(Source: How Brakes Work! Mark Lam University of British Columbia
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1 (Source: How Brakes Work! Mark Lam University of British Columbia
2 (Source: Give me a place to stand on, and I will move the Earth. - Archimedes
3 Mechanical Advantage What are some situations where force must be amplified to carry out a task?
4 Mechanical Advantage What are some situations where force must be amplified to carry out a task? (Source:
5 Mechanical Advantage What are some situations where force must be amplified to carry out a task? (Source: (Source:
6 Mechanical Advantage What are some situations where force must be amplified to carry out a task? (Source: (Source: (Source:
7 Mechanical Advantage What are some situations where force must be amplified to carry out a task? (Source: (Source: (Source: (Source:
8 Mechanical Advantage
9 Mechanical Advantage Mechanical advantage is a measure of the the force amplification of a system
10 Mechanical Advantage Mechanical advantage is a measure of the the force amplification of a system MA = Fout/Fin
11 Example Using a pallet jack, I am able to raise 2000 lbs (~9000N) of goods with effort of 25N What is the Mechanical Advantage of the Pallet Jack? (Source:
12 Example (Source:
13 Example MA = Fout/Fin (Source:
14 Example MA = Fout/Fin Fout = 9000 N (Source:
15 Example MA = Fout/Fin Fout = 9000 N Fin = 25 N (Source:
16 Example MA = Fout/Fin Fout = 9000 N Fin = 25 N MA = 9000/25 = 360 (Source:
17 Mechanical Advantage in Cars
18 Mechanical Advantage in Cars We will discuss 2 ways in which force is amplified in order to stop a car:
19 Mechanical Advantage in Cars We will discuss 2 ways in which force is amplified in order to stop a car: 1) Leverage (which you already may already be familiar with)
20 Mechanical Advantage in Cars We will discuss 2 ways in which force is amplified in order to stop a car: 1) Leverage (which you already may already be familiar with) 2) Hydraulics (probably new to most of you)
21 Leverage Archimedes discovered that with the use of levers, one is able to amplify forces The Law of the Lever can be expressed as Findin = Foutdout
22 Leverage Findin = Foutdout Fout = Fin (din/dout) MA = din/dout
23 Example - Scissors (Source:
24 Example - Bolt Cutters (Source:
25 Example - Bolt Cutters (Source:
26 Leverage in Brakes The brake pedal and the master cylinder are connected the same rod When the force is applied to the brake pedal, that force is amplified and transmitted to the master cylinder. (Source:
27 Pressure What is the difference between Force and Pressure?
28 Pressure Pressure is the ratio of the Force to the Area over which the force is distributed P = F/A (P is pressure, F = force, A = area) Force tells you how an object will accelerate Pressure tells you how it will feel
29 (Source:
30 (Source:
31 Example Which is greater: the force of Zdeno Chara standing with running shoes or Katy Perry standing with high heels? (Source: (Source:
32 Example Which is greater: the pressure of Zdeno Chara standing with running shoes or Katy Perry standing with high heels? (Source: (Source:
33 Example (Source:
34 Example PChara = Fweight/Ashoes (Source:
35 Example PChara = Fweight/Ashoes m = 115kg => Fweight = 1150N (Source:
36 Example PChara = Fweight/Ashoes m = 115kg => Fweight = 1150N Ashoes = 2 x 500cm 2 = 0.1m 2 (Source:
37 Example PChara = Fweight/Ashoes m = 115kg => Fweight = 1150N Ashoes = 2 x 500cm 2 = 0.1m 2 PChara = 1150/0.1 = 11,500 Pa (Source:
38 Example (Source:
39 Example PKaty = Fweight/Ashoes (Source:
40 Example PKaty = Fweight/Ashoes m = 55kg => Fweight 550N (Source:
41 Example PKaty = Fweight/Ashoes m = 55kg => Fweight 550N Ashoes = 2 x 25cm 2 = 0.005m 2 (Source:
42 Example PKaty = Fweight/Ashoes m = 55kg => Fweight 550N Ashoes = 2 x 25cm 2 = 0.005m 2 PKaty = 550/0.005 = 110,000 Pa (Source:
43 Example PChara = 11,500 Pa PKaty = 110,000 Pa Who would you rather step on your foot? (Source: (Source:
44 Pressure in Liquids We can also talk about pressure in liquids In a body of water, the water pressure increases with depth because as you descend, the weight of water above you increases (Source:
45 Pressure in Liquids If we have a container with a piston on one end, we can apply pressure in the liquid by exerting a force on the piston
46 Pascal s Principal Pressure applied to an enclosed fluid is transmitted undiminished to every portion of the fluid and to the walls of the containing vessel. What does this mean? How can we apply this?
