Transcript Document 7569371
Lecture 8
Goals: Differentiate between Newton’s 1 Use Newton’s 3 rd st , 2 nd and 3 Law in problem solving rd Laws Assignment: HW4, (Chapters 6 & 7, due 10/1, Wednesday) Finish Chapter 7 1 st Exam Thursday, Oct. 2 nd from 7:15-8:45 PM Chapters 1-7 Physics 207: Lecture 8, Pg 1
Inclined plane with “Normal” and Frictional Forces
1.
2.
Static Equilibrium Case
Dynamic Equilibrium (see 1) 3.
Dynamic case with non-zero acceleration “Normal” means perpendicular
Normal Force
S
F
= 0 F x = 0 = mg sin q – f F y = 0 = mg cos q – N f Friction Force with mg sin q = f ≤ m S N if mg sin q > m S N, must slide Critical angle m k
= tan
q q mg sin q mg cos q q q
Block weight is mg
Physics 207: Lecture 8, Pg 2
y x
Inclined plane with “Normal” and Frictional Forces 1.
Static Equilibrium Case
2.
Dynamic Equilibrium
Friction opposite velocity (down the incline) S
F
= 0 F x = 0 = mg sin q – f k F y = 0 = mg cos q – N f K Friction Force f k = m k N = m k mg cos q F x = 0 = mg sin q – m k mg cos q m k
= tan
q (only one angle)
v
“Normal” means perpendicular
Normal Force
q mg sin q mg cos q q q
mg
Physics 207: Lecture 8, Pg 3
y x
Inclined plane with “Normal” and Frictional Forces 3.
Dynamic case with non-zero acceleration
Result depends on direction of velocity Normal Force Friction Force Sliding Down
v
q mg sin q f k Sliding q Up F x = ma x = mg sin q ± f k F y = 0 = mg cos q – N F x = ma x f k = m k N = = mg sin q ± m m k k mg cos q mg cos q a x = g sin q ± m k g cos q Weight of block is mg Physics 207: Lecture 8, Pg 4
Friction in a viscous medium Drag Force Quantified With a cross sectional area,
A
1.0 (most objects), (in m 2 ), coefficient of drag of sea-level density of air, and velocity,
v
(m/s), the drag force is: D
=
½
C
A v 2
c
A v 2
c = ¼ kg/m 3 in Newtons 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 m 2 exerts ~30 Newtons Minimizing drag is often important Physics 207: Lecture 8, Pg 8
Fish Schools
Physics 207: Lecture 8, Pg 9
By swimming in synchrony in the correct formation, each fish can take advantage of moving water created by the fish in front to reduce drag. Fish swimming in schools can swim 2 to 6 times as long as individual fish. Physics 207: Lecture 8, Pg 10
“Free” Fall
Terminal velocity reached when
F
drag =
F
grav (=
mg
) For 75 kg person with a frontal area of 0.5 m 2 ,
v
term 50 m/s, or 110 mph which is reached in about 5 seconds, over 125 m of fall Physics 207: Lecture 8, Pg 11
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,
F
NET =
S
F = ma
Law 3
: Forces occur in pairs:
F
A , B = - F B , A
(For every action there is an equal and opposite reaction.) Read: Force of B on A Physics 207: Lecture 8, Pg 13
Newton’s Third Law: If object 1 exerts a force on object 2 (
F 2,1
) then object 2 exerts an equal and opposite force on object 1 (
F
1,2 )
F
1,2 = -
F
2,1 For every “action” there is an equal and opposite “reaction” IMPORTANT: Newton’s 3 rd law concerns force pairs which act on
two different objects
(not on the same object) !
Physics 207: Lecture 8, Pg 14
Gravity
Newton also recognized that
objects.
gravity is an attractive, long-range force between any two
When two objects with masses
m
1 and
m
2 are separated by distance as follows:
r
, each object “pulls” on the other with a force given by Newton’s law of gravity, Physics 207: Lecture 8, Pg 15
Cavendish’s Experiment
F = m 1 g = G m 1 m 2 / r 2
g = G m
2
/ r
2 If we know big G , little g and r then will can find m 2 the mass of the Earth!!!
Physics 207: Lecture 8, Pg 16
Example (non-contact) Consider the forces on an object undergoing projectile motion F B,E = - m B
g
F E,B = m B
g
EARTH F F B,E E,B = - m = m B B
g 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 17
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 18
m
g
Ball Falls Example
The Free Body Diagram
F
B,T =
N
m
g
For Static Situation
N
= m
g
Physics 207: Lecture 8, Pg 19
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).
F B,T F T,B Main goal at this point : Identify force pairs and apply Newton’s third law Physics 207: Lecture 8, Pg 20
Example
First: Free-body diagram Second: Action/reaction pair forces
F B,E = -m
g
F E,B = m
g F
B,T =
N F
T,B = -
N
F B,E = -m
g
F E,B = m
g
Physics 207: Lecture 8, Pg 21
Exercise
Newton’s Third Law
A fly is deformed by hitting the windshield of a speeding bus.
v
The force exerted by the bus on the fly is, A.
greater than B.
equal to C.
less than that exerted by the fly on the bus.
Physics 207: Lecture 8, Pg 22
Exercise 2
Newton’s Third Law
Same scenario but now we examine the accelerations 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 A.
greater than B.
equal to C.
less than that of the fly.
Physics 207: Lecture 8, Pg 23
Exercise 2
Newton’s Third Law
Solution By Newton’s third law these two forces form an interaction pair which are equal ( but in opposing directions ).
