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NEWTON’S LAWS, PART 2 Pearland High School Physics

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Acknowledgements © 2013 Mark Lesmeister/Pearland ISD This work is licensed under the Creative Commons Attribution- ShareAlike 3.0 Unported License. To view a copy of this license, visit or send a letter to Creative Commons, 444 Castro Street, Suite 900, Mountain View, California, 94041, USA. Selected graphics and problems from OpenStax College. (2012, June 12). College Physics. Retrieved from the Connexions Web site: Selected questions from LearnAPPhysics.com, © 2009 Richard White Select problems from Serway and Faughn, Holt Physics, © 2002 Holt Rinehart Winston

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NEWTON’S 2 ND LAW

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Newton’s Second Law The acceleration experienced by an object is directly proportional to and in the same direction as the net force that acts on it, and inversely proportional to the mass of the object.

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Newton’s Second Law

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Newton’s Second Law in Component Form Force and acceleration are vectors, which can be broken into components. Newton’s Second Law can be applied in each component direction separately.

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Practice problem 1 Space shuttle astronauts experience accelerations of about 35 m/s 2 during takeoff. What magnitude of force does a 75 kg astronaut experience during an acceleration of this magnitude? (answer in the correct significant figures) Answer: 2600 N

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Practice problem 2 A 7.5 kg bowling ball initially at rest is dropped from the top of an 11 m building. It hits the ground 1.5 s later. Find the net force on the bowling ball, including direction and magnitude. Down is negative. Answer: -74 N

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Practice Problems 3 A 2.0 kg otter starts from rest at the top of a muddy incline 85 cm long and slides to the bottom in 0.50 s. What net external force acts on the otter?

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Mass and Weight Mass is the amount of stuff in an object Weight can be defined as the magnitude of the force of gravity acting on an object Weight = The weight always acts downwards, toward the center When the mass of an object and the acceleration due to gravity are known, the weight of an object can be calculated using the equation:

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Example: Free Fall Show that the acceleration of free-fall is the same as the factor for converting mass to weight. F G =mg

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BASIC PROBLEM SOLVING WITH NEWTON’S LAWS

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Constant Force Model vs. Equilibrium Model Equilibrium ∑F = 0. Object will be at rest or move with constant velocity. Position vs. time graph- Constant Force ∑ F = constant. Object will accelerate in the direction of the net force. Position vs. time graph- x t x t or x t

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Problem Solving Tips Often the problem will be equilibrium in one direction and constant acceleration in another. You may have to use the constant acceleration model either before or after finding the acceleration.

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Solving Equilibrium Problems Givens and Unknowns: Sketch and label a diagram of the object and its surroundings. Enclose your object in a boundary to help identify outside forces. Draw a free body diagram. If there is motion, choose one axis in the direction of motion. Identify all forces that act on the object, and draw them on the diagram. Model: Equilibrium Method Apply Newton’s 1 st Law in component form. F net = 0 so ΣF x = 0 and ΣF y =0

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Solving Constant Force (Acceleration) Problems Givens and Unknowns: Sketch and label a diagram of the object and its surroundings. Enclose your object in a boundary to help identify outside forces. Draw a free body diagram. If there is motion, choose one axis in the direction of motion. Identify all forces that act on the object, and draw them on the diagram. Model: Constant Force Method Apply Newton’s 1 st and 2 nd Laws in component form. F net = ma so ΣF x = ma x and ΣF y = ma y

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Figure 4.38 Braces are used to apply forces to teeth to realign them. Shown in this figure are the tensions applied by the wire to the protruding tooth. The total force applied to the tooth by the wire, F app, points straight toward the back of the mouth.

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Example 1: Equilibrium Find the tension in each rope, as a function of the angle. 5.0 kg 30 o

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Solution to Example 1 F g =mg F 2x =F 2 cos F 1 =? F 2y =F 2 sin F 1x =F 1 cos F 1y =F 1 sin

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Solution to Example 1, Continued Implement From the horizontal equation From the vertical equation F g =mg F 1 =? F 2x =F 2 cos F 2y =F 2 sin F 1x =F 1 cos F 1y =F 1 sin

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Example 2: Constant Force A truck is pulling a trailer with a force of 2200 N. Resistive forces opposed to the motion of the trailer total 1200 N. The trailer has a mass of 500 kg. How long will the trailer take to reach 20 m/s?

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Constant Force Example Givens and Unknown- as in diagram to the right, and m = 500 kg v 0 = 0 and v = 20 m/s t = ? Model: Equilibrium in y, constant force (so constant acceleration) in x. Method: F g =mg FNFN F Truck-Trailer = 2200 N F friction-Trailer = 1200 N

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Example 2 Solution Implement F g =mg FNFN F Truck-Trailer = 2200 N F friction-Trailer = 1200 N

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Additional Practice, Newton’s Laws From OpenSTAX College Physics: Normal, Tension and Other Forces, Exercises 2, 4 and 6. Problem Solving Strategies, Exercises 2,4, 10,

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NEWTON’S LAW LAB

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Force, Mass and Acceleration Lab Notes On the LabQuest unit, select “File”, “Open” and choose the file labeled “N2L”. In the trials where the force is varied, Carefully raise the ramp until the force is what you want it to be. Use the same cart each trial. In the trials where the mass is varied, The total mass includes the mass of the cart, fan and added mass.

