Upload
jack-mathews
View
215
Download
0
Embed Size (px)
Citation preview
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