5. Application of Newtons Laws

8/19/2019 5. Application of Newtons Laws

1/45

5.

Newton’s Laws and FrictionPhysics 50

Peter Beyersdorf

1

F o x t r o t b y

B i l l A m e n d


2/45

5.

Class Outline

Problem solving strategy

Example Problems

Description of Friction

Static

Kinetic

Example Problems

2


3/45

5.

Solving “Force” Problems

1. Identify the object of interest

2. Draw a free-body diagram for that object

3. Add up all forces in x, in y

4. Use F=ma, in x and in y directions

5. Solve for quantity of interest

6. Repeat with another object if necessary tosolve for another unknown

3


4/45

5.

Objects in Equilibrium

Equilibrium implies no net force on object

Forces in x-direction balance

Forces in y-direction balance

4


5/45

5.

Equilibrium Example 1

5

A car engine is suspended from a chain linked at O to two

other chains. Which of the following forces should be

included in the free-body diagram for the engine?

1. tension T 1

2. tension T 2

3. tension T 3

4. two of the above

5. all of T 1, T 2, and T 3

Q5.1

bit.ly/phys51


6/45

5.

Equilibrium Example 2

6

bit.ly/phys51

A cable attached to the

car holds the car at

rest on the frictionless

ramp (angle ! ).

1. n = w

2. n > w

3. n < w

4. not enough information given to decide

Q5.2

The ramp exerts a normal force on the car. How does the magnitude

n of the normal force compare to the weight w of the car?


7/455.

Modified Atwood’s Machine Example

7

bit.ly/phys51

A cart (mass m1, weight w1)

is attached by a lightweight

cable to a bucket (mass m2,

weight w2) as shown. The

ramp is frictionless.

Q5.3

When released, the cart accelerates up the ramp.

Which of the following is a correct free-body diagram for the cart?

T

w1

n

T

w1

n

T

w1

n

m a1

w1

n

T

m a1

1. 2. 3. 4.


8/45

5.

A cart (mass m1, weight w1)

is attached by a lightweight

cable to a bucket (mass m2,

weight w2) as shown.

Q5.4

The ramp is frictionless. The pulley is frictionless and does

not rotate. When released, the cart accelerates up the ramp

and the bucket accelerates downward.

1. T = w2

2. T > w2

3. T < w2

4. not enough information given to decide

Which statement about the cable tension (magnitude T ) is correct?

Modified Atwood’s Machine Example

8

bit.ly/phys51


9/45

5.

Friction Example

9

bit.ly/phys51

You are pushing a 1.00-kg

food tray through the

cafeteria line with a

constant 9.0-N force. As

the tray moves, it pushes ona 0.50-kg milk carton.

1. the weight of the milk carton

2. the force of the milk carton on the tray

3. the force you exert with your hand

4. the acceleration of the milk carton

5. more than one of the above

Q5.5

Which of the following forces should be included in a free-body

diagram for the milk carton?


10/45

5.

Blocks A and C are connected

by a string as shown. When

released, block A accelerates

to the right and block C

accelerates downward.

1. block B accelerating to the right

2. block B accelerating to the left3. block B moving at constant speed to the right

4. block B moving at constant speed to the left

5. none of the above

Q5.6

Although there is friction between blocks A and B,there is not enough to prevent block B from slipping.

If you stood next to the table during the time that

block B is slipping on top of block A, you would see

Modified Atwood’s Machine Example 2

10

bit.ly/phys51


11/45

5.

You are driving your car

(mass m) at a constant speed v

around a flat, unbanked curve

of radius R.

Which of the following forcesshould be included in a free-

body diagram for the car?

1. an outward centrifugal force of magnitude mv2/ R

2. an inward centripetal force of magnitude mv2/ R

3. the force of the cars acceleration

4. two of the above

5. none of the above

Q5.7

R

Unbanked Curve Example

11

bit.ly/phys51


12/45

5.

You are driving your car

(mass m) at a constant speed v

around a banked curve of

radius R and bank angle !

(measured from thehorizontal).

1. a normal force that points vertically upward

2. a normal force that points at an angle ! from the vertical3. a normal force that points at an angle ! from the horizontal

4. an outward centrifugal force of magnitude mv2/ R

5. more than one of the above

Q5.8

Which of the following forces should be included in a free-body

diagram for the car?

!

R

Banked Curve Example

12

bit.ly/phys51


13/45

5.

Equilibrium example

An engine of mass M hangs inan autoshop as shown in thediagram. What is the tensionin each of the three chains?

