, caused by the 1 kg mass if we
now have replace the joint with a pulley? Assume the strings are vertical.
a) Each 4.9 N upward
b) Each 4.9 N downward
c) Each 9.8 N upward
T
d) Each 9.8 N downward
e) Tension on the left string
is 9.8 N upward and the
tension in the right string is
9.8 N downward
mg
Clicker Question:
5) What is each piece of the string’s tension, T, caused by the 1 kg mass if we
now have replace the joint with a pulley? Assume the strings are vertical.
a) Each 4.9 N upward
b) Each 4.9 N downward
c) Each 9.8 N upward
T
d) Each 9.8 N downward
e) Tension on the left string
is 9.8 N upward and the
tension in the right string is
9.8 N downward
mg
Centripetal Force
Roads designed for high-speed travel are banked to give the normal force a
component towards the center of the curve. A properly designed bank will
permit a car to go around a circular curve without any need for frictional
forces (steering).
---- same reason applies for why airplanes bank when they turn.
What angle should a road with a 300 m radius of curvature be banked for
travel at 30 m/s?
θ
N
Fnet
θ
X:
r
mg
N sin θ = Fnet = mv2/r
Y: N cos θ = mg
Fnet = mv2/r = (mg/cos θ) sin θ
v2/r = g tan(θ)
Clicker Question:
7) Now lets consider what happens if we rely on friction rather than banking.
Lets say Jimmy Johnson enters a turn of 250 m (~0.15 mil) radius of
curvature, at 70 m/s (155 mph). If we assume a NASCAR car (+ driver) has a
mass of 1500 kg and a µs of 1.0 will the car be able to track the curve without
wiping out in a massive crash?
a) yes
b) no
Clicker Question:
7) Now lets consider what happens if we rely on friction rather than banking.
Lets say Jimmy Johnson enters a turn of 250 m (~0.15 mil) radius of
curvature, at 70 m/s (155 mph). If we assume a NASCAR car (+ driver) has a
mass of 1500 kg and a µs of 1.0 will the car be able to track the curve without
wiping out in a massive crash?
a) yes
b) no
Newton’s 3rd Law
For a force there is always an equal
and opposite reaction: or the forces
of two bodies on each other are
always equal and are directed in
opposite directions.
-Newton
- forces between two objects always
come in pairs
- forces equal in magnitude
- directed in opposite directions
- forces act of different objects
Fr-f
Ff-r
Newton’s 3rd Law: Examples
Hero of Alexandria:
Circa:
time of Christ
reaction
- Invented first
steam engine:
Action
- Invented syringe
- Invented forerunner of motion picture
- Invented the vending machine
- First known use of wind power to power an instrument
- Many others
Newton’s 3rd Law
- Newton’s 3rd law: Fup = -Fdown
ame
= Fup/m
me
mme = 100 kg
a⊕
m⊕ = 6.0 x 1024 kg
m⊕
= Fdown/