Explanation Given data: Refer to the Figure P5.73 in the textbook for the holding position of the book by the person with his fingers. The normal force exerted by the fingers of the person onto the book is 6.0 N. The coefficient of static friction between the fingers and the book ( μ s ) is 0.80. Formula used: Write the expression for equilibrium condition for the forces in the y -component as follows. ∑ F y = 0 (1) Here, ∑ F y is the sum of the forces in the y -component (vertical component). Write the expression for force due to weight as follows. w = m g (2) Here, m is the mass of the book, and g is the acceleration due to gravity, which is 9.8 m s 2 . Write the expression for static friction force for the given system as follows. f s = μ s n (3) Here, μ s is the coefficient of static friction between the fingers and the book, which is 0.80, and n is the normal force of the fingers on the block. Explanation: From the given data, the following are forces act on the book. 1. The force due to weight of the book, which acts in the downward direction, 2. The static friction force of each finger on each side, which acts in the upward direction, 3. The normal force exerted by the fingers on the book, and 4. The force of the book on the fingers (reaction force for the normal force exerted by the fingers on the book). The normal force exerted by the fingers on the book and force of the book on the fingers are equal in magnitude and opposite in direction. From the analysis draw the figure for various forces on the book along with its free-body diagram as shown in Figure 1. From the free-body diagram in Figure 1, write the expression for the weight of the book by using the equation (1) as follows. w = f s + f s = 2 f s From equation (2), substitute ( m g ) for w

4th Edition

ISBN: 9780134779218

Author: Knight

Publisher: PEARSON

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Question

Chapter 5, Problem 73GP

To determine

To find: The greatest mass book that can hold by the person with his fingers.

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Chapter 5, Problem 72GP

Chapter 5, Problem 74GP

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