13


14/45

5.

Equilibrium example

Free-bodydiagram for

Engine

Free-bodydiagram for

suspension ring

14


15/45

5.

Equilibrium example

T1=mg T2=T3 cos 60° = mg/tan(60°) T3 = mg/sin(60°)

15


16/45

5.

Elevator example

An elevator acceleratesdownwards at 2m/s2. Awoman of mass M is on ascale, what weight does the

scale read?

Before stopping thedownward traveling elevatordecelerates at 2m/s2, what

weight does the scale read?

16


17/45

5.

Elevator example

An elevator acceleratesdownwards at 2m/s2. Awoman of mass M is on ascale, what weight does the

scale read?

N=m 7.8 m/s^2 = 0.8 Mg

Before stopping thedownward traveling elevatordecelerates at 2m/s2, whatweight does the scale read?

N=m 11.8m/s^2 = 1.2 Mg

17


18/45

5.

Elevator example 2

An elevator of mass m iscounterbalanced by a weight withmass M. If the motor and brakeare not engaged what will be the

acceleration of the elevator?

m M

18


19/45

5.

Elevator example 2

An elevator of mass m iscounterbalanced by a weight withmass M. If the motor and brakeare not engaged what will be the

acceleration of the elevator?

a=(m-M)/(M+m) g

m M

19


20/45

5.

Example

A tightrope walker has a massof 50 kg. The rope has abreaking strength of 2.5 kN. Ifthe rope is 10m long, What is the

least amount it can sag when theacrobat is in the middle of therope?

20


21/45

5.

2(T sin θ)−mg = may

y

21

L/2

d

p d 2 +

( L/ 2 ) 2

mg

TT

in y-direction

x

0

d

sin θ = d

p d2 + (L/2)2

2Td = mgp d2 + (L/2)2

2Tdp d2 + (L/2)2

= mg

(2Td)2 = (mg)2 d2 + (L/2)2

d2

(2T )2 − (mg)2

= (mg)2(L/2)2

d2 = (mgL/2)2

((2T )2 − (mg)2)

d = (mgL/2)

p (2T )

2−

(mg)2

d = L

2

1p

(2T/mg)2 − 1

when T=Tmax=2500 N

d = m

2 p

(2(2500 N)/(50 kg)(9.8 m/s2))2 − 1

d = m

2 p

(2(2500 N)/(50 kg)(9.8 m/s2))2 − 1

= 0.

50 m


22/45

5.

Frictional forces

Friction is a contact force

Opposes motion

Proportional to the Normal force

Depends on the materials in contact

22


23/45

5.

Friction as a contact force

Comes from roughness ofsurfaces

Opposes motion

Friction ranges from zero toa maximum value for“stationary objects”

Friction has a constant value

for” moving objects”

23


24/45

5.

Friction depends on theNormal force

Max value proportional tothe normal force

Constant of

proportionality dependson the materials incontact

direction of force vectoropposes (would-be) motion

N

N

-N

24


25/45

5.

Some values for ! s

25


26/45

5.

Free-body diagram

N

mg

f

26


27/45

5.

Free-body diagram

N

mg sinf

mg cos

27


28/45

5.

Static friction example

A brass and an aluminum block sit on an inclined planemade of steel. Over what range of inclination angleswill one block stay stationary while the other slidesdown the slope?

Which one will slide down the slope?

28


29/45

5.

Static friction example

A brass and an aluminum block sit on an inclined planemade of steel. Over what range of inclination angleswill one block stay stationary while the other slidesdown the slope?

Which one will slide down the slope?

The angle when a block will slide is =tan1() so for aluminum-steel = 0.61

brass-steel=0.51 27°


30/45

5.

Kinetic friction

Constant force opposingmotion for a “movingobject”

Constant of proportionalitydepends on the materials incontact

Is always less or equal tomaximum static friction

direction of force vectoropposes motion

30


31/45

5.

Some values for ! k

31


32/45

5.

Kinetic friction example

A brass and an aluminum block sit on an inclined planemade of steel. When the plane is inclined just enoughso that they both begin to slide, what is theiracceleration?

How much further apart are they are sliding for 1second.

32


33/45

5.

Kinetic friction example

A brass and an aluminum block sit on an inclined planemade of steel. When the plane is inclined just enoughso that they both begin to slide, what is theiracceleration?

How much further apart are they are sliding for 1second.

With =31° (from earlier problem) both will slide

Brass (k=0.44) so a=1.35m/s^2

Aluminum (

k=0.47) a=1.10 m/s^2

and x(1 s)=12.5 cm 33


34/45

5.

