common forces...iagrams common forces: •gravity (fg) •normal (fn) •friction (ff ... always...

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ch4 Newton's Laws NOTES.notebook 1 December 02, 2013 F ree B ody D iagram s Common Forces: Gravity (F G ) Normal (F N ) Friction (F F ) Tension (F T ) Spring (F S ) Always points down. Always away from surface. Opposes motion. Rope, chain, string, cable, etc. away from object. Away or towards object.

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ch4 Newton's Laws NOTES.notebook

1

December 02, 2013

Free Body Diagrams Common Forces:

• Gravity (FG)

• Normal (FN)

• Friction (FF)

• Tension (FT)

• Spring (FS)

Always points down.

Always      away from surface.

Opposes motion.

Rope, chain, string, cable, etc. away from object.

Away or towards object. 

ch4 Newton's Laws NOTES.notebook

2

December 02, 2013

ch4 Newton's Laws NOTES.notebook

3

December 02, 2013

Chapter 4 Newton's Laws (Forces)

Every object continues at a state of rest or constant velocity unless acted upon by an unbalanced force.

Inertia

Acceleration is inversely proportional to mass.

F = ma

F = ma

F = ma

Newton's First Law

Newton's Second Law

frictionless

F10 kg

a = 3.5 m/s2

The 20 kg crate accelerates to the right at 0.1 m/s2. What is the force F?

ch4 Newton's Laws NOTES.notebook

4

December 02, 2013

Newton's Third Law

Forces come in pairs...action and reaction

not on the same object.....

Air pushed down

Helicopter pushed up

ch4 Newton's Laws NOTES.notebook

5

December 02, 2013

NSL Problems

30 N10 N4 kg

40 N

30 Nm

a = ?

10 N

a = 2 m/s2

a = 1.45 m/s28 kg

90 N

m = ?

F = ?

F

10,000 N

9,800 N

500 N200 N

m = 1000 kg

FNET= ?

a = ?

ch4 Newton's Laws NOTES.notebook

6

December 02, 2013

Normal

Weight & Normal

11 kg

8 kg

11 kgThe Moon

ch4 Newton's Laws NOTES.notebook

7

December 02, 2013

FNET = ΣF = maΣFx = max

Multiple Forces

ΣFy = may

7.5 kgF = 150 Nf = ­70 N

T = ?

a = 0.85 m/s2

75 kg

4 kg

[1.23 m/s2 at 4.1o]

sumup

6 kg

a = 2.5 m/s2

F = 20 N

What must be the friction force?

[­5 N]

ch4 Newton's Laws NOTES.notebook

8

December 02, 2013

Vertical Motion (Elevators & Pulleys)

SCALE READS THE NORMAL FORCE

scale

mgirl = 43 kg

Scale reads 370 N

a = ?

Find F if:

at rest.....

a = ­0.8 m/s2

12 kg

F

What will the scale read if...

accelerates up at 1.2 m/s2

accelerates down at 4.5 m/s2m = 70 kg

What is the force of motor A and motor B?

[­1.2 m/s2]

[14.7 N]

[13.5 N]

[770 N]

[371 N]

[A: 49 N]

[B: 98 N]

20 kg

ch4 Newton's Laws NOTES.notebook

9

December 02, 2013

Friction

FF = μFN μ = coefficient  of friction(mu)

• Opposes motion• 2 Types (static & kinetic)

A 17 kg  rubber tire is at rest on wet concrete• What force is required to move it?

at rest

moving

FBD

FBD

(wet)

(wet)

• What force is required to accelerate it 3.0 m/s2 to the right?

ch4 Newton's Laws NOTES.notebook

10

December 02, 2013

Solving Problems• Draw FBD (calculate F's)• Forces Cancel?• Set ΣF = ma• Solve

Curlingv0 = 4 m/s

What will be the acceleration of the stone?

How far will it travel?

μK = 0.04

[­0.39 m/s2]

[20 m]

ch4 Newton's Laws NOTES.notebook

11

December 02, 2013

T = 60 N8 kg

μk=0.25

What will be the acceleration?

ch4 Newton's Laws NOTES.notebook

12

December 02, 2013

Inclines

F||

F Τ

θ

w = mg

θ

N

Starts at rest...how long to reach the bottom if frictionless?

1.7 m

How long to reach the bottom if μK = 0.22?

25o

[0.91 s]

[1.25 s]

If μS=0.38 what is the maximum angle it can be so it won't slip?

θ

ch4 Newton's Laws NOTES.notebook

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December 02, 2013

Multiple Objects• Draw FBD (Calculate Forces)• Forces Cancel?• Ignore Tensions...at first.• FNET = ma (which way is a?)• Solve......bubble to get tensions..

Atwood's Machine...Find the acceleration and tension in the rope.

2 kg

a = 4.2 m/s2T = 28 N

3 kg

Find the a, then the T...

Find the a and T...

6 kg 13 kg

4 kg

5 kg

a = 1.4 m/s2T = 34 N

a = 3.6 m/s2T = 81 N

ch4 Newton's Laws NOTES.notebook

14

December 02, 2013

Calculate a and T if it is frictionless.

T3.5 kg

2 kg

a = 3.56 m/s2T = 12.5 N

Calculate a and T if μK = 0.33

T

2 kg

3.5 kg

a = 1.51 m/s2T = 16.6 N

Find m.....

m

4 kg

a = 2 m/s2

m = 1.28 kg

ch4 Newton's Laws NOTES.notebook

15

December 02, 2013

Calculate a, T1, and T2 if frictionless.

Calculate a, T1, and T2 if μK = 0.3

T1

T2

T2T1

Calculate a, T1, and T2 if μK = 0.2

T1

T2

μK = 0.19

friction

less

a = 1.37 m/s2T1 = 88.2 NT2 = 32.3 N 

a = 3 43 m/s2T1 = 6 37 NT2 = 19 1 N 

a = 4 9 m/s2T1 = 4 9 NT2 = 14 7 N 

ch4 Newton's Laws NOTES.notebook

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December 02, 2013

Find a, T1, and T2. No friction...

3 kg

10 kg

7 kg

T1T2

Find a, T1, and T2, μK = 0.35

4 kg

T2 6 kgT1

9.5 kg

a = 1.96 m/s2T1 = 55 NT2 = 35 N

a = 1.71 m/s2T1 = 46 NT2 = 77 N

ch4 Newton's Laws NOTES.notebook

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December 02, 2013

100 N5 kg

8 kg

Find the acceleration and the tension....

μK = 0.1

μK = 0.2

Stacked Problems.....

3 kg

 

μ = 0.1

μ = 0.2T

 

Acceleration of the 5 kg?

Find T

ch4 Newton's Laws NOTES.notebook

18

December 02, 2013

24o

mB = 3.5 kgmA = 0.5 kg

μAB = 0.2

μincline = 0.045

Find the acceleration....

a = 2.14 m/s2