5. Newton's Laws Applications

1. Using Newton’s 2nd Law

2. Multiple Objects

3. Circular Motion

4. Friction

5. Drag Forces

Why doesn’t the roller coaster fall its loop-the loop track?

Ans. The downward net force is just enough to make it move in a circular path.

5.1. Using Newton’s 2nd Law

Example 5.1. Skiing

A skier of mass m = 65 kg glides down a frictionless slope of angle = 32. Find

(a)The skier’s acceleration

(b) the force the snow exerts on him.

net g m F n F ax g x x

y g y y

n F m a

n F m a

0 , ynn

sin , cosg m g F

, 0xaa

sinxa a g

sinyn n m g

29.8 / sin 32m s 25.2 /m s

265 9.8 / cos 32kg m s 540 N

Example 5.2. Bear Precautions

Mass of pack in figure is 17 kg.

What is the tension on each rope?

0m a1 2net g F T T F 0asince

1 1 cos , sinT T

1 2T T T

2 2 cos , sinT T

2 sin

m gT

1 2cos cos 0T T

0 ,g m g F

1 2sin sin 0T T m g

217 9.8 /

2 sin 22

kg m sT

220 N

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

A roofer’s toolbox rests on a frictionless 45 ° roof,

secured by a horizontal rope.

Is the rope tension

(a)greater than,

(b)less than, or

(c)equal to

the box’s weight?

GOT IT? 5.1.

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

r c cm a m g m a

c

r c

ma g

m m

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

What are

(a)the rope tension and

(b)the force exerted by the hook on the rope?

1N

1N

GOT IT? 5.1.

5.3. Circular Motion

2nd law:2

net

vF m a m

r

Uniform circular motion

centripetal

Example 5.5. Whirling a Ball on a String

Mass of ball is m. String is massless.

Find the ball’s speed & the string tension.

g m T F a

cos , sinT T

0 ,g m g F

cosT m a

, 0aa

sin 0T m g

sin

m gT

cosT

am

cotg

v a r cot cosg L

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

0.3189

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

5.4. Friction

Some 20% of fuel is used to overcome friction inside an engine.

The Nature of Friction

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

Application of Friction

Walking & driving require static friction.

No slippage:

Contact point is momentarily at rest

static friction at work

foot pushes ground

ground pushes you

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

Application: Antilock Braking Systems (ABS)

Skidding wheel:kinetic friction

Rolling wheel:static friction

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 nvm

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

Example 5.10. Avalanche!

Storm dumps new snow on ski slope.

s between new & old snow is 0.46.

What’s the maximum slope angle to which the new snow can adhere?

g f m n F f a

0 , nn

sin , cosg m g F

0a

, 0f s n f

sin 0sm g n cos 0n m g

tan s

1tan s 1tan 0.46 25

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

Is the frictional force

(a)less than, (b) equal to , or (c) greater than

the weight multiplied by the coefficient of friction?

GOT IT? 5.4

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.

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

Drag & Projectile Motion