Friction in real life

What force causes a car to turn?

What is the maximum value for that force?

How can that be increased? (i.e. how do you design a

car that corners better?)

34


35/45

5.

Example #1

On the “Gravitron” riderstravel in a circle atconstant speed. Theperiod of revolution is

4.0s, the radius is 5m.What is the minimumcoefficient of frictionnecessary to keep themfrom sliding down the

walls when the floordrops out?

35


36/45

5.

Example #1

On the “Gravitron” riderstravel in a circle atconstant speed. Theperiod of revolution is

4.0s, the radius is 5m.What is the minimumcoefficient of frictionnecessary to keep themfrom sliding down the

walls when the floordrops out?

circumference=31.4m

speed=7.9 m/s a=12.3 m/s^2 s>0.79

36


37/45

5.

Example #2

Draw free body diagramsfor the following situations.

Under what conditions willthe masses slide?

37


38/45

5.

Example #2

38

x y

mgMg

N1

N2

f 2

f 1

Mg cos

m g

c o s

M g s i n

m g s i n

Fy=0:

N2-Mg cos=0

N1-mg cos=0

Fx=0:

f 2-Mg sin=0

f 1-mg sin=0

For static friction f 2"sN2

f 1"sN1

s(Mg cos)-Mg sin=0

s(mg cos)-mg sin=0 s(cos)-sin=0 =tan-1

When friction is at its maximum value

Beyond this the blocks will slide, i.e. they will slide for>tan-1


39/45

5.

Example #2

39

x y

mg Mg

N1

N2

f 2

f 1

Mg cos

m g

c o s

M g s i n

m g s i n

Fy=0:

N1-m1g cos=0

N2=m1g cos

Fx=0:

T-f 1-m1g sin=0

For static friction: f 1"sN1

m1g sin+sm1g cos"T

T

Tf 1N1

Block 1:

thus:


40/45

5.

Example #2

40

x y

mg Mg

N1

N2

f 2

f 1

Mg cos

m g

c o s

M g s i n

m g s i n

Fy=0:

N2 —N1-m2g cos=0

N2 —m1g cos-m2g cos=0

N2=(m1+m2)g cos

Fx=0:

T+f 1+f 2-m2g sin=0

For static friction: f 2"sN2

T+s(m1g cos)+s(m1+m2)g cos-m2g sin"0

T

Tf 1N1

Block 2:

thus:

T"-s(m1g cos)-s(m1+m2)g cos+m2g sin


41/45

5.

Example #2

41

x y

mg Mg

N1

N2

f 2

f 1

Mg cos

m g

c o s

M g s i n

m g s i n

T

Tf 1N1

Block 2: T"-s(m1g cos)-s(m1+m2)g cos+m2g sin

m1g sin+sm1g cos"T Block 1:

m1g sin+sm1g cos"-s(m1g cos)-s(m1+m2)g cos+m2g sin

m1g sin-m2g sin"-s(m1g cos)-s(m1+m2)g cos-sm1g cos

(m1-m2)g sin "s(-m1-(m1+m2)-m1)g cos

tanθ ≤ µs

−3m1 + m2

m1 −m2

θ ≤ tan−1µs

3m1 + m2

m2 −m1


42/45

5.

Example 3

A crate of mass m rests on a horizontal floor withcoefficients of friction µs and µk. A woman pushes on itwith a force of magnitude F at an angle below the

horizontal. What magnitude of force F is required to

keep it moving at a steady rate.

Show that there is a critical value for µs beyond whichshe cannot get the crate to move no matter how hardshe pushes.

42


43/45

5.

Example 3

A crate of mass m rests on a horizontal floor withcoefficients of friction µs and µk. A woman pushes on itwith a force of magnitude F at an angle below the

horizontal. What magnitude of f is required to keep it

moving at a steady rate.

Show that there is a critical value for µs beyond whichshe cannot get the crate to move no matter how hard

she pushes.

f = k(mg) /(cos -k sin)

for k=tan-1, f #. Beyond this the crate cannot be moved!

43


44/45

5.

Summary

Problems involving forces can be solved systematically

In equilibrium

horizontal forces balance

vertical forces balance

Friction is a contact force that opposes motion

static friction is less than or equal to some maximum value

kinetic friction is a constant

coefficient of friction is a constant of proportionality betweenthe maximum frictional force and the normal force at thesurface

44


45/45

5. Application of Newtons Laws

Documents

Transcript of 5. Application of Newtons Laws