Explanation Given data: Refer to the Figure P5.79 in the textbook for the identical ball system. Mass of the each identical block is 2 kg. The coefficient of static friction between the blocks ( μ s ) is 0.35. Formula used: Write the expression for static friction force between the two blocks as follows. f s = μ s m g (1) Here, μ s is the coefficient of static friction between the blocks, m is the total mass of the blocks, and g is the acceleration due to gravity, which is 9.8 m s 2 . Write the expression for mass of the two blocks as follows. m = m 1 + m 2 (2) Here, m 1 is the upper block (consider), which is 2.0 kg, and m 2 is the mass of the lower block (consider), which is 2.0 kg. Explanation: From the given data, the force that can be applied to the lower without slipping the upper block is equal to the maximum friction between the two blocks. Therefore, write the expression to find the force that can be applied to the lower without slipping the upper block as follows. F = f s From equation (1), substitute ( μ s m g ) for f s . F = μ s m g From equation (2), substitute ( m 1 + m 2 ) for m . F = μ s ( m 1 + m 2 ) g Substitute 0

College Physics: A Strategic Approach Plus Mastering Physics with Pearson eText -- Access Card Package (4th Edition) (What's New in Astronomy & Physics)

4th Edition

ISBN: 9780134641492

Author: Randall D. Knight (Professor Emeritus), Brian Jones, Stuart Field

Publisher: PEARSON

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Chapter 5, Problem 79GP

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To find: The maximum force that can be applied to the lower block without the slipping of upper block.

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Chapter 5, Problem 77GP

Chapter 5, Problem 80GP

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College Physics: A Strategic Approach Plus Mastