Physics 207: Lecture 8, Pg 1 Lecture 8 l Goals: Differentiate between Newton’s 1 st, 2 nd and 3...
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Transcript of Physics 207: Lecture 8, Pg 1 Lecture 8 l Goals: Differentiate between Newton’s 1 st, 2 nd and 3...
Physics 207: Lecture 8, Pg 1
Lecture 8
Goals:Goals: Differentiate between Newton’s 1st, 2nd and 3rd Laws
Use Newton’s 3rd Law in problem solving
Assignment: HW4, (Chapters 6 & 7, due 2/18, Wednesday)
Finish Chapter 7
1st Exam Wed., Feb. 18th from 7:15-8:45 PM Chapters 1-7
in room 2103 Chamberlin Hall
Physics 207: Lecture 8, Pg 2
Inclined plane with “Normal” and Frictional Forces
1. Static Equilibrium Case
2. Dynamic Equilibrium (see 1)
3. Dynamic case with non-zero acceleration
Block weight is mg
NormalForce
Friction Force
“Normal” means perpendicular
mg cos
f
x
y
mg sin
F = 0
Fx= 0 = mg sin – f
Fy= 0 = mg cos – N
with mg sin = f ≤ S N
if mg sin > S N, must slide
Critical angle s = tan c
Physics 207: Lecture 8, Pg 3
Inclined plane with “Normal” and Frictional Forces
1. Static Equilibrium Case
2. Dynamic Equilibrium
Friction opposite velocity
(down the incline)
mg
NormalForce
Friction Force
“Normal” means perpendicular
mg cos
fK
x
y
mg sin
F = 0
Fx= 0 = mg sin – fk
Fy= 0 = mg cos – N
fk = k N = k mg cos
Fx= 0 = mg sin – k mg cos
k = tan (only one angle)
v
Physics 207: Lecture 8, Pg 4
Inclined plane with “Normal” and Frictional Forces
3. Dynamic case with non-zero acceleration
Result depends on direction of velocity
Weight of block is mg
NormalForce
Friction ForceSliding Down
vmg sin fk
Sliding Up
Fx= max = mg sin ± fk
Fy= 0 = mg cos – N
fk = k N = k mg cos
Fx= max = mg sin ± k mg cos
ax = g sin ± k g cos
Physics 207: Lecture 8, Pg 7
Friction in a viscous mediumDrag Force Quantified
With a cross sectional area, A (in m2), coefficient of drag of 1.0 (most objects), sea-level density of air, and velocity, v (m/s), the drag force is:
D = ½ C A v2 c A v2 in Newtons
c = ¼ kg/m3
In falling, when D = mg, then at terminal velocity Example: Bicycling at 10 m/s (22 m.p.h.), with projected area
of 0.5 m2 exerts ~30 Newtons Minimizing drag is often important
Physics 207: Lecture 8, Pg 8
“Free” Fall Terminal velocity reached when Fdrag = Fgrav (= mg) For 75 kg person with a frontal area of 0.5 m2,
vterm 50 m/s, or 110 mph
which is reached in about 5 seconds, over 125 m of fall
Physics 207: Lecture 8, Pg 10
Newton’s Laws
Law 1: An object subject to no external forces is at rest or moves with a constant velocity if viewed from an inertial reference frame.
Law 2: For any object, FFNET = FF = ma a
Law 3: Forces occur in pairs: FFA , B = - FFB , A
(For every action there is an equal and opposite reaction.)
Read: Force of B on A
Physics 207: Lecture 8, Pg 11
Newton’s Third Law:
If object 1 exerts a force on object 2 (F2,1 ) then object 2
exerts an equal and opposite force on object 1 (F1,2)
F1,2 = -F2,1
IMPORTANT: Newton’s 3rd law concerns force pairs whichact on two different objects (not on the same object) !
For every “action” there is an equal and opposite “reaction”
Physics 207: Lecture 8, Pg 12
Gravity
Newton also recognized that gravity is an attractive, long-range force between any two objects.
When two objects with masses m1 and m2 are separated by distance r, each object “pulls” on the other with a force given by Newton’s law of gravity, as follows:
Physics 207: Lecture 8, Pg 13
Cavendish’s Experiment
F = m1 g = G m1 m2 / r2
g = G m2 / r2
If we know big G, little g and r then will can find m2 the mass of the Earth!!!
Physics 207: Lecture 8, Pg 14
Example (non-contact)
Consider the forces on an object undergoing projectile motion
FB,E = - mB g
EARTH
FE,B = mB g
FB,E = - mB g
FE,B = mB g
Question: By how much does g change at an altitude of 40 miles? (Radius of the Earth ~4000 mi)
Physics 207: Lecture 8, Pg 15
Example
Consider the following two cases (a falling ball and ball on table),
Compare and contrast Free Body Diagram and
Action-Reaction Force Pair sketch
Physics 207: Lecture 8, Pg 16
Example
The Free Body Diagram
mg
mg
FB,T= N
Ball FallsFor Static Situation
N = mg
Physics 207: Lecture 8, Pg 17
Normal Forces
Certain forces act to keep an object in place. These have what ever force needed to balance all others
(until a breaking point).
FT,B
FB,T
Main goal at this point : Identify force pairs and apply Newton’s third law
Physics 207: Lecture 8, Pg 18
Example
First: Free-body diagramSecond: Action/reaction pair forces
FB,E = -mg FB,T= N
FE,B = mg
FB,E = -mg
FE,B = mg
FT,B= -N
Physics 207: Lecture 8, Pg 19
The flying bird in the cage
You have a bird in a cage that is resting on your upward turned palm. The cage is completely sealed to the outside (at least while we run the experiment!). The bird is initially sitting at rest on the perch. It decides it needs a bit of exercise and starts to fly.
