1. 5. Newton's Laws Applications
Using Newton’s 2nd Law
Multiple Objects
Circular Motion
Friction
Drag Forces

2. 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.

3. 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
The skier’s acceleration
The force the snow exerts on him.  y
x : n a
y :  x  Fg

4. Example 5.2. Bear Precautions
Mass of pack in figure is 17 kg.
What is the tension on each rope?
since
x :  y
y : T2 T1   x Fg

5. 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?
since y n
x :  
y :  x Fh Fg

6. Alternative Approach Net force along slope (x-direction) : y n   FhFg

7. GOT IT? 5.1. A roofer’s toolbox rests on a frictionless 45 ° roof,
secured by a horizontal rope.
Is the rope tension
greater than,
less than, or
equal to
the box’s weight? n
x :   T
Smaller   smaller T Fg x

8. 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.
What’s his acceleration?
How much time does he have before the rock goes over the edge?
Neglect mass of the rope. 

9.  
Tension
T = 1N throughout

10. GOT IT? 5.1. What are the rope tension and
the force exerted by the hook on the rope? 1N 1N

11. 5.3. Circular Motion
Uniform circular motion
2nd law:
centripetal

12. 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.
x :
y : y  T  a x Fg

13. 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)? y
x :
y : n   x a Fg

14. 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?
Minimum speed  n = 0

15. 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 + m g = m a
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 ).

16. 5.4. Friction The Nature of Friction
Some 20% of fuel is used to overcome friction inside an engine.
The Nature of Friction

17. Frictional Forces Pushing a trunk:
Nothing happens unless force is great enough.
Force can be reduced once trunk is going.
Static friction
s = coefficient of static friction
Kinetic friction
k = coefficient of kinetic friction
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

18. 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

19. Example 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),
determine the minimum stopping distance.
the stopping distance with the wheels fully locked (car skidding).
(a)  = s : 
(b)  = k :

20. Application: Antilock Braking Systems (ABS)
Skidding wheel:
kinetic friction
Rolling wheel:
static friction

21. 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 ).
(a)
(b)

22. 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? y n
x :
y : fs   x Fg

23. 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? y
y :
x : n T fs  x Fg

24. GOT IT? 5.4 Reason: Chain is pulling downward, thus increasing n.
Is the frictional force
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.

25. 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.
Sky-diver varies falling speed by changing his cross-section.
Drag & Projectile Motion