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Chapter 4, Part 1 Newton’s Laws of Motion Copyright © 2010 Pearson Education, Inc. Force Force: push or pull Force is a vector – it has magnitude and direction Copyright © 2010 Pearson Education, Inc. Newton’s First Law of Motion If you stop pushing an object, does it stop moving? Only if there is friction! In the absence of any net external force, a moving object will keep moving at a constant speed in a straight line. This is also known as the law of inertia: objects in motion tend to stay in motion; objects at rest tend to stay at rest. In order to change the velocity of an object – magnitude or direction – a net force is required. Copyright © 2010 Pearson Education, Inc. Newton’s Second Law of Motion Acceleration is directly proportional to force: Copyright © 2010 Pearson Education, Inc. Newton’s Second Law of Motion An object may have several forces acting on it; the acceleration is due to the net force: Copyright © 2010 Pearson Education, Inc. Forces and the Laws of Motion • Problem – A 873-kg (1930-lb) dragster, starting from rest, attains a speed of 26.3m/s (58.9 mph) in 0.59 s. What is the magnitude of the average net force on the dragster during this time? Copyright © 2010 Pearson Education, Inc. Forces and the Laws of Motion • Solution m = 873kg v f − vi 26.3m / s − 0m / s vf = 26.3m/s a = = = 44.58m / s 2 t 0.59s vi = 0m/s t = 0.59s 2 4 F = ma = (873kg )(44.58m / s ) = 3.9 x10 N Copyright © 2010 Pearson Education, Inc. Newton’s Third Law of Motion For every action, there is an equal and opposite reaction. Forces always come in pairs, acting on different objects. These forces are called action-reaction pairs. Copyright © 2010 Pearson Education, Inc. Newton’s Third Law of Motion Some action-reaction pairs: Copyright © 2010 Pearson Education, Inc. Free-body diagrams: A free-body diagram shows every force acting on an object. • Isolate the object of interest • Sketch the forces as vectors Copyright © 2010 Pearson Education, Inc. Example of a free-body diagram: Copyright © 2010 Pearson Education, Inc. Weight The weight of an object on the Earth’s surface is the gravitational force exerted on it by the Earth. (g is the absolute value of Earth’s gravitational acceleration, –9.81 m/s2) Copyright © 2010 Pearson Education, Inc. Normal Forces The normal force is the force exerted by a surface on an object. Copyright © 2010 Pearson Education, Inc. Normal Forces The normal force may be equal to, greater than, or less than the weight. Copyright © 2010 Pearson Education, Inc. Copyright © 2010 Pearson Education, Inc. Normal Forces The normal force is always perpendicular to the surface. Copyright © 2010 Pearson Education, Inc. Forces and the Laws of Motion • Example Problem If two horizontal forces of 225N and 165N are exerted in opposite directions on a crate, what is the net force on the crate? Copyright © 2010 Pearson Education, Inc. Forces and the Laws of Motion • Solution 225N Copyright © 2010 Pearson Education, Inc. 165N Forces and the Laws of Motion • Solution 60N Copyright © 2010 Pearson Education, Inc. Forces and the Laws of Motion • Problem On Earth, a scale shows that Ben weighs 712N. What is Ben’s mass? Copyright © 2010 Pearson Education, Inc. Forces and the Laws of Motion • Solution F = 712N g = a = 9.81m/s2 F 712N m= = = 72.6kg(160lb) 2 g 9.81m / s Copyright © 2010 Pearson Education, Inc. Forces and the Laws of Motion • Problem The acceleration of gravity on the moon is –1.60m/s2. What would the scale indicate that Ben weighs if he were on the moon and his mass is the same as in the previous question? (72.6kg) Copyright © 2010 Pearson Education, Inc. Forces and the Laws of Motion • Solution m = 72.6kg gmoon = 1.60m/s2 2 F = mgmoon = (72.6kg)(1.60m / s ) = 116N Copyright © 2010 Pearson Education, Inc. Forces and the Laws of Motion • Problem – Male lions and human sprinters can both accelerate at about 10.0 m/s2. If a typical lion weighs 170 kg and a typical sprinter weighs 75.0 kg, what is the difference in the force exerted on the ground during a race between these two species? Copyright © 2010 Pearson Education, Inc. Forces and the Laws of Motion • Solution ml = 170 kg mh = 75.0kg a = 10.0m/s2 Flion − Fhuman mlion a − mhumana (170kg)(10.0m / s 2 ) − (75.0kg)(10.0m / s 2 ) = 950N Copyright © 2010 Pearson Education, Inc. Homework pp. 126-127 Multiple Choice 1-11 (odd) pp. 128-132 9, 11, 17, 29, 39, 43 Copyright © 2010 Pearson Education, Inc.