Physics 100 Lecture 5, Slide 1
PHYS 100: Lecture 5
NEWTON’S SECOND LAW
NMg
T
θ
u
v
tot iF F ma= =∑� �
�
Physics 100 Lecture 5, Slide 2
Music
Who is the Artist?
A) David Benoit
B) Oscar Peterson
C) Thelonius Monk
D) Dr. John
E) Mose Allison
SORRY.. DON’T HAVE MY IPOD TODAY!
My theme for today was to have been Mardi Gras!
Which of the above piano players is most closely associated with New Orleans?
classics
BB
Physics 100 Lecture 5, Slide 3
How To Learn• Lectures
– Study the Lecture slides after lecture. Ask yourself: What was the Point of the lecture? Try to understand the logic of what was asked.
• Homework– Do Interactive Examples first: Use as much HELP as you need !
• Do the Follow-Ups: they test whether you understand what you just did!– Your initial score on the delayed feedback questions is your first indicator of your understanding. If you do not answer those correctly, try to understand why!
• Discussion– Study the solutions after discussion. What was the point of asking those questions?
KEYS FOR PROBLEM SOLVING
AVOID EQUATION HUNTING
THINK FIRST !!• What concepts apply to the problem?• What are the relevant variables?
• Which are known? Which are unknown?• Develop a plan (first in English)• Write down equations for relevant variables• Solve
Physics 100 Lecture 5, Slide 4
THE BIG IDEAS
1. An object accelerates only when the total force on it is non-zero.2. The weight of an object is always equal to Mg3. The Normal Force and Tension forces must be determined from F=ma
NOTE: THE BIG IDEAS ARE ALWAYS GIVEN IN THE LAST SLIDE
Physics 100 Lecture 5, Slide 5
Thrown BallA ball is thrown straight up in the air. Take up to be the positive direction
What are the velocity (v) and acceleration (a) of the ball at its highest point?
Just before the ball reaches its highest point: what is its velocity?
(A) v = 0a > 0
(B) v = 0a < 0
(C) v = 0a = 0
(D) v ∫ 0a = 0
(E) v ∫ 0a ∫ 0
v
v1
Just after the ball reaches its highest point: what is its velocity? v2
Just as the ball reaches its highest point: what is its acceleration?
dt
vda
�
�
≡12
vvva����
−=∆∝v2
+
-v1
Just as the ball reaches its highest point: what is its velocity? v = 0
=
a
+
BB
Physics 100 Lecture 5, Slide 6
CheckPoint 1A ball is thrown straight up in the air.
How many forces are acting on the ball at its highest point?
You said:
• The ball is not moving at its highest point so there are no forces acting on the ball at that moment.
• weight is the only force acting on the ball
• the gravitational force and the force exerted to throw it into the air
(A) 0 (B) 1 (C) 2 (D) 3
0
10
20
30
40
50
60
70
A B C D E
(E) Greater than 3
We know a = g throughout the flight of the ball ! The net force at all times then must be given by:F = ma = mg (the weight !!)
BB
Physics 100 Lecture 5, Slide 7
Pulling a block
WHAT DO YOU KNOW??
In both cases shown, forces of equal magnitude, F, are exerted on identical blocks supported by a frictionless floor. In Case I, the force is applied at an angle θ wrt horizontal while in Case II, the force is applied horizontally.
Compare aI to aII:
(A) aI < aII
(B) aI = aII
(C) aI > aII
M M
Case I Case II
FFθ
amFFitot
�
��
==∑
WHAT DO YOU DO FIRST?
Draw Free Body Diagram !!
M M
FFθ
NI NIIMg Mg
APPLY NEWTON’S SECOND LAW IN HORIZONTAL
DIRECTION
IMaF =θcos II
MaF =
θcosM
Fa
I=
M
Fa
II=θcos
IIIaa =
i) ΣFxi = max
ii) ΣFyi = may
BB
Physics 100 Lecture 5, Slide 8
Pulling a block
WHAT DO YOU KNOW??
In both cases shown, forces of equal magnitude, F, are exerted on identical blocks supported by a frictionless floor. In Case I, the force is applied at an angle θ wrt horizontal while in Case II, the force is applied horizontally.
Compare NI to NII:
(A) NI < NII
(B) NI = NII
(C) NI > NII
M M
Case I Case II
FFθ
amFFitot
�
��
==∑
WHAT DO YOU DO FIRST?
Draw Free Body Diagram !!
M M
FFθ
NI NIIMg Mg
APPLY NEWTON’S SECOND LAW IN VERTICAL
DIRECTION
MgNFI
=+θsin MgNII
=
θsinFMgNI
−= θsinFNNIII
−=
BB
Physics 100 Lecture 5, Slide 9
CheckPoint 2
0
10
20
30
40
50
60
70
A B C
You said:
• Case 1 has gravity acting against it at angle theta while case 2 is acting at a 90 degree angle. In case 1 part of the force is used to counteract that of gravity while case 2 all of the force is used to move the box
• Acceleration = Force/mass, so since the blocks have the same mass and each block has the same force acting upon it, the acceleration is equal.
• It takes a greater acceleration to push the object up a incline than to push it straight across a surface.
