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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
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