Transcript 05._UsingNewtonsLaws

5. Newton's Laws Applications
1.
2.
3.
4.
5.
Using Newton’s 2nd Law
Multiple Objects
Circular Motion
Friction
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.
Fnet  n  Fg  m a
n   0 , ny 

nx  Fg x  m ax
ny  Fg y  m a y
a   ax , 0

Fg  m g  sin  ,  cos  
y
x:
2
a  ax   9.8 m / s 2  sin  32   5.2 m / s
n
a
y:


Fg
m g sin   m ax
x
n y  m g cos   0
n  n y   65 kg   9.8 m / s 2  cos  32   540 N
Example 5.2. Bear Precautions
Mass of pack in figure is 17 kg.
What is the tension on each rope?
Fnet  T1  T2  Fg  m a  0
T1  T1  cos , sin  
T2  T2   cos , sin  
x : T1 cos   T2 cos   0
y
T2
y : T1 sin   T2 sin   m g  0
T1


x
Fg
T
17 kg   9.8 m / s 2 
2 sin 22
a0
since
Fg   0 , m g 

 220 N
T1  T2  T
T
mg
2 sin 
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?
Fnet  Fh  n  Fg
ma0
Fh    Fh , 0 
n  n  sin  , cos  
y
n


x:
 Fh  n sin   0
y:
n cos   m g  0
x
Fh
Fh 
Fg
a0
since
Fg   0 , m g 

Fh  n sin 
n
mg
cos 
mg
2
sin    60 kg   9.8 m / s  tan 30  340 N
cos 
Alternative Approach
Net force along slope (x-direction) :
 Fh cos   Fg sin   0
Fh  Fg tan 
y
n
  60 kg   9.8 m / s 2  tan 30  340 N


Fh
x
Fg
 m g sin 
GOT IT? 5.1.
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?
n

x:

T cos   mg sin   0
T  mg tan 
T
Smaller   smaller T
Fg
x
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.
Frock  Tr  Fg r  n  mr a r
ac  ar  a
Fclimber  Tc  Fg c  mc ac
Tc  Tr  T
Tr  Tr , 0
Fg r   0 ,  mr g 
Tc   0 , Tc 
Fg c   0 , mc g 
Tr  mr ar
mr g  n  0
Tc  mc g  mc ac
n  0 , n
ar   ar , 0
ac   0 ,  ac 

T  mr a
mr g  n  0
T  mc g  mc a
T  mr a
mr g  n  0

T  mc g  mc a
a

mc
g
mr  mc
70 kg
9.8 m / s 2 

940 kg  70 kg
 0.679 m / s 2
x  x0  v0 t 
x  x0  51 m
1 2
at
2

t
2  x  x0 

a
v0  0
Tension
T = 1N throughout
2  51 m 
0.679 m / s 2
 12 s
GOT IT? 5.1.
What are
1N
(a) the rope tension and
1N
(b) the force exerted by the hook on the rope?
5.3. Circular Motion
Uniform circular motion
2nd
law:
v2
Fnet  m a  m
r
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.
T  Fg  m a
T  T  cos , sin  
a   a , 0
Fg   0 ,  m g 
x:
T cos   m a
y : T sin   m g  0
y

T
T
T
a  cos   g cot 
m

a
v2

r
x
v
Fg
mg
sin 
ar 
g L cot  cos
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)?
n  Fg  m a
n  n  sin  , cos 
Fg   0 ,  m g 
y
v2
x : n sin   m
r
n

y : n cos   m g  0
25 m / s 

v2

tan  
r g  200 m   9.8 m / s 2 
2

a
Fg
 v2 
a   , 0
 r

x
 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?
n  Fg  m a
n  0 ,  n
Fg   0 ,  m g 

v2 
a  0 , 

r 

v2
n  m g  m
r
Minimum speed  n = 0
v
gr 
9.8 m / s   6.3 m
2
 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
Consider hair as mass point connected to head by
massless string.
Then
T+mg=ma
where T is tension on string.
Thus,
T = n.
Since n is downward, so is T.
This means hair points upward
( opposite to that shown in cartoon ).
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
f s  s n
v0
s = coefficient of static friction
Kinetic friction
f k  k n
v0
k = coefficient of kinetic friction
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.
foot pushes
ground
ground
pushes you
No slippage:
Contact point is momentarily at rest
 static friction at work
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).
a  a , 0
n  Fg  f f  m a
n  0 , n
Fg   0 ,  m g 
 n  m a
a
v  v  2 a  x  x0 
2
v0
2
0
v02
 x  
2a
n
m
(a)  = s :
(b)  = k :

f f    n , 0

nm g 0
  g
x 
x 
v02
 25 m / s 

2  0.89   9.8 m / s 2 
 36 m
v02
 25 m / s 

2  0.61  9.8 m / s 2 
 52 m
2s g
2k g
2
2
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 ).
 v2

a ,0 
 r

n  Fg  f f  m a
n  0 , n
Fg   0 ,  m g 
v2
s n  m
r
v
s r n
m
f f   s n , 0

nm g 0

s r g
(a)
v
 0.88  73 m 9.8 m / s 2 
 25 m / s  90 km / h
(b)
v
 0.21  73 m 9.8 m / s 2 
 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?
n  Fg  f f  m a
n  0 , n
a0
f f    s n , 0
Fg  m g  sin  ,  cos 
y
n
x:
fs
m g sin   s n  0
y:
n  m g cos   0
tan   s

Fg


x
  tan 1 s  tan 1 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?
n  Fg  f f  T  m a
a0
n  0 , n
f f    k n , 0
Fg   0 ,  m g 
y
x:
n
T
fs

x
Fg
T  T  cos , sin 
k n  T cos   0
n
T
T
k
mg
cos 
k
 sin 
y:
T
cos 
k



n  m g  T sin   0
cos   m g  T sin   0
m g k
cos    k sin 
GOT IT? 5.4
Is the frictional force
(a) less than, (b) equal to , or (c) greater than
the weight multiplied by the coefficient of friction?
Reason: Chain is pulling downward, thus increasing n.
5.5. Drag Forces
Drag force: frictional force on moving objects in fluid.
Depends on fluid density, object’s cross section area, & speed.
Terminal speed: max speed of free falling object in fluid.
Parachute: vT ~ 5 m/s.
Ping-pong ball: vT ~ 10 m/s.
Golf ball: vT ~ 50 m/s.
Drag & Projectile Motion
Sky-diver varies falling speed by
changing his cross-section.