Short Version : 5. Newton's Laws Applications

Example 5.3. Restraining a Ski Racer

A starting gate acts horizontally to restrain a 60 kg ski racer on a frictionless 30 slope.

What horizontal force does the gate apply to the skier?

0m anet h g F F n F 0asince

, 0h hF F

sin , cosn n 0 ,g m g F

sinhF n

cos

m gn

sin 0hF n

cos 0n m g

sincosh

m gF

260 9.8 / tan 30kg m s 340 N

x

y

Fg

n

Fh

x :

y :

Alternative Approach

cos sin 0h gF F

tanh gF F

260 9.8 / tan 30kg m s 340 N

x

y

Fg

n

Fh

Net force along slope (x-direction) :

sinm g

5.2. Multiple Objects

Example 5.4. Rescuing a Climber

A 70 kg climber dangles over the edge of a frictionless ice cliff.

He’s roped to a 940 kg rock 51 m from the edge.

(a)What’s his acceleration?

(b)How much time does he have before the rock goes over the edge?

Neglect mass of the rope.

rock r g r F T F n

, 0r rTT

r r rT m a

0 ,g c cm g F

r rm a

climber c g c F T F c cm a

c ra a a

0 , nn 0 ,g r rm g F

0 ,c cTT

, 0r raa

0 ,c ca a

0rm g n

c c c cT m g m a

c rT T T

rT m a 0rm g n

c cT m g m a

rT m a 0rm g n

c cT m g m a

c

r c

ma g

m m

2709.8 /

940 70

kgm s

kg kg

20.679 /m s

20 0

1

2x x v t a t

0 51x x m

0 0v

02 x x

ta

2

2 51

0.679 /

m

m s 12 s

Tension

T = 1N throughout

5.3. Circular Motion

2nd law:2

net

vF m a m

r

Uniform circular motion

centripetal

Example 5.6. Engineering a Road

At what angle should a road with 200 m curve radius be banked for travel at 90 km/h (25 m/s)?

g m n F a

sin , cosn n

0 ,g m g F

2

, 0v

r

a

2

sinv

n mr

cos 0n m g

2

tanv

r g

2

2

25 /

200 9.8 /

m s

m m s

x

y

n

Fg

a

x : y :

0.318877... 0.32

17.74... 18

Example 5.7. Looping the Loop

Radius at top is 6.3 m.

What’s the minimum speed for a roller-coaster car to stay on track there?

g m n F a

0 , n n

0 ,g m g F

2

0 ,v

r

a

Minimum speed n = 0

2vn m g m

r

v g r 29.8 / 6.3m s m 7.9 /m s

Conceptual Example 5.1. Bad Hair Day

What’s wrong with this cartoon showing riders of a loop-the-loop roller coaster?

From Eg. 5.7: n m g = m a = m v2 / r( a g )

Consider hair as mass point connected to head by massless string.

Then T m g = m awhere T is tension on string.

Thus, T = m ( g a ) 0 . ( downward )

This means hair points upward( opposite to that shown in cartoon).

Frictional Forces

Pushing a trunk:

1.Nothing happens unless force is great enough.

2.Force can be reduced once trunk is going.

Static friction s sf n

s = coefficient of static friction

0v

Kinetic friction k kf n

k = coefficient of kinetic friction

0v

k s

k : < 0.01 (smooth), > 1.5 (rough)

Rubber on dry concrete : k = 0.8, s = 1.0

Waxed ski on dry snow: k = 0.04

Body-joint fluid: k = 0.003

Example 5.11. Dragging a Trunk

Mass of trunk is m. Rope is massless. Kinetic friction coefficient is k.

What rope tension is required to move trunk at constant speed?

g f m n F f T a

0 , nn

0 ,g m g F

0a

, 0f k n f

cos 0k n T sin 0n m g T

cos , sinT T

cosk

Tn

cos sin 0

k

Tm g T

cossin

k

m gT

cos sin

k

k

m g

x

y

T

Fg

fs

n

x : y :

Rolling wheel: X R

Skidding wheel 滑動的輪子 : kinetic friction 動摩擦

k 0.8

Rolling wheel 滾動的輪子 :static friction 靜摩擦

s 1

Rolling friction 滾動摩擦r 0.01

Dynamics of Wheels

F

fsfr

Example 5.8. Stopping a Car

k & s of a tire on dry road are 0.61 & 0.89, respectively.

If the car is travelling at 90 km/h (25 m/s),

(a) determine the minimum stopping distance.

(b) the stopping distance with the wheels fully locked (car skidding).

g f m n F f a

0 , nn 0 ,g m g F

, 0aa

, 0f n f

n m a 0n m g

na

m

g

2 20 02v v a x x

20

2

vx

a 0v

(a) = s : 20

2 s

vx

g

2

2

25 /

2 0.89 9.8 /

m s

m s 36 m

(b) = k : 20

2 k

vx

g

2

2

25 /

2 0.61 9.8 /

m s

m s 52 m

Steering

Bicycle turning to the left. Car turning to the left.

More details

Example 5.9. Steering

A level road makes a 90 turn with radius 73 m.

What’s the maximum speed for a car to negotiate this turn when the road is

(a) dry ( s = 0.88 ).

(b) covered with snow ( s = 0.21 ).

g f m n F f a

0 , nn 0 ,g m g F

2

, 0v

r

a

, 0f s nf

2

s

vn m

r 0n m g

s r nv

m

s r g

(a)

20.88 73 9.8 /v m m s 25 /m s 90 /km h

(b)

20.21 73 9.8 /v m m s 12 /m s 44 /km h

5.5. Drag Forces

Terminal speed: max speed of free falling object in fluid.

Drag force: frictional force on moving objects in fluid.

Depends on fluid density, object’s cross section area, & speed.

Parachute: vT ~ 5 m/s.

Ping-pong ball: vT ~ 10 m/s.

Golf ball: vT ~ 50 m/s.

Sky-diver varies falling speed by changing his cross-section.

Drag & Projectile Motion

d F v

Simple Machines