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Warm-up Quiz Question 1: Force, Mass and Acceleration The graph on the right most likely shows A) Acceleration vs. mass. B) Acceleration vs. force

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Warm-up Quiz Question 2: Force, Mass and Acceleration The graph on the right most likely shows: A) Acceleration vs. mass. B) Acceleration vs. force

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Warm-up Quiz Question 3: Force, Mass and Acceleration Which of the following statements agrees with the results of the lab? A) Acceleration is directly proportional to force and mass. B) Acceleration is inversely proportional to force and mass. C) Acceleration is directly proportional to force and inversely proportional to mass. D) Acceleration is inversely proportional to force and directly proportional to mass.

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Acceleration and Net Force Acceleration is directly proportional to net force.

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Acceleration and Mass Acceleration is inversely proportional to mass.

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Combining Our Results The acceleration of an object is directly proportional to the net external force and inversely proportional to the object’s mass.

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Challenge Problem: p. 155, #63 Three blocks are in contact with each other on a frictionless horizontal surface. A horizontal force of 180N is applied to the right. Find the acceleration of the blocks. Find the net force on the 2 kg block. Find the force of the 2 kg block on the 3 kg block. 2 kg 3 kg 4 kg

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SPECIAL SITUATIONS WITH NEWTON’S LAWS

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Systems of Objects This device is called Atwood’s Machine. Problem: Find the acceleration in terms of the masses and g. m1m1 m2m2

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Systems of Objects: Practice Problem Find T 1 and T 2 in terms of m 1, m 2 and the acceleration a. m1m1 m2m2 T1T1 T2T2

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Accelerating Reference Frames The “apparent weight” is what the scale reads, i.e. the normal force. ©2012 OpenSTAX College

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Accelerating reference frames: Practice Problem The man has a mass of 80 kg. g = 10 m/s 2. What does the scale read when The elevator is moving up with a constant velocity of 30 m/s? The elevator is moving down with a constant velocity of 30 m/s? The elevator is moving up and gaining speed at 2 m/s 2 ? The elevator is moving up and slowing down from 10 m/s to 0 m/s in 2 s? ©2012 OpenSTAX College

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Additional Practice Systems of Objects: OpenSTAX College, Problem Solving Strategies, Exercises 5,6 and7 OpenSTAX College, Problem Solving Strategies, Exercises 5,6 and7 Accelerated Reference Frames OpenSTAX College, Further Applications of Newton’s Laws, Exercises 10 and 12 OpenSTAX College, Further Applications of Newton’s Laws, Exercises 10 and 12

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FRICTION

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Resistive Forces Resistive forces, such as friction and drag forces, are forces which oppose the relative motion of two surfaces, or a surface and a fluid. Friction occurs when two solid surfaces interact. Drag forces occur in fluids, i.e. liquids and gases.

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Friction Friction is really just one component of the force between two surfaces. Friction is the force of interaction parallel to the surfaces. The perpendicular component is the normal force. We generally treat friction and the normal force as separate forces. FNFN F

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Static and Kinetic Friction The amount of friction between an object and a surface depends on whether the object is moving relative to the surface. Static friction (F s ) applies when the object is at rest relative to the surface. Kinetic friction (F k ) applies when the object is moving relative to the surface.

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Static Friction The force is equal to the applied force, until it exceeds a certain maximum. Once the maximum force of static friction is exceeded, the object breaks free and begins to slide.

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Kinetic Friction Kinetic friction applies when the object is moving across the surface. The force of kinetic friction is typically less than the corresponding maximum force of static friction. In other words, once things are moving they are easier to keep moving. FNFN FkFk FgFg

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Accelerating a car. The force the road applies to accelerate a car (speed up, slow down or change direction) is a frictional force. If the wheels do not slip, the friction can be treated as static friction. If the wheels slip, kinetic friction is more appropriate. F This tire will accelerate to the right because of the force of friction applied by the road.

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Kinetic Friction and the Normal Force The force of kinetic friction is given by the formula F N is the normal force between the surfaces. μ K is the coefficient of kinetic friction, which depends on the nature of the surfaces in contact. FNFN FkFk FgFg

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Maximum Force of Static Friction The maximum force of static friction is given by μ s the coefficient of static friction μ s depends on the surfaces in contact. It is usually larger than the corresponding μ k. FNFN FsFs FgFg F App

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Practice Problem 1:

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Practice Problem 2:

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A 10 Kg block is at rest on a level surface. It accelerates from rest to 51.2 m/s in 8 seconds when you exert a horizontal force of 100 N on it. Calculate the coefficient of friction? Practice Problem 3:

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