1. B1.8 - Braking Chapter B1

2. Factors Affecting Braking Reaction distance is affected most by the person driving the car. Braking distance is affected by other factors: Weather Road conditions The vehicle – The average vehicle can slow to a stop at a deceleration rate of 5.85 m/s 2

3. Force of Friction A contact force between two surfaces that oppose the motion of one surface past the other. In the case of driving, the force of friction occurs between the tires and the road. The bigger the force of friction, the better the grip of the tires on the road and the faster the car will stop. Too small of a force of friction (e.g. on icy roads) & the car will likely slide through the intersection. Like all forces, friction is measured in newtons, & is a vector quantity.

4. Force of Friction At slow speeds, the friction between the tires and the road is enough to slow the car down without applying the brakes. At faster speeds, additional friction will occur between the brakes and the wheels.

5. Net Force The vector sum of all forces acting on an object In driving, the net force is the force moving the car forward, minus the forces of air resistance, road resistance and the force applied by the braking system Because we are dealing with vector quantities signs (+ or – ) of the forces are important Net force if +, the car is accelerating if 0, the car is traveling at a constant speed (coasting) if –, the car is slowing down Positive forces move the car forward e.g. the force of the engine on the wheels Opposing forces oppose the car’s motion e.g. brakes air resistance road resistance

6. The Effect of Mass The mass of a vehicle (a scalar quantity, measured in kg) also affects the rate of deceleration. E.g. A transport truck has a greater contact with the road than a motorbike.

7. Newton’s Second Law of Motion An object will accelerate in the direction of the net force In this case, the object will slow down as a result of the net force acting on the vehicle F = m a Example: A vehicle traveling at 12.5m/s and a mass of 1250kg brakes to stop at an intersection. Assume it has a typical deceleration of -5.85m/s 2. What is the net braking force acting on the vehicle? v i = 12.5 m/s (extraneous information) v f = 0 m/s (extraneous information) m = 1250 kg a = – 5.85 m/s 2 F = m a = (1250 kg)( -5.85 m/s 2 ) = -7312.5 N = -7.31x10 3 N

8. Practice Problem #3, page 227 A toboggan travels the level terrain at the bottom of a hill with a velocity of 6.0 m/s, east. The riders dig in their boots and bring the toboggan to rest in 3.0 s. The mass of the toboggan and the riders is 185 kg. a)Sketch a diagram illustrating the braking process. Be sure to add arrows to represent the vector quantities of velocity, acceleration, and braking force. a)Calculate the acceleration of the toboggan. a)Calculate the braking force on the toboggan.

9. Practice Problem #3, page 227 a)Sketch a diagram illustrating the braking process. Be sure to add arrows to represent the vector quantities of velocity, acceleration, and braking force. WEST  EAST – + v a F braking

10. Practice Problem #3, page 227 b)Calculate the acceleration of the toboggan. c)Calculate the braking force on the toboggan. b) v i = + 6.0 m/s [E] v f = 0 m/s ∆t = 3.0 s a = v f - v i = 0 – 6.0 m/s ∆t 3.0 s = -2.0 m/s 2 c) the braking force is the net force F = ma = (185kg)(-2.0m/s 2 ) = -370 kgm/s 2 = -3.7 x 10 2 N [W]