Newton s 3 Laws of Motion - PDF Free Download

Newton s 3 Laws of Motion 1. If F = 0 No change in motion 2. = ma Change in motion Fnet 3. F = F 1 on 2 2 on 1

Newton s First Law (Law of Inertia) An object will remain at rest or in a constant state of motion unless acted upon by external net forces.

Statics Problem Find the tensions in the wires. +y +x T 1 43 55 T 2 W

Newton s 2nd Law F = ma a = F net m The acceleration of object is directly related to the net forces acting on it and inversely proportional to its mass.

Problem Starting from rest, Sally pulls Billy on the sled (total mass = 60kg) with a total force of 100 N at an angle of 40 degrees above the horizontal, as shown. After 5 seconds, how fast is the sled moving and how far has it traveled from where it started? F x = ma x F cos θ = ma x F cosθ ax = v f = v0 + at m Fcosθ 100Ncos 40 v f = v0 + at = t = 5 s = 6.38 m / s m 60kg

Frictional Forces Friction always opposes the applied force and is in the opposite direction of motion. The greater the normal force the greater the frictional force. f s = µ N s f k = µ k N

Statice vs Kinetic Friction f s > f k Fig. 5.16, p.131

Problem The magnitude of F 1 is 75.0N and F 2 is 50.0N. The coefficient of friction between the block and the floor is 0.04. What is the acceleration of the block?

a) First of all, which way is the box going to go? You have to figure that out first. To do that compare the x-component of F2 to the x-component of F1. Which is larger? F = F cosθ = 75N cos 65 = 31.7N < 50N = F 1x 1 2 So the box is going to end up moving to the left!!! So I m going to make LEFT the positive x direction in my FBD. Also, since F1 is pushing down, the normal force is going to be larger than the weight!!! F2 should be longer than F1x too. Apply Newton s Second Law: a x Fx F2 F1x 50N 31.7N = = = = 3.66 m/ s m m 5kg b) Now there is friction which will slow it down. You have to know which direction the box is going without friction first because friction ALWAYS acts in the opposite direction to slow it down. You can draw the friction vector at the end of F1x or from the center, your choice. The frictional force is proportional to the normal force. To find that, apply Newton s second law in the y direction: F = N = mg + F = kg m s + N = f = N = N = N y 2 0 1y (5 )(9.8) / 75 sin 65 117 N, µ 0.04(117 ) 4.68 2 a x Fx F2 F1x f 50N 31.7N 4.68 = = = = 2.72 m/ s m m 5kg 2

At an instant when a 4.0-kg object has an acceleration equal to (5i + 3j) m/s2, one of the two forces acting on the object is known to be (12i + 22j) N. Determine the magnitude of the other force acting on the object. a. 2.0 N b. 13 N c. 18 N d. 1.7 N e. 20 N

Newton s 3rd Law F hand on wall = F wall on hand To every force there is an equal but opposite reaction force.

Newton s 3rd Law F hand on wall = F wall on hand You can t TOUCH without being TOUCHED back!!

Newton s 3rd Law F hand on wall = F wall on hand To every force there is an equal but opposite reaction force.

Newton s 3rd Law F hand on wall = F wall on hand This is an INTERACTIVE Universe.

Gravity is an Interaction F Earth on Rock = F Rock on Earth

Gravity is an Interaction The Earth pulls on you, you pull on the Earth. You fall to the Earth, the Earth Falls to you. You accelerate towards the Earth with g =9.8m/s 2. With what acceleration is the Earth falling towards you? F This is your weight: Earth on You mg a = F = M a = mg You on Earth E M E (65 kg)(9.8 m / s 2 ) a = = 1.1x10 22 m/ s 2 E 5.98x10 24 kg E E

Force is not Acceleration Force is the Same! Acceleration is NOT! F Earth on You = F You on Earth a Earth to You = a You to Earth

An interaction requires a pair of forces acting on two objects. kick Gun Pushes Bullet out. Bullet Pushes back on Gun (& Man)

Action Reaction Pairs kick Gun Pushes Bullet out. Bullet Pushes back on Gun (& Man)

Rocket Thrust Rocket Pushes Gas Out. Gas Pushes Back on Rocket.

Newton s 3 rd Law Exploding Systems

Reading Question 7.3 The propulsion force on a car is due to A. Static friction. B. Kinetic friction. C. The car engine. D. Elastic energy. 2013 Pearson Education, Inc. Slide 7-14

Reading Question 7.3 The propulsion force on a car is due to A. Static friction. B. Kinetic friction. C. The car engine. D. Elastic energy. 2013 Pearson Education, Inc. Slide 7-15

In order to get an object moving, you must push harder on it than it pushes back on you. A) True B) False

Question You push a heavy car by hand. The car, in turn, pushes back with an opposite but equal force on you. Doesn t this mean the forces cancel one another, making acceleration impossible? How is it that the car moves? The System Action-Reaction forces act on different objects. For F = ma, the forces must act on ONE object: the system.