Question: How does the weight of the cage plus bird vary when the bird is flying up, when the bird is flying sideways, when the bird is flying down?
So, what is holding the airplane up in the sky?
Physics 207: Lecture 8, Pg 20
Exercise Newton’s Third Law
A. greater than
B. equal to
C. less than
A fly is deformed by hitting the windshield of a speeding bus.
v
The force exerted by the bus on the fly is,
that exerted by the fly on the bus.
Physics 207: Lecture 8, Pg 21
Exercise 2Newton’s Third Law
A. greater than
B. equal to
C. less than
A fly is deformed by hitting the windshield of a speeding bus.
v
The magnitude of the acceleration, due to this collision, of the bus is
that of the fly.
Same scenario but now we examine the accelerations
Physics 207: Lecture 8, Pg 22
Exercise 2Newton’s Third Law
Solution
By Newton’s third law these two forces form an interaction pair which are equal (but in opposing directions).
However, by Newton’s second law Fnet = ma or a = Fnet/m.So Fb, f = -Ff, b = F0
but |abus | = |F0 / mbus | << | afly | = | F0/mfly |
Thus the forces are the same
Answer for acceleration is (C)
Physics 207: Lecture 8, Pg 23
Exercise 3Newton’s 3rd Law
A. 2
B. 4
C. 6
D. Something else
a b
Two blocks are being pushed by a finger on a horizontal frictionless floor.
How many action-reaction force pairs are present in this exercise?
Physics 207: Lecture 8, Pg 24
6
Exercise 3Solution:
a bFFa,f FFf,a FFb,a FFa,b
FFg,ag,a
FFa,ga,g
FFg,bg,b
FFb,gb,g
FFE,aE,a
FFa,Ea,E
FFE,bE,b
FFb,Eb,E
Physics 207: Lecture 8, Pg 25
Example: Friction and Motion
A box of mass m1 = 1 kg is being pulled by a horizontal string having tension T = 40 N. It slides with friction
(k= 0.5) on top of a second box having mass m2 = 2 kg, which in turn slides on a smooth (frictionless) surface. What is the acceleration of the second box ?
Central question: What is force on mass 2?
(A) a = 0 N (B) a = 5 N (C) a = 20 N (D) can’t tell
mm22
T mm11slides with friction (k=0.5)
slides without frictiona = ?
v
Physics 207: Lecture 8, Pg 26
ExampleSolution
First draw FBD of the top box:
m1
N1
m1g
T fk = KN1 = Km1g
v
Physics 207: Lecture 8, Pg 27
Newtons 3rd law says the force box 2 exerts on box 1 is equal and opposite to the force box 1 exerts on box 2.
m1ff1,2 = Km1g = 5 N
m2
ff2,1 == -ff1,2
As we just saw, this force is due to friction:
ExampleSolution
ActionReaction
(A) a = 0 N (B) a = 5 N (C) a = 20 N (D) can’t tell
Physics 207: Lecture 8, Pg 28
Now consider the FBD of box 2:
m2 ff2,1 = km1g
m2g
N2
m1g
ExampleSolution
Physics 207: Lecture 8, Pg 29
Finally, solve Fx = ma in the horizontal direction:
m2 ff2,1 = Km1g
K m1g = m2a
ExampleSolution
= 2.5 m/s2
kg 2
N 5ga
2
k1
m
m
Physics 207: Lecture 8, Pg 30
Home Exercise Friction and Motion, Replay
A box of mass m1 = 1 kg, initially at rest, is now pulled by a horizontal string having tension T = 10 N. This box (1) is on top of a second box of mass m2 = 2 kg. The static and kinetic coefficients of friction between the 2 boxes are s=1.5 and k= 0.5. The second box can slide freely (frictionless) on an smooth surface.
Compare the acceleration of box 1 to the acceleration of box 2 ?
mm22
T mm11
friction coefficients s=1.5 and k=0.5
slides without frictiona2
a1
Physics 207: Lecture 8, Pg 31
Home Exercise Friction and Motion, Replay in the static case
A box of mass m1 = 1 kg, initially at rest, is now pulled by a horizontal string having tension T = 10 N. This box (1) is on top of a second box of mass m2 = 2 kg. The static and kinetic coefficients of friction between the 2 boxes are s=1.5 and k= 0.5. The second box can slide freely on an smooth surface (frictionless).
If no slippage then the maximum frictional force between 1 & 2 is
(A) 20 N (B) 15 N (C) 5 N (D) depends on T
mm22
T mm11friction coefficients
s=1.5 and k=0.5
slides without frictiona2
a1
Physics 207: Lecture 8, Pg 32
Home ExerciseFriction and Motion
fs = 10 N and the acceleration of box 1 is
Acceleration of box 2 equals that of box 1, with |a| = |T| / (m1+m2) and the frictional force f is m2a
(Notice that if T were in excess of 15 N then it would break free)
mm2 2
T mm11
friction coefficients s=1.5 and k=0.5
slides without frictiona2
a1
TTm1 gg
NN
fS
fS S N = S m1 g = 1.5 x 1 kg x 10 m/s2
which is 15 N (so m2 can’t break free)
Physics 207: Lecture 8, Pg 33
Lecture 8 Recap
Assignment: HW4, (Chapters 6 & 7, due 2/18, Wednesday)
Finish Chapter 7
1st Exam Wed., Feb. 18th from 7:15-8:45 PM Chapters 1-7
in room 2103 Chamberlin Hall