Lecture 8

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 10/1, Wednesday)

Finish Chapter 7

1st Exam Thursday, Oct. 2nd from 7:15-8:45 PM Chapters 1-7


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 k = tan



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


fk = k N = k mg cos

Fx= 0 = mg sin – k mg cos

k = tan (only one angle)

v



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


fk = k N = k mg cos

Fx= max = mg sin ± k mg cos

ax = g sin ± k g cos


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


Fish Schools


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.


“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


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


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”


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:


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!!!


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)


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


Example

The Free Body Diagram

mg

mg

FB,T= N

Ball FallsFor Static Situation

N = mg


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


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


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.


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


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?


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


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 ?

(This is what I solved for in class!) But first, 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


ExampleSolution

First draw FBD of the top box:

m1

N1

m1g

T fk = KN1 = Km1g

v


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 = 0 N (B) a = 5 N (C) a = 20 N (D) can’t tell


Now consider the FBD of box 2:

m2 ff2,1 = km1g

m2g

N2

m1g

ExampleSolution


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


Example 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


ExampleFriction 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 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

mm22

T mm11friction coefficients

s=1.5 and k=0.5


a1


Exercise 4Friction and Motion, Replay in the static case

A. 20 N

B. 15 N

C. 5 N

D. depends on T

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 there is no slippage, what is the maximum frictional force between 1 & 2 is

mm22

T mm11



a1


Exercise 4Friction 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 raised to 15 N then it would break free)

mm2 2

T mm11



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)


Exercise Tension example

A. Ta = ½ Tb

B. Ta = 2 Tb

C. Ta = Tb

D. Correct answer is not given

Compare the strings below in settings (a) and (b) and their tensions.


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 10/1, Wednesday)

Finish Chapter 7

1st Exam Thursday, Oct. 2nd from 7:15-8:45 PM Chapters 1-7

Lecture 8

Documents

Transcript of Lecture 8