Thus the forces are the same However, by Newton’s second law
F
net = m
a
So
F
b, f = -
F
f, b =
F
0 or
a
=
F
net /m .
but |
a
bus | = |
F
0 / m bus | << |
a
fly | = |
F
0 /m fly | Answer for acceleration is (C) Physics 207: Lecture 8, Pg 24
Exercise 3
Newton’s 3rd Law
Two blocks are being pushed by a finger on a horizontal frictionless floor. How many action-reaction force pairs are present in this exercise?
a b
A.
2 B.
4 C.
6 D.
Something else Physics 207: Lecture 8, Pg 25
Exercise 3
Solution:
6
F a,f F f,a F b,a F a,b F g,a
a
F E,a F g,b
b
F E,b F a,g F b,g F a,E F b,E
Physics 207: Lecture 8, Pg 26
Example
Friction and Motion
A box of mass
m 1 = 1
kg is being pulled by a horizontal string having tension
T = 40
N . It slides with friction ( m
k =
0.5
) on top of a second box having mass
m 2
slides on a smooth (frictionless) surface.
= 2
kg , which in turn What is the acceleration of the second box ?
(This is what I solved for in class!)
(A)
But first, what is force on mass 2?
a =
0 N (B)
a
= 5 N (C)
a
= 20 N (D) can’t tell
T
a = ?
m 2 m 1 v
slides with friction ( m k
=0.5
) slides without friction Physics 207: Lecture 8, Pg 27
Example
Solution
First draw FBD of the top box:
T v m 1 N 1 f k =
m K
N 1 =
m K
m 1 g m 1 g
Physics 207: Lecture 8, Pg 28
Example
Solution
Newtons 3 rd law says the force
box 2 exerts on box 1
equal and opposite to the force
box 1 exerts on box 2
.
is As we just saw, this force is due to friction: Reaction
f
2,1
= -
f
1,2
a =
0 N
(B) a = 5 N
m 1 m 2
Action
f 1,2 =
m K
m 1 g = 5 N
(C)
a
= 20 N (D) can’t tell Physics 207: Lecture 8, Pg 29
Example
Solution
Now consider the FBD of box 2:
N 2
f
2,1 =
m k
m 1 g m 2 m 2 g m 1 g
Physics 207: Lecture 8, Pg 30
Example
Solution
Finally, solve
F x = ma
m K
m 1 g = m 2 a
in the horizontal direction: a
m
1 m k g
m
2 5 N 2 kg
= 2.5 m/s 2
f
2,1 =
m K
m 1 g m 2
Physics 207: Lecture 8, Pg 31
Example
Friction and Motion , Replay A box of mass
m 1 = 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
m 2 = 2 kg .
The static and kinetic coefficients of friction between the 2 boxes are m s
=1.5
and m 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 ?
a 1
T
a 2 m 2 m 1
friction coefficients m s
=1.5
and m k
=0.5
slides without friction Physics 207: Lecture 8, Pg 32
Example
Friction and Motion , Replay in the static case A box of mass
m 1 = 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 m k
= 0.5.
m 2 = 2 kg .
The static coefficients of friction between the 2 boxes are m s and
=1.5
and The second box can slide freely on an smooth surface (frictionless).
kinetic If there is no slippage then maximum frictional force between 1 & 2 is (A) 20 N (B) 15 N (C) 5 N (D) depends on T
T
a 1 m 1
friction coefficients m s
=1.5
and m k
=0.5
a 2 m 2
slides without friction Physics 207: Lecture 8, Pg 33
Exercise 4
Friction and Motion , Replay in the static case A box of mass
m 1 = 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 m k
= 0.5.
m 2 = 2 kg .
The static coefficients of friction between the 2 boxes are m s and
=1.5
and The second box can slide freely on an smooth surface (frictionless).
kinetic If there is no slippage, what is the maximum frictional force between 1 & 2 is
T
a 1 m 1
friction coefficients m s
=1.5
and m k
=0.5
A.
20 N
a 2 m 2
slides without friction B.
15 N C.
5 N D.
depends on T Physics 207: Lecture 8, Pg 34
Exercise 4
Friction and Motion
f
S m S
N =
m S
m 1 g = 1.5 x 1
kg
x 10
which is 15 N (so m 2 can’t break free) m/s 2
N
f
S
T
m 1 g f s
= 10 N and the acceleration of box 1 is Acceleration of box 2 equals that of box 1, with |
a
| = |
T
| / (
m 1 +m 2
) and the frictional force
f
is
m 2 a
(Notice that if T were raised to 15 N then it would break free)
a 1
T
m 1
friction coefficients m s
=1.5
and m k
=0.5
a 2 m 2
slides without friction Physics 207: Lecture 8, Pg 35
Exercise
Tension example Compare the strings below in settings (a) and (b) and their tensions.
A.
B.
C.
D.
T a = ½ T b T a = 2 T b T a = T b Correct answer is not given Physics 207: Lecture 8, Pg 36
Lecture 8
Goals: Differentiate between Newton’s 1 st , 2 nd Use Newton’s 3 rd and 3 Law in problem solving rd Laws Assignment: HW4, (Chapters 6 & 7, due 10/1, Wednesday) Finish Chapter 7 1 st Exam Thursday, Oct. 2 nd from 7:15-8:45 PM Chapters 1-7 Physics 207: Lecture 8, Pg 37