Interacting Objects If object A exerts a force on object B, then object B exerts a force on object A. The pair of forces, as shown, is called an action/reaction pair. Slide 7-23

QuickCheck 7.11 Block A is accelerated across a frictionless table. The string is massless, and the pulley is both massless and frictionless. Which is true? A. Block A accelerates faster in case a than in case b. B. Block A has the same acceleration in case a and case b. C. Block A accelerates slower in case a than in case b. 2013 Pearson Education, Inc. Slide 7-80

Problem m = 2 kg, m = 3 kg, m = 5kg 1 2 3 The three blocks are pushed across a rough surface by a 40-N force. If the coefficient of kinetic friction between each of the blocks and the surface is 0.20, determine the magnitude of the force exerted by m2 on m3. a) 20 N b) 30 N c) 10 N d) 15 N e) 25 N

QuickCheck 7.6 Boxes A and B are sliding to the right on a frictionless surface. Hand H is slowing them. Box A has a larger mass than B. Considering only the horizontal forces: A. F B on H = F H on B = F A on B = F B on A B. F B on H = F H on B > F A on B = F B on A C. F B on H = F H on B < F A on B = F B on A D. F H on B = F H on A > F A on B Slide 7-67

Reading Question 7.4 Is the tension in rope 2 greater than, less than, or equal to the tension in rope 1? A. Greater than rope 2. B. Less than rope 2. C. Equal to rope 2. 2013 Pearson Education, Inc. Slide 7-16

QuickCheck 7.8 The two masses are at rest. The pulleys are frictionless. The scale is in kg. The scale reads A. 0 kg. B. 5 kg. C. 10 kg. Slide 7-73

QuickCheck 7.8 The two masses are at rest. The pulleys are frictionless. The scale is in kg. The scale reads A. 0 kg. B. 5 kg. C. 10 kg. Slide 7-74

Tension Forces Tension forces are transmitted undiminished through the rope. Different T Same T

QuickCheck 7.7 All three 50-kg blocks are at rest. The tension in rope 2 is A. greater than the tension in rope 1. B. equal to the tension in rope 1. C. less than the tension in rope 1. 2013 Pearson Education, Inc. Slide 7-71

QuickCheck 7.7 All three 50-kg blocks are at rest. The tension in rope 2 is A. greater than the tension in rope 1. B. equal to the tension in rope 1. C. less than the tension in rope 1. Each block is in static equilibrium, with. 2013 Pearson Education, Inc. Slide 7-72

Find the acceleration and tension of the system. 2013 Pearson Education, Inc. Slide 7-75

QuickCheck 7.5 Boxes A and B are being pulled to the right on a frictionless surface. Box A has a larger mass than B. How do the two tension forces compare? A. T 1 > T 2 B. T 1 = T 2 C. T 1 < T 2 D. Not enough information to tell. 2013 Pearson Education, Inc. Slide 7-61

Pulleys Block B drags block A across a frictionless table as it falls. The string and the pulley are both massless. There is no friction where the pulley turns on its axle. Therefore, T A on S = T B on S. 2013 Pearson Education, Inc. Slide 7-69

Pulleys Since T A on B = T B on A, we can draw the simplified freebody diagram on the right, below. Forces and act as if they are in an action/reaction pair, even though they are not opposite in direction because the tension force gets turned by the pulley. 2013 Pearson Education, Inc. Slide 7-70

Pulleys, Masses, Strings What is the acceleration of the system? (If they are connected, it is the same for both masses!) What is the tension in the string? 1. If it falls from rest 2. If it is dragged to the left 3. If the string is cut FIRST: Draw free-body diagrams for each mass!!! 2013 Pearson Education, Inc.

Problem A force F = 40 N pulls the two masses. If the table is frictionless, find the tension in the string. m = 3 kg, m = 1.5kg a) 13 N b) 36 N c) 23 N d) 15 N e) 28 N 1 2

QuickCheck 7.10 The top block is accelerated across a frictionless table by the falling mass m. The string is massless, and the pulley is both massless and frictionless. The tension in the string is A. T < mg. B. T = mg. C. T > mg. 2013 Pearson Education, Inc. Slide 7-78

Force Vector Diagrams Draw free-body diagrams for every object! Note: T and a are the same! Ropes connected by ideal pulleys have the same tension everywhere!

Problem A constant force F pulls the system as shown. The pulleys are frictionless. The coefficient of kinetic friction between the block and the table is µ. a) Draw free body diagrams for both masses. b) Find an expression for the acceleration in terms of the given variables.

HW Problem In the figure shown, the coefficient of kinetic friction between the block and the incline is 0.29. What is the magnitude of the acceleration of the suspended block as it falls? Disregard any pulley mass or friction in the pulley. Draw the free body diagrams for each mass. Derive a general solution for the acceleration in terms of M, and g, box it, then put the numbers in and get a numerical value then box that too. Then find a numerical value for the tension in the string. Box that. Show all your work and make it pretty! Use 3 significant figures. 2M 30 M

Tension Two 10 N weights are pulling on the spring scale as shown (the right side is attached to a hook, the left side is attached to the body of the scale) What does the scale read? a) 0 N b) 10 N c) 20 N

Force Vector Diagrams Align axes to simplify the problem!

QuickCheck 7.1 A mosquito runs head-on into a truck. Splat! Which is true during the collision? A. The mosquito exerts more force on the truck than the truck exerts on the mosquito. B. The truck exerts more force on the mosquito than the mosquito exerts on the truck. C. The mosquito exerts the same force on the truck as the truck exerts on the mosquito. D. The truck exerts a force on the mosquito but the mosquito does not exert a force on the truck. E. The mosquito exerts a force on the truck but the truck does not exert a force on the mosquito. Slide 7-39

QuickCheck 7.2 A mosquito runs head-on into a truck. Which is true during the collision? A. The magnitude of the mosquito s acceleration is larger than that of the truck. B. The magnitude of the truck s acceleration is larger than that of the mosquito. C. The magnitude of the mosquito s acceleration is the same as that of the truck. D. The truck accelerates but the mosquito does not. E. The mosquito accelerates but the truck does not. Slide 7-41