47 Hydraulic Systems Hydraulic systems are made of 3 main parts: an incompressible fluid a container pistons free to move at the ends of the container (Source:
48 Hydraulic Systems Hydraulic systems are made of 3 main parts: an incompressible fluid a container pistons free to move at the ends of the container (Source:
49 Example 5kg r=2m 100kg r=10m (Source: Consider the system on the above. What do you expect to happen?
50 Hydraulic Systems (Source:
51 Hydraulic Systems (Source:
52 Hydraulic Systems Pout = Pin (Source:
53 Hydraulic Systems Pout = Pin Fout/Aout = Fin/Ain (Source:
54 Hydraulic Systems Pout = Pin Fout/Aout = Fin/Ain Fout = Fin (Aout/Ain) (Source:
55 Hydraulic Systems Pout = Pin Fout/Aout = Fin/Ain Fout = Fin (Aout/Ain) MA = Aout/Ain (Source:
56 Example 5kg r=2m 100kg r=10m (Source:
57 Example MAhydraulics = Aout/Ain 5kg 100kg r=2m r=10m (Source:
58 Example MAhydraulics = Aout/Ain 5kg 100kg r=2m r=10m A=100π m 2 (Source:
59 Example MAhydraulics = Aout/Ain 5kg 100kg r=2m r=10m A=4π m 2 A=100π m 2 (Source:
60 Example MAhydraulics = Aout/Ain MA = 100π/4π = 25 5kg 100kg r=2m r=10m A=4π m 2 A=100π m 2 (Source:
61 Example MAhydraulics = Aout/Ain MA = 100π/4π = 25 MA = Fout/Fin 5kg 100kg r=2m r=10m A=4π m 2 A=100π m 2 (Source:
62 Example MAhydraulics = Aout/Ain MA = 100π/4π = 25 MA = Fout/Fin 5kg 100kg r=2m r=10m A=4π m 2 A=100π m 2 Fout = MA x Fin (Source:
63 Example MAhydraulics = Aout/Ain MA = 100π/4π = 25 MA = Fout/Fin 5kg=>50N 100kg r=2m r=10m A=4π m 2 A=100π m 2 Fout = MA x Fin (Source:
64 Example MAhydraulics = Aout/Ain MA = 100π/4π = 25 MA = Fout/Fin 5kg=>50N 100kg r=2m r=10m A=4π m 2 A=100π m 2 Fout = MA x Fin Fout = 25 x 50 = 1250N (Source:
65 Example MAhydraulics = Aout/Ain MA = 100π/4π = 25 MA = Fout/Fin 5kg=>50N 100kg=>1000N r=2m r=10m A=4π m 2 A=100π m 2 Fout = MA x Fin Fout = 25 x 50 = 1250N (Source:
66 Example MAhydraulics = Aout/Ain MA = 100π/4π = 25 MA = Fout/Fin 5kg=>50N 100kg=>1000N r=2m r=10m A=4π m 2 A=100π m 2 Fout = MA x Fin Fout = 25 x 50 = 1250N 1250N is enough to lift a mass of ~125kg. The 5kg mass will lower and the 100kg mass will rise (Source:
67 Example MAhydraulics = Aout/Ain MA = 100π/4π = 25 MA = Fout/Fin 5kg=>50N 100kg=>1000N r=2m r=10m A=4π m 2 A=100π m 2 Fout = MA x Fin Fout = 25 x 50 = 1250N 1250N is enough to lift a mass of ~125kg. The 5kg mass will lower and the 100kg mass will rise (Source:
68 Hydraulic Systems (Source:
69 Hydraulic Systems In hydraulic systems, there is often a trade-off of distance for force (Source:
70 Hydraulic Systems In hydraulic systems, there is often a trade-off of distance for force To the right, the area of the input piston is 1π in 2 and the area of the output piston is 9π in 2 (9x larger) (Source:
71 Hydraulic Systems In hydraulic systems, there is often a trade-off of distance for force To the right, the area of the input piston is 1π in 2 and the area of the output piston is 9π in 2 (9x larger) Since the fluid is incompressible, the input piston must move 9 in for the output piston to move 1 in (Source:
72 Hydraulic Systems In hydraulic systems, there is often a trade-off of distance for force To the right, the area of the input piston is 1π in 2 and the area of the output piston is 9π in 2 (9x larger) Since the fluid is incompressible, the input piston must move 9 in for the output piston to move 1 in (Source:
73 Hydraulic Systems (Source:
74 Hydraulic Systems Knowing this, we are able to lift heavy objects like cars (Source:
75 Hydraulic Systems Knowing this, we are able to lift heavy objects like cars Not only can we lift cars - we are able to stop cars as well (Source:
76 Hydraulics in Brakes
77 Hydraulics in Brakes When the brake pedal is pressed, the amplified force is transmitted to the master cylinder
78 Hydraulics in Brakes When the brake pedal is pressed, the amplified force is transmitted to the master cylinder The master cylinder then translates this force into hydraulic fluid pressure.