In both cases shown, forces of equal magnitude, F, are exerted on identical blocks. The forces are directed parallel to the frictionless surfaces
Compare aI to aII:
(A) aI < aII
(B) aI = aII
(C) aI > aII
BB
Physics 100 Lecture 5, Slide 10
CheckPoint 2
In both cases shown, forces of equal magnitude, F, are exerted on identical blocks. The forces are directed parallel to the frictionless surfaces
Compare aI to aII:
(A) aI < aII
(B) aI = aII
(C) aI > aII
WHAT DO YOU DO FIRST?
Draw Free Body Diagram !!
Same kind of forces, but differences in orientation will cause differences in magnitude !
APPLY NEWTON’S SECOND LAW IN DIRECTION OF
ACCELERATION
IIMaF =
NIIMg
Mgθ
IMaMgF =− θsin
θsingM
Fa
I−=
M
Fa
II=
θsingaaIII
−=
x
ya
Physics 100 Lecture 5, Slide 11
Follow-Up
In both cases shown, forces of equal magnitude, F, are exerted on identical blocks. The forces are directed parallel to the frictionless surfaces
Compare NI to NII:
(A) NI < NII
(B) NI = NII
(C) NI > NII
WHAT DO YOU DO FIRST?
Draw Free Body Diagram !!
APPLY NEWTON’S SECOND LAW PERPENDICULAR TO
THE PLANE
NIIMg
Mgθ
0cos =− θMgNI
θcosIII
NN =
0=− MgNII
θcosMgNI
= MgNII
=
BB
Physics 100 Lecture 5, Slide 12
Elevators
Both elevators shown carry identical crates and are moving down. In Case I, the elevator is moving at constant velocity v, while in Case II the elevator is moving at constant velocity 2v.
Compare NI to NII:
(A) NI < NII (B) NI = NII (C) NI > NII
2v
WHAT DO YOU DO FIRST?
Draw Free Body Diagram !!2v
NIIMgNIMg
APPLY NEWTON’S SECOND LAW IN VERTICAL
DIRECTION(a = 0 !!)
0=− MgNII0=− MgN
I
IIINN =
MgNI
= MgNII
=
BB
Physics 100 Lecture 5, Slide 13
CheckPoint 3
0
10
20
30
40
50
60
A B C
You said:
• The acceleration is positive in case two where it is negative in case one, the overall force of case two is greater.
• I would say the normal forces are equal in both cases because its pushing against identical crates.
• N1 has more force opposing the fall, this is why it is slowing down. N2 has less force opposing the fall, this is why it speeds up
Both elevators shown carry identical crates and are moving down. In Case I, the elevator is slowing down, while in Case II the elevator is speeding up.
Compare NI to NII:
(A) NI < NII (B) NI = NII (C) NI > NII
BB
Physics 100 Lecture 5, Slide 14
CheckPoint 3
Both elevators shown carry identical crates and are moving down. In Case I, the elevator is slowing down, while in Case II the elevator is speeding up.
Compare NI to NII:
(A) NI < NII (B) NI = NII (C) NI > NII
WHAT DO YOU DO FIRST?
Draw Free Body Diagram !!
NIIMgNIMg
WHAT IS ACCELERATION ?? aI aII
APPLY NEWTON’S SECOND LAW IN VERTICAL
DIRECTION(take direction of a to be
positive)
IIIIMaNMg =−
IIMaMgN =−
)(II
agMN += )(IIII
agMN −=
IIINN >
Physics 100 Lecture 5, Slide 15
Follow Up
Both elevators shown carry identical crates and are moving up. In Case I, the elevator is slowing down, while in Case II the elevator is speeding up.
Compare NI to NII:
(B) NI = NII (C) NI > NII
WHAT DO YOU DO FIRST?
Draw Free Body Diagram !!
WHAT IS ACCELERATION ?? aI aII
APPLY NEWTON’S SECOND LAW IN VERTICAL
DIRECTION(take direction of a to be
positive)
IIIIMaMgN =−
IMaNMg =−
)(II
agMN −= )(IIII
agMN +=
(A) NI < NII
IIINN <
v v
NIIMgNIMg
v v
BB
Physics 100 Lecture 5, Slide 16
CheckPoint 4
0
20
40
60
80
100
A B C
You said:
• In case I, the tension equals to the mass of the object (except the signs are different). In case II, the x and the y components of the tensions have to be considered in order to find the value of TL or TR.
• The similarity in angles calls for the tension to be equal.
• the sum of TL and TR will equal T since both cases are at equilibrium, therefore T must be greater than TL
In both cases shown identical signs are suspended from the ceiling. One string is used in Case I, while two strings, each making an angle of θ = 20o with the horizontal, are used in Case II.
Compare T to TL:
(A) T < TL (B) T = TL (C) T > TL
LET’S DO THE EXPERIMENT !!
BB
Physics 100 Lecture 5, Slide 17
CheckPoint 4
In both cases shown identical signs are suspended from the ceiling. One string is used in Case I, while two strings, each making an angle of θ = 20o with the horizontal, are used in Case II.
Compare T to TL:
(A) T < TL (B) T = TL (C) T > TL
WHAT DO YOU DO FIRST?
Draw Free Body Diagram !!T
Mg Mg
APPLY NEWTON’S SECOND LAW IN VERTICAL DIRECTION
(Symmetry of Case II insures ax = 0)(i.e., TL = TR)
MgT = MgTL
=θsin2
θsin2
MgT
L=
θsin2
TT
L=
TL > T if sinθ < 1/2 i.e., θ < 30o
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