# 5. Newton's Laws Applications

## Presentation on theme: "5. Newton's Laws Applications"— Presentation transcript:

5. Newton's Laws Applications
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 The skier’s acceleration The force the snow exerts on him. y x : n a y : x Fg

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

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

Alternative Approach Net force along slope (x-direction) : y n   Fh
Fg

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

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.

Tension T = 1N throughout

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

5.3. Circular Motion Uniform circular motion 2nd law: 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. x : y : y T a x Fg

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

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

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 ).

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

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

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 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 :

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 ). (a) (b)

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

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

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.

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