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