More Newton’s Laws Applications

Unit 3 Presentation 2

Revisit Newton’s 3rd Law

“Every action force has an equal in magnitude but opposite in direction reaction force”

The action and reaction forces act on DIFFERENT objects

Newton’s 3rd Law Examples

Action Force Reaction Force

A gun is fired, propelling a bullet forward.

The gun fired backwards, in the opposite direction of the bullet.

A person pushes on a wall. The wall pushes back on the person.

A person walks down a street, exerting a frictional force against the ground.

The ground exerts a normal force against the person, propelling them forward.

Newton’s 3rd Law Example

A block-block system, like the one pictured below, is pushed by a force of 15 N from the left to the right. Calculate(a) The acceleration of the block-block system(b) The magnitude of the action-reaction force in between the two

blocks

2 kg

5 kg15 N

First, break the two blocks into individual problems and draw two free-body diagrams:

2 kg block 5 kg block

Gravity Gravity

Normal ForceNormal Force

15 NAction/ Reaction Force (same on both blocks)

2 kg 5 kg

AR Force AR

Force15 N

Newton’s 3rd Law Example (cntd)

Now, consider the free-body diagrams. Remembering to keep the x and y variables separate, set up some equations.

Do we need to consider the y direction forces? NO! The motion is only in the x direction in this example.

amFN ar 115

Now, using Newton’s 2nd Law on each block in the x-direction only:

amFar 2

?

?

5

2

2

1

arF

a

kgm

kgmNotice, we have 2 equations with 2 unknowns. Lets use substitution and substitute the right-hand equation into the left-hand equation:

asm

a

aa

amamN

2

12

/14.2

715

2)5(15

)(15 Next, solve for the AR Force:

NF

F

ar

ar

7.10

)14.2(5

Another Newton’s 3rd Law ExampleA block-block system, like the one pictured below, is pushed and accelerates at 3.2 m/s2 from the left to the right. Calculate:

(a) The force on the 8 kg block(b) The magnitude of the action-reaction force in between the two blocks

8 kg

3 kg

3.2 m/s2

FAction/ Reaction Force (same on both blocks)

8 kg 3 kg

First, break the two blocks into individual problems and draw two free-body diagrams:

8 kg block 3 kg block

Gravity Gravity

Normal ForceNormal Force

AR Force AR

ForceF

Another Newton’s 3rd Law Example (cntd)

Now, consider the free-body diagrams. Remembering to keep the x and y variables separate, set up some equations.

Do we need to consider the y direction forces? NO! The motion is only in the x direction in this example.

)/2.3( 21 smmFF ar

Now, using Newton’s 2nd Law on each block in the x-direction only:

)/2.3( 22 smmFar

?

?

/2.3

3

8

2

2

1

F

F

sma

kgm

kgm

ar

Now, solve the right-hand equation to determine the magnitude of the action-reaction force:

NF

NNF

smkgsmkgF

2.35

6.256.9

)/2.3(8)/2.33( 22

Next, solve for the Force:

NF

smkgF

ar

ar

6.9

)/2.3(3 2

The Force of Friction

Friction is a force that always opposes motion Opposite in direction to the motion of the

object Friction is directly related to the normal

force:

Nf

f

Ff

frictionf

= Coefficient of Friction (pure number between 0 and 1)

=0 No Friction (frictionless surface)

=1 Highest frictional force possible

Types of Friction

Static Friction: The force of friction that opposes motion from beginning from rest. Also known as sticking friction.

Kinetic Friction: The force of friction that opposes motion as it is occurring. This only occurs after motion has begun.

Example: When pushing a large block, static friction prevents you from moving the block at all. When you finally start pushing it, kinetic friction takes over and slows the motion.

Always…..static friction is stronger than or equal to kinetic friction in magnitude.

Coefficients of Friction

s =Coefficient of Static Friction

k =Coefficient of Kinetic Friction

Always….

ks

Coefficients of Friction for Common Surfaces

Materials s value k valueSteel on Steel 0.74 0.57

Aluminum on Steel 0.61 0.47

Copper on Steel 0.53 0.36

Rubber on Concrete ~1.0 0.8

Wood on Wood 0.25 – 0.50 0.2

Glass on Glass 0.94 0.4Waxed wood on wet snow 0.14 0.10Waxed wood on dry snow 0.01 0.04

Metal on Metal (lubricated) 0.15 0.06

Ice on Ice 0.10 0.03

Teflon on Teflon 0.04 0.04Synovial joints in Humans 0.01 0.003

Frictional Forces Example

Calculate the static and kinetic frictional forces of a 5 kg block of aluminum on a steel table if a force were applied to move it to the right.

First, find the coefficients from the table on the previous slide:

2/8.9

5

47.0

61.0

smg

kgmk

s

Gravity (Weight)

Normal Force

f (friction)F (applied)

Noting that there is no acceleration in the y direction and using Newton’s Second Law:

NsmkgmgF

mgF

n

n

49)/8.9(5

02

Frictional Forces Example (cntd)

Now, lets calculate the static frictional force:

Now, lets calculate the kinetic frictional force:

NNFf nss 9.2949*61.0

NNFf nss 0.2349*47.0