79 Hydraulics in Brakes When the brake pedal is pressed, the amplified force is transmitted to the master cylinder The master cylinder then translates this force into hydraulic fluid pressure. The pressure is transmitted to the slave pistons which come out to apply the brake pads on the rotor of each wheel
80 Hydraulics in Brakes When the brake pedal is pressed, the amplified force is transmitted to the master cylinder The master cylinder then translates this force into hydraulic fluid pressure. The pressure is transmitted to the slave pistons which come out to apply the brake pads on the rotor of each wheel By friction, the kinetic energy is transformed into thermal energy as the car slows down
81 Hydraulic Braking System (Source:
82 The Caliper (Source: (Source:
83 Example
84 Example Consider a the following braking system: The distance from the brake pedal to the pivot is 12 inches The distance from the master cylinder to the pivot is 2 inches The master cylinder has a radius of 1 inch The master cylinder is connected to 8 slave cylinders (2 per wheel) Each slave cylinder has a radius of 2 inches
85 Example Consider a the following braking system: The distance from the brake pedal to the pivot is 12 inches The distance from the master cylinder to the pivot is 2 inches The master cylinder has a radius of 1 inch The master cylinder is connected to 8 slave cylinders (2 per wheel) Each slave cylinder has a radius of 2 inches What is the mechanical advantage of this system?
86 Example Consider a the following braking system: The distance from the brake pedal to the pivot is 12 inches The distance from the master cylinder to the pivot is 2 inches The master cylinder has a radius of 1 inch The master cylinder is connected to 8 slave cylinders (2 per wheel) Each slave cylinder has a radius of 2 inches What is the mechanical advantage of this system? If I apply 100N of force on the brake pedal, what is the total force applied to the rotors?
