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(a)
To identify an action-reaction pair of forces.
(a)
Answer to Problem 1Q
The force the person exerts on the wall and the force the wall exerts on the person is an action-reaction pair.
Explanation of Solution
Action – reaction pair of forces always acts on separate object.
When a person pushes on a wall, the wall pushes back on the person. Here the person exerts force on the wall and wall in turn exerts force on the person. The force exerted by the person on the wall is equal in magnitude and opposite in direction to the force exerted by the wall on the person and they acts on different objects.
Conclusion:
Therefore, the force the person exerts on the wall and the force the wall exerts on the person is an action-reaction pair.
(b)
To identify an action-reaction pair of forces, book resting on the table.
(b)
Answer to Problem 1Q
The gravitational force exerted the book on the table and the normal force exerted by the table on the book is action-reaction pair force.
Explanation of Solution
The gravitational force exerted by the book on the table and the normal force exerted by the table on the book is action-reaction pair force and they acts on different objects. The gravitational force exerted by the book on the table is equal in magnitude and opposite in direction to the force exerted by the table on the book.
Conclusion:
Therefore, the gravitational force exerted the book on the table and the normal force exerted by the table on the book is action-reaction pair force.
(c)
To identify an action-reaction pair of forces, a hockey puck sliding across an icy surface.
(c)
Answer to Problem 1Q
The
Explanation of Solution
The frictional force exerted by the moving hockey puck on the ice and the frictional force exerted by the ice on the hockey puck is action-reaction pair force and they acts on different objects. The frictional force exerted by the moving hockey puck on the ice is equal in magnitude and opposite in direction to the frictional force exerted by the ice on the hockey puck.
Conclusion:
Therefore, the frictional force exerted by the moving hockey puck on the ice and the frictional force exerted by the ice on the hockey puck is action-reaction pair force.
(d)
To identify an action-reaction pair of forces, a car accelerating from rest.
(d)
Answer to Problem 1Q
The frictional force exerted by the tire on the road and the force exerted by the road on the tire is action-reaction pair force.
Explanation of Solution
The frictional force exerted by the tire on the road and the force exerted by the road on the tire is action-reaction pair force and they acts on different objects. The frictional force exerted by the rotating tire on the road is equal in magnitude and opposite in direction to the frictional force exerted by the road on the tire.
Conclusion:
Therefore, the frictional force exerted by the tire on the road and the force exerted by the road on the tire is action-reaction pair force.
(e)
To identify an action-reaction pair of forces, an object undergoing free fall in a vacuum.
(e)
Answer to Problem 1Q
The gravitational force exerted on the object by the object on which the object is falling and gravitational force on the falling object on which the object is falling is an action-reaction pair.
Explanation of Solution
The gravitational force exerted on the object by the object on which the object is falling is equal in magnitude and opposite in direction to gravitational force on the falling object on which the object is falling is an action-reaction pair and they acts on different objects.
Conclusion:
Therefore, the gravitational force exerted on the object by the object on which the object is falling and gravitational force on the falling object on which the object is falling is an action-reaction pair.
(f)
To identify an action-reaction pair of forces, a basketball player jumping to dunk a basketball.
(f)
Answer to Problem 1Q
The force exerted by the basketball player against the floor and the force the floor exerts on the basket ball are action-reaction pairs.
Explanation of Solution
The force exerted by the basketball player against the floor while he dunks the ball and the force the floor exerts on the basket ball are equal in magnitude and opposite in direction and they acts on different objects. The force exerted by the basketball player against the floor while and the force the floor exerts on the basket ball are action-reaction pairs.
Conclusion:
Therefore, the force exerted by the basketball player against the floor while and the force the floor exerts on the basket ball are action-reaction pairs.
(g)
To identify an action-reaction pair of forces, a person throwing a baseball.
(g)
Answer to Problem 1Q
The force the person exerts against the basket ball to accelerate it and the force the basketball exerts on the person is action-reaction pair.
Explanation of Solution
The force the person exerts against the basket ball to accelerate it and the force the basketball exerts on the person is equal in magnitude and opposite in direction and they acts on different objects. The force the person exerts against the basket ball to accelerate it and the force the basketball exerts on the person is action-reaction pair.
Conclusion:
Therefore, the force the person exerts against the basket ball to accelerate it and the force the basketball exerts on the person is action-reaction pair.
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Chapter 3 Solutions
College Physics, Volume 1
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Principles of Physics: A Calculus-Based TextPhysicsISBN:9781133104261Author:Raymond A. Serway, John W. JewettPublisher:Cengage Learning