applied natural sciences - porous mediahigh-rise syndrome in cats,whitney wo, mehlhaff cj. am vet...

Applied Natural Sciences

Leo Pele‐mail: [email protected]://tiny.cc/3NAB0

Het basisvak Toegepaste Natuurwetenschappen

http://www.phys.tue.nl/nfcmr/natuur/collegenatuur.html

Copyright © 2012 Pearson Education Inc.

PowerPoint® Lectures forUniversity Physics, Thirteenth Edition

– Hugh D. Young and Roger A. Freedman

Lectures by Wayne Anderson

Chapter 5

Applying Newton’s Laws

LEARNING GOALS

• How to use Newton’s first law to solve problems

involving the forces that act on a body in equilibrium.

• How to use Newton’s second law to solve problems

involving the forces that act on an accelerating body.

• The nature of the different types of friction forces—static

friction, kinetic friction, rolling friction, and fluid

resistance—and how to solve problems that involve

these forces.

• How to solve problems involving the forces that act on a

body moving along a circular path.

3

Summary

aF

m

ABBA FF

1 th law: inertia

2nd law: acceleration

3th law: action = -reaction

0 constant F v

Free body diagram

4

Content (part 1)

The three laws of Newton

Type of forces and properties

Free body diagram

Circular motion and netto force

Friction

5

Properties of friction

Friction depends on the normal force

Friction depends on the material properties

6

Magnitude of the frictional force

7


8


T = fs9


10


T = fs

T = fs = μsn

11


T = fs fk= constant (< T)

T = fs = μsn

12




Static friction variesbetween 0 and a maximum value (=μsN)

14

http://phet.colorado.edu/en/simulation/ramp-forces-and-motion

http://www.phys.tue.nl/nfcmr/natuur/

ramp

15

The static frictional force keeps an object from starting to move when a force is applied. The static frictional force has a maximumvalue, but may take on any value from zero to the maximum, depending on what is needed to keep the sum of forces zero.

Static Friction

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Static friction variesbetween 0 and a maximum value (=μsN)

s k

Fs s N s: static friction coefficient

Fk = k N k: kinetic friction coefficientinequality

Please note: no vectors18

Static Friction

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Kinetic Friction

Fpush fk = kn

20

Friction force

I push a crate at constant speed on a floor. I then decides to turn the crate, so that the crate with a two times lower surface is in contact with the floor. If I want to push the crate with the same speed on the same floor, I must exert a force that

2. 2x as small.

1. 2x as large.

3. the same.

Answer: 3. The frictional force is dependent on the normal force ( which remained the same), and the kinetic coefficient of friction (independent of the size of the area)

DEMO surface area does not matter

Example friction?

Calculate the kinetic friction coefficient at constant speed going down

Constant speed, a = 0fk = w sin n = w cos and : fk = μk n

So: μk = fk/n = tan

23

Watch for this common error

A good “road sign” is to be sure that the normal force comes out

perpendicular to the surface.

24

Example friction?

Calculate the acceleration for a kinetic friction coefficient

w sin – μk n = ma

n = w cos

so: a = g (sin – μk cos )

fk

25

High-rise syndrome in cats,Whitney WO, Mehlhaff CJ.

Am Vet Med Assoc, 1987 Dec 1;191(11):1399-403.

In a 1987 study of 132 cats brought to a New YorkCity emergency veterinary clinic after falls fromhigh-rise buildings, 90% of treated cats survived andonly 37% needed emergency treatment to keepthem alive. One that fell 32 stories onto concretesuffered only a chipped tooth and a collapsed lungand was released after 48 hours.

27

Terminal velocity

Fluid resistance

Cat 60 km/h : survives

Human 120 km/h : dies

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Unbanked and banked curves

The required horizontal force on the car to allow him to maintain the circular path is provided by the frictional force.

What is the maximum speed for friction coefficient μ?

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What is the maximum speed for friction coefficient μ?

1) Static friction 2) Dynamic friction



What is the maximum speed for friction coefficient μs?

31

Calculation

F ma

so:

0x x

y

F maF

2vf mR

n mg

max sf n

2

maxmaxs s

vmg m v gRR

and:

Ex: μs = 0.87, R = 230 m, dan

vmax= 160 km/h (44m/s)

32


Here it is the horizontal component of the normal force.

No friction (not required). What is required angle β at speed v?

34

Calculation

F ma

so:

0x x

y

F maF

2

sin

cos

vn mR

n mg

2

tan vgR

Ex: v = 88 km/h (25 m/s), R = 230 m, dan β = 15o.

hence:

35

Examples

www.phys.tue.nl/nfcmr/natuur

See chap 5

Summary

38

Summary

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applied natural sciences - porous mediahigh-rise syndrome in cats,whitney wo, mehlhaff cj. am vet...

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