87 Example
88 Example MAlever = dpedal/dmaster cylinder
89 Example MAlever = dpedal/dmaster cylinder = 12/2
90 Example MAlever = dpedal/dmaster cylinder = 12/2 = 6
91 Example MAlever = dpedal/dmaster cylinder = 12/2 = 6 MAhydraulics = Aslave cylinders/amaster cylinder
92 Example MAlever = dpedal/dmaster cylinder = 12/2 = 6 MAhydraulics = Aslave cylinders/amaster cylinder = 8 x 2 2 π/1 2 π
93 Example MAlever = dpedal/dmaster cylinder = 12/2 = 6 MAhydraulics = Aslave cylinders/amaster cylinder = 8 x 2 2 π/1 2 π = 32
94 Example MAlever = dpedal/dmaster cylinder = 12/2 = 6 MAhydraulics = Aslave cylinders/amaster cylinder = 8 x 2 2 π/1 2 π = 32 MAtotal = MAlever x MAhydraulics
95 Example MAlever = dpedal/dmaster cylinder = 12/2 = 6 MAhydraulics = Aslave cylinders/amaster cylinder = 8 x 2 2 π/1 2 π = 32 MAtotal = MAlever x MAhydraulics = 6 x 32
96 Example MAlever = dpedal/dmaster cylinder = 12/2 = 6 MAhydraulics = Aslave cylinders/amaster cylinder = 8 x 2 2 π/1 2 π = 32 MAtotal = MAlever x MAhydraulics = 6 x 32 = 192
97 Example MAlever = dpedal/dmaster cylinder = 12/2 = 6 MAhydraulics = Aslave cylinders/amaster cylinder = 8 x 2 2 π/1 2 π = 32 MAtotal = MAlever x MAhydraulics = 6 x 32 = 192
98 Example MAlever = dpedal/dmaster cylinder = 12/2 = 6 MAhydraulics = Aslave cylinders/amaster cylinder = 8 x 2 2 π/1 2 π = 32 MAtotal = MAlever x MAhydraulics = 6 x 32 = 192 MA = Fout/Fin
99 Example MAlever = dpedal/dmaster cylinder = 12/2 = 6 MAhydraulics = Aslave cylinders/amaster cylinder = 8 x 2 2 π/1 2 π = 32 MAtotal = MAlever x MAhydraulics = 6 x 32 = 192 MA = Fout/Fin Ftotal = MAtotal x Fbrake pedal
100 Example MAlever = dpedal/dmaster cylinder = 12/2 = 6 MAhydraulics = Aslave cylinders/amaster cylinder = 8 x 2 2 π/1 2 π = 32 MAtotal = MAlever x MAhydraulics = 6 x 32 = 192 MA = Fout/Fin Ftotal = MAtotal x Fbrake pedal = 192 x 100
101 Example MAlever = dpedal/dmaster cylinder = 12/2 = 6 MAhydraulics = Aslave cylinders/amaster cylinder = 8 x 2 2 π/1 2 π = 32 MAtotal = MAlever x MAhydraulics = 6 x 32 = 192 MA = Fout/Fin Ftotal = MAtotal x Fbrake pedal = 192 x 100 = 19,200 N
102 Example MAlever = dpedal/dmaster cylinder = 12/2 = 6 MAhydraulics = Aslave cylinders/amaster cylinder = 8 x 2 2 π/1 2 π = 32 MAtotal = MAlever x MAhydraulics = 6 x 32 = 192 MA = Fout/Fin Ftotal = MAtotal x Fbrake pedal = 192 x 100 = 19,200 N Enough force to lift the car!
103 Demonstration
104 Demonstration
105 Demonstration What is the mechanical advantage of this system?
106 Demonstration What is the mechanical advantage of this system? MA = Aslave cylinders/amaster cylinder
107 Demonstration What is the mechanical advantage of this system? MA = Aslave cylinders/amaster cylinder MA = (2 x Asyringe)/Asyringe
108 Demonstration What is the mechanical advantage of this system? MA = Aslave cylinders/amaster cylinder MA = (2 x Asyringe)/Asyringe MA= 2
109 Demonstration What is the mechanical advantage of this system? MA = Aslave cylinders/amaster cylinder MA = (2 x Asyringe)/Asyringe MA= 2 What changes could be made to make this fit to stop a car?
110 Summary
111 Summary Mechanical Advantage is a measure of the force amplification of a system
112 Summary Mechanical Advantage is a measure of the force amplification of a system Brakes combine 2 forms of mechanical advantage to multiply the force applied by your foot:
113 Summary Mechanical Advantage is a measure of the force amplification of a system Brakes combine 2 forms of mechanical advantage to multiply the force applied by your foot: Leverage
114 Summary Mechanical Advantage is a measure of the force amplification of a system Brakes combine 2 forms of mechanical advantage to multiply the force applied by your foot: Leverage Hydraulics
115 Summary Mechanical Advantage is a measure of the force amplification of a system Brakes combine 2 forms of mechanical advantage to multiply the force applied by your foot: Leverage Hydraulics With these, we can multiply our force over 200x
116 Leaks (Source:
117 Leaks (Source:
118 Power Brakes (Source:
119 Power Brakes (Source:
120 Anti-Lock Brakes (ABS) Skidding wheels have less traction than non-skidding wheels This means that you have more control of your vehicle when braking if your wheels are still rotating To keep your wheels from locking up, pressure is released when necessary The result is a pulsing that keeps the wheel rotating as the car is moving
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