lecture 12: newton’s 3 law (cont.) this week’s...
TRANSCRIPT
Week 6 Assignments: - HW #5 Chp 4: Q11, Q22, P9, P33, P83 || Q17, Q21 P13, P55, P82
- MasteringPhysics: - Assignment #5
Week 6 Reading: Chapter 1-5 - Giancoli
This Week’s Announcements:
Lecture 12: Newton’s 3rd Law (Cont.)
Due: 9/29
- Good luck and study hard :-)
* Midterm class period (09/29) is the last day to turn in late HW# 1 - 4
* Midterm I will be next class period
* Late HW available for pick up
* I will have an extra office ‘hour’ next monday 2-4pm
- 20 questions (~10 MC-TF-SA / 10 Word problems)
* Midterm I is tuesday 09/29:
- one 3x5” card (one side) of notes
- 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
- Material to be tested: - 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
* Midterm class period (09/29) is the last day to turn in late HW# 1 - 4
- 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
Newton’s 3rd Law
* Newton’s 3rd law: Fcart-you = -Fyou-cart
Fyou-cart Fcart-you
(|Fyou-cart| = |Fcart-you|)
* Even when cart is accelerating
Naming convention: “force on you from cart” = Fyou-cart
object of interest (FBD)
Clicker Question:
2) You go to a hockey game and witness a fight (how unusual). While on the ice (assume no friction between ice and players) hockey player 1 (m1 = 100 kg) punches hockey player 2 (m2 = 75 kg) in the face, sending him accelerating to the left at 2 m/s2 (while fist is in contact with the face). What happens to hockey player 1?
a) Accelerates to the left at 2 m/s2 also
b) Accelerates to the right at 2 m/s2
e) Accelerates to the right at 1.5 m/s
d) Accelerates to the left at 1.5 m/s2
c) Nothing (stands there motionless)
2 1
Clicker Question:
2) You go to a hockey game and witness a fight (how unusual). While on the ice (assume no friction between ice and players) hockey player 1 (m1 = 100 kg) punches hockey player 2 (m2 = 75 kg) in the face, sending him accelerating to the left at 2 m/s2 (while fist is in contact with the face). What happens to hockey player 1?
a) Accelerates to the left at 2 m/s2 also
b) Accelerates to the right at 2 m/s2
e) Accelerates to the right at 1.5 m/s
d) Accelerates to the left at 1.5 m/s2
c) Nothing (stands there motionless)
2 1
F21 N2
m2g
F12 N1
m1g
Clicker Question:
3) After the fist and the hockey player’s face separate, what happens to the player punched? (Again assume the ice is frictionless).
a) Continues accelerating to the left at 2 m/s2
b) Accelerates to the right at 2 m/s2
d) Decelerates to a stop
c) Continues to left at constant velocity
2 1
Clicker Question:
3) After the fist and the hockey player’s face separate, what happens to the player punched? (Again assume the ice is frictionless).
a) Continues accelerating to the left at 2 m/s2
b) Accelerates to the right at 2 m/s2
d) Decelerates to a stop
c) Continues to left at constant velocity
2 1
T
mpg
Npe
Fnet
* What is the tension T of a cable ?
T
meg Nep
Npe – mpg = mp anet
Fnet
Naming convention: “force on 2 from 1” = F21
Clicker Question:
4) Elevator Example: Which of the following are ‘Newton’s 3rd Law pairs’?
a) T and meg
b) T and Npe
d) T and Nep
c) Npe and Nep
e) Nep and Fnet
T
mpg
Npe
Fnet
T
meg Nep
person elevator
Clicker Question:
4) Elevator Example: Which of the following are ‘Newton’s 3rd Law pairs’?
a) T and meg
b) T and Npe
d) T and Nep
c) Npe and Nep
e) Nep and Fnet
T
mpg
Npe
Fnet
T
meg Nep
person elevator
T
mpg
Npe
Fnet
* What is the tension T of a cable ?
T
meg Nep
Npe – mpg = mp anet
T - Nep - meg = meanet
Fnet
|Npe| = |Nep|
T = (me + mp)(g + anet)
Naming convention: “force on 2 from 1” = F21
. Block 2: . Fp
Fnet
F12 Block 1:
Block 1: Block 2:
F21
Fnet
* Newton’s 3rd law: F21 = -F12
* Newton’s 2nd law: F = m a
Naming convention: “force on 2 from 1” = F21
m1 = 1 kg m2 = 2 kg Fp=12 N
(frictionless)
N1
m1g
N2
m2g
. Block 3: . Fp
Fnet1
F12 Block 1:
Block 1 Block 2
F32
Fnet3
* Newton’s 3rd law: F12 = -F21
* Newton’s 2nd law: F = m a a = 12N/6 kg = 2 m/s2
Block 3
F23 = -F32
F21
Fnet2
F23 Block 2: .
m1 = 1 kg m2 = 2 kg Fp=12 N
m3 = 3 kg
Naming convention: “force on 2 from 1” = F21 (Neglecting balanced N and mg’s)
Block 1 Block 2 Block 3
m1 = 1 kg m2 = 3 kg
Clicker Question:
b) It would increase
a) It would decrease
c) It would remain the same
5) If I switched the mass of blocks 2 & 3, how would the new F32 change? (see figure).
F32
F23 (previous)= 6 N
Fp=12 N m3 = 2 kg
Block 1 Block 2 Block 3
m1 = 1 kg m2 = 3 kg
Clicker Question:
b) It would increase
a) It would decrease
c) It would remain the same
5) If I switched the mass of blocks 2 & 3, how would the new F32 change? (see figure).
F32
F23 (previous)= 6 N
Fp=12 N m3 = 2 kg