chapter 4, part 1 newton s laws of motion

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Copyright © 2010 Pearson Education, Inc. Chapter 4, Part 1 Newtons Laws of Motion

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Page 1: Chapter 4, Part 1 Newton s Laws of Motion

Copyright © 2010 Pearson Education, Inc.

Chapter 4, Part 1 Newton’s Laws of Motion

Page 2: 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

Page 3: Chapter 4, Part 1 Newton s Laws of Motion

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.

Page 4: Chapter 4, Part 1 Newton s Laws of Motion

Copyright © 2010 Pearson Education, Inc.

Newton’s Second Law of Motion Acceleration is directly proportional to force:

Page 5: Chapter 4, Part 1 Newton s Laws of Motion

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:

Page 6: Chapter 4, Part 1 Newton s Laws of Motion

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?

Page 7: Chapter 4, Part 1 Newton s Laws of Motion

Copyright © 2010 Pearson Education, Inc.

Forces and the Laws of Motion

•  Solution m = 873kg vf = 26.3m/s vi = 0m/s t = 0.59s

a =vf − vit

=26.3m / s− 0m / s

0.59s= 44.58m / s2

NxsmkgmaF 42 109.3)/58.44)(873( ===

Page 8: Chapter 4, Part 1 Newton s Laws of Motion

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.

Page 9: Chapter 4, Part 1 Newton s Laws of Motion

Copyright © 2010 Pearson Education, Inc.

Newton’s Third Law of Motion Some action-reaction pairs:

Page 10: Chapter 4, Part 1 Newton s Laws of Motion

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

Page 11: Chapter 4, Part 1 Newton s Laws of Motion

Copyright © 2010 Pearson Education, Inc.

Example of a free-body diagram:

Page 12: Chapter 4, Part 1 Newton s Laws of Motion

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)

Page 13: Chapter 4, Part 1 Newton s Laws of Motion

Copyright © 2010 Pearson Education, Inc.

Normal Forces

The normal force is the force exerted by a surface on an object.

Page 14: Chapter 4, Part 1 Newton s Laws of Motion

Copyright © 2010 Pearson Education, Inc.

Normal Forces The normal force may be equal to, greater than, or less than the weight.

Page 15: Chapter 4, Part 1 Newton s Laws of Motion

Copyright © 2010 Pearson Education, Inc.

Page 16: Chapter 4, Part 1 Newton s Laws of Motion

Copyright © 2010 Pearson Education, Inc.

Normal Forces

The normal force is always perpendicular to the surface.

Page 17: Chapter 4, Part 1 Newton s Laws of Motion

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?

Page 18: Chapter 4, Part 1 Newton s Laws of Motion

Copyright © 2010 Pearson Education, Inc.

Forces and the Laws of Motion

•  Solution

225N 165N

Page 19: Chapter 4, Part 1 Newton s Laws of Motion

Copyright © 2010 Pearson Education, Inc.

Forces and the Laws of Motion

•  Solution

60N

Page 20: Chapter 4, Part 1 Newton s Laws of Motion

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?

Page 21: Chapter 4, Part 1 Newton s Laws of Motion

Copyright © 2010 Pearson Education, Inc.

Forces and the Laws of Motion

•  Solution F = 712N g = a = 9.81m/s2

m =Fg=

712N9.81m / s2

= 72.6kg(160lb)

Page 22: Chapter 4, Part 1 Newton s Laws of Motion

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)

Page 23: Chapter 4, Part 1 Newton s Laws of Motion

Copyright © 2010 Pearson Education, Inc.

Forces and the Laws of Motion

•  Solution m = 72.6kg gmoon = 1.60m/s2

F =mgmoon = (72.6kg)(1.60m / s2 ) =116N

Page 24: Chapter 4, Part 1 Newton s Laws of Motion

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?

Page 25: Chapter 4, Part 1 Newton s Laws of Motion

Copyright © 2010 Pearson Education, Inc.

Forces and the Laws of Motion

•  Solution ml = 170 kg mh = 75.0kg a = 10.0m/s2

(170kg)(10.0m / s2 )− (75.0kg)(10.0m / s2 ) = 950N

humanlion FF −

amam humanlion −

Page 26: Chapter 4, Part 1 Newton s Laws of Motion

Copyright © 2010 Pearson Education, Inc.

Homework

pp. 126-127 Multiple Choice 1-11 (odd) pp. 128-132 9, 11, 17, 29, 39, 43