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Principles of Technology Waxahachie High School Resistancein Mechanical Systems PIC Chapter 4.1 Resistancein Mechanical Systems PIC Chapter 4.1 PT TEKS.

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Presentation on theme: "Principles of Technology Waxahachie High School Resistancein Mechanical Systems PIC Chapter 4.1 Resistancein Mechanical Systems PIC Chapter 4.1 PT TEKS."— Presentation transcript:

1 Principles of Technology Waxahachie High School Resistancein Mechanical Systems PIC Chapter 4.1 Resistancein Mechanical Systems PIC Chapter 4.1 PT TEKS

2 Resistance in Mechanical Systems : Objectives:   State Newton’s second law of motion and use it to solve problems involving force, mass, and acceleration.   Calculate an object’s weight, given its mass.   Explain the difference between static and kinetic friction.   Use the linear model to calculate the force of friction between two surfaces.   Explain how lubrication and rolling reduce friction. : Objectives:   State Newton’s second law of motion and use it to solve problems involving force, mass, and acceleration.   Calculate an object’s weight, given its mass.   Explain the difference between static and kinetic friction.   Use the linear model to calculate the force of friction between two surfaces.   Explain how lubrication and rolling reduce friction.

3 Newton’s 1st law - Every object will remain at rest, or will continue to move in a straight line with constant speed unless the object is acted on by a net force. Resistance in Mechanical Systems

4 A book sitting on a table is in equilibrium. The weight of the book is the force of gravity acting downward. The table pushes upward on the book with an equal and opposite force. The vector sum of the two forces is zero and it remains at rest. Resistance in Mechanical Systems

5 If you push and exert a net force in a horizontal direction from left to right the book will begin to move, or be accelerated, in the direction of the net force. If you push it with twice the force the acceleration will double, triple the force, acceleration will triple. Resistance in Mechanical Systems

6 In other words, the acceleration of the book is directly proportional to the force acting on the book. Resistance in Mechanical Systems

7 Suppose you use the same net force to accelerate more than one book. For an equal force, two books are accelerated at one-half rate of one book. Three books are accelerated at one-third the rate, and so on. The acceleration of the books is inversely proportional to the mass. Resistance in Mechanical Systems

8 The relationship between acceleration, net force, and mass is: Newton’s 2nd law - The acceleration of an object is directly proportional to the net force acting on the object and inversely proportional to the mass of the object. Resistance in Mechanical Systems

9 Newton’s 2nd law equation = Net force = mass x acceleration F net = m x a Units of Force = Newtons (N) 1N = (1 kg) (1 m/s 2 ) = 1kg. m/s 2 Resistance in Mechanical Systems

10 A 1200 kg race car accelerates 0 to 100 m/s in 5 seconds. What is the average force on the car? First find the Acceleration = (v f - v i ) / t a ave = (100 m/s – 0 m/s) / 5 s a ave = 100 m/s / 5 s = 20 m/s 2 F ave = m x a ave F ave = 1200 kg x 20 m/s 2 F ave = 24,000 kg. m/s 2 or N Resistance in Mechanical Systems

11 You can use Newton's second law to relate an object's weight to its mass. In this case, the acceleration is that experienced by the object in the Earth's gravitational field. We use the symbol Fg to represent gravitational force, or weight. Resistance in Mechanical Systems

12 You have experienced gravitational acceleration if you have ever stepped off a diving board. While standing on the board, your weight Fg acting downward is balanced by an upward force exerted on you by the board. Resistance in Mechanical Systems

13 When you step off, the upward force is removed and a net force Fg is acting downward. You accelerate in the direction of this force, toward the surface of the pool. Resistance in Mechanical Systems

14 When an object is in a gravitational field and gravity is the only force acting on the object, it accelerates in the direction of the force. This acceleration is called gravitational acceleration (g). Force of gravity = F g Force of gravity = mass x gravity F g = m x g Resistance in Mechanical Systems

15 A 65 kg person stands on a scale, what is the force of gravity on the person? F g = m x g g = 9.8 m/s 2 (gravitational acceleration) F g = 65 kg x 9.8 m/s 2 F g = 637 kg. m/s 2 or N Resistance in Mechanical Systems

16 When you push an object across a surface, the opposing force is called friction. To accelerate an object, the force must be greater than the frictional force. The force of friction resists movement. Resistance in Mechanical Systems

17 Friction is a result of irregularities in the surfaces of objects. Resistance in Mechanical Systems

18 The force required to overcome friction is called the static friction force. The force needed to keep a constant speed is called the kinetic friction force. Resistance in Mechanical Systems

19 Imagine your car broke down and you have to push it. Which takes the most force? To get it started rolling or to keep it rolling? Resistance in Mechanical Systems

20 To get it started Static friction is greater than kinetic friction. Resistance in Mechanical Systems

21 Friction depends on 3 things: The friction force depends on whether or not the surfaces are moving The friction force depends on the materials of which the surfaces are made of. Resistance in Mechanical Systems

22 3. 3. The friction force depends on how hard the surfaces are pressed together. This is called the Normal Force between the surfaces. Friction opposes motion. In other words, the direction of the friction force is opposite the direction of the applied force. Resistance in Mechanical Systems

23 The magnitude of the friction force is proportional to the normal force (N). The constant of proportionality is called the coefficient of friction (μ, the Greek letter mu). The coefficient of static friction is μ s and the coefficient of kinetic friction is μ k. Resistance in Mechanical Systems

24 Coefficients of Friction Equations Static force = Coefficient of static x Normal Force (Weight) F s = μ s x N Kinetic Force = Coefficient of kinetic x Normal Force (Weight) F k = μ k x N Resistance in Mechanical Systems

25 The crate below is made of wood and weighs 85 lb. To slide the crate across a concrete floor: What force is required to get it moving? What force is required to keep the crate moving? Resistance in Mechanical Systems

26 A crate made of wood weighs 85 lb is on a concrete floor. What force is required to get it moving? F s = μ s x N μ s =.6 F s =.6 x 85 lb = 51 lb To get the crate moving, the applied force must exceed 51 lb. Resistance in Mechanical Systems

27 The same crate is now moving, what force is needed to keep it moving at a constant speed? F k = μ k x N μ k =.4 F k =.4 x 85 lb = 34 lb To keep the crate moving at a constant speed, the applied force must equal 34 lb. Resistance in Mechanical Systems

28 Lubricants reduce friction by keeping the two sliding surfaces apart with a thin layer of fluid. Friction is no longer rubbing the irregularities of the surfaces, but instead rubbing the lubricant. Resistance in Mechanical Systems

29 For example, in a car engine, a piston ring and a cylinder wall do not touch each other when moving because they are lubricated with oil. Resistance in Mechanical Systems

30 Another way to reduce friction is to roll an object over a surface, instead of sliding it. This is called rolling frictional force instead of kinetic frictional force. Resistance in Mechanical Systems

31 The force of rolling friction is much less than the force of sliding friction because there is less movement between surfaces. The surfaces don’t scrape past one another as they do in sliding friction. Resistance in Mechanical Systems


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