– Static friction

Chapter 6. Force and motion II

• Friction

– Static friction

– Sliding (Kinetic) friction

• Circular motion

Summary of last lecture

• Newton’s First Law:

The motion of an object does not change unless it is acted on by a net force

• Newton’s Second Law: F

= ma

• Newton’s Third Law: F

= - F

• Types of Forces

– Gravity

– Normal Force – Friction

– Tension

Static Friction (정지쓸림힘)

• Static Friction: a force between two surfaces that prevents motion

• f_s < μ_sF_{N =}F_s,max (size is whatever is needed to prevent motion) μ_s = coefficient of static friction

a property of the two surfaces 1 ≥ μ_s ≥ 0

• direction is whatever direction needed to prevent motion

W F

f

F

Kinetic Friction (운동쓸림힘)

• Kinetic (sliding) Friction: a force between two surfaces that opposes motion

• f

= μ

F

• μ

= coefficient of sliding friction

a property of the two surfaces 1 ≥ μ

≥ 0

• direction is opposite to motion W

F

F f

direction of motion

Static Kinetic

f_s ≤ μ_sn , μ_s : coefficient of static friction f_k = μ_kn , μ_k : coefficient of kinetic friction

Example 1

v

=8 m/s μ

= 0.2

Find stopping distance

Answer: 16.3 m

F = μ

F

=

F = ma, V

= V

- 2as

mg F_N

Example 5-1

Initial velocity v₀ = 20 m/sec.

Stop at l = 115 m

Find the friction coef. : μ_k

n

F = −μ _k = −μ _k mg = ma g

a = −μ _k

g t v

t g v

t a v

v

k

k

= μ

=

⋅ μ

−

=

⋅ +

=

0

0

0

0

( )

177 115 0

8 9 2

20 2

2

2 2 2

0

2 2 0

2 1 0

2 2

1 0

m . sec

/ m .

sec /

m gl

v

g gt v

t v at

t v l

k

k k

× =

= ×

= μ

= μ μ

−

= +

=

Example 1-2

Example 2

T=50 N

M = 5 kg μ

= 0.2

Find acceleration of block

Answer: 8.04 m/s

F = T- μ

F

= ma

Example 3

T=50 N

M = 5 kg μ

= 0.2

Find acceleration of block θ=50

Answer: 6.0 m/s

F = T cosθ - μ

F

= ma

Example 4

a b

Which case requires the lesser force to overcome static friction?

a) a b) b

c) the same

F₁ = T - μ_km₁g = m₁a F₂ = m₂g - T = m₂a

+)

m₂g - μ_km₁g = m₁a +m₂a _{m m} ^g m

a m ^k

2 1

1 2

+ μ

= − Example 5-2

=μ_km₁g

Example 5 : Find a.

m₁

m₂

m₁

m₂

Uniform Circular Motion

(circular motion with constant speed)

vv RR

a v

= R²

centripetal acceleration

• Instantaneous velocity is tangent to circle

• Instantaneous acceleration is radially inward

• There must be a force to provide the acceleration a

Centripetal force (구심력) and Centrifugal force (원심력)

Angular velocity

dt d θ

≡

ω ω

= 2 π T

x y

θ

v

T = F_r = mrω²

r m v

= 2

Centripetal force Centrifugal force?

Question

Consider the following situation: You are driving a car with constant speed around a horizontal circular track. How many forces are

acting on the car?

1 2 3 4 5

W

f F

correct

The net force on the car is

1. Zero

2. Pointing radially inward 3. Pointing radially outward

W

f F

ΣF = ma = mv

/R

a=v²/R R

The car has a constant speed so acceleration is zero

A car that is driving in a circle is accelerating towards the center of the circle. According to Newton's second law the net force must be pointing toward the center of the circle.

when I sit in this car I can feel the FORCE...pulling me to the dark side...away from the center of the circle

Question

correct

Suppose you are driving through a valley whose bottom has a circular shape. If your mass is m, what is the magnitude of the

normal force F_N exerted on you by the car seat as you drive past the bottom of the hill?

1. F_N < mg 2. F_N = mg 3. F_N > mg

v

mg F_N

R

ΣF = ma = mv²/R F_N - mg = mv²/R

a=v²/R

You feel like you are being pulled down in your seat so the mg must be greater then the normal force -- wrong!

F_N is always equal to mg --- wrong!

Question

correct

Going over top of hill….

1. F

< mg 2. F

= mg 3. F

> mg

mg F

a=v²/R R

ΣF = ma = mv

/R mg - F

= mv

/R F

= mg - mv

/R

v

Question

correct

Example: Circular motion and gravity

• Newton’s Law of Gravity: F = GM_Em/R²

• Newton’s 2nd law: F=ma

• Centripetal acceleration: a=v²/R

• Thus…GM_Em/R² = mv²/R

• v = √GM_E/R

• T = time for one orbit = 2πR/v

M_E R m

R

T ₌ 2 π

Find the period for one orbit.

Example: Circular motion and gravity

E 3/2

GM R

T ₌ 2 π

1. R_moon-earth = 3.85 x 10⁸ m Plug in and calculate: T=27.5 days

2. R_shuttle = R_E = 6.38 x 10⁶ m Plug in and calculate: T= 84.5 minutes 3. Rsynchronous = 4.23 x 10⁷ m Plug in and calculate: T=1 day

M_E R m

Example: Carnival Ride

g

Find minimum coefficient of static friction so that you don’t fall.

v

R

Answer: μ_S > gR/v²

Example : Find the angle θ.

Tsinθ Tcosθ

θ

= 0

− θ

= T cos mg

F : ˆy _y

= θ

cos T mg

r m v ma

sin T F

: xˆ _x

= 2

= θ

=

r tan mv

mg

= 2

θ θ θ

tan sin

2 2

gL v gr

v =

=

Example: Find the maximum velocity on the track.

r m v ma

f

F _s

= 2

=

=

mg f _{s ,max} = μ _s

m gr r

v _max = f ^{s ,max} = μ _s

Example : Find the normal forces at bottom, top, and at A-point.

r mg mv

n _bot

= 2

At Bottom: − ^mg

r

n _bot = mv ² +

r mg mv

n _top

= 2

+ mg

r

n _top = mv ² −

At Top:

mv ² nG A

Terminal speed (종단속력)

V _terminal when a = 0

v b

R G G

−

=

a m v

b g

m

F G G G G

=

−

=

ma bv

mg − =

= 0

−

= v

m g b

a

Terminal speed

b g

v

= m

2 2

1

D Av R = ρ

Drag Coefficient

Air density Cross section

ma Av

D mg

R mg

F = − = − ¹₂ ρ ² =

2

2 v

m A g D

a ρ

−

= _{= 0} : For terminal velocity

v_t mg

= 2 ρ

r 3

∝

r v ∝

Terminal speed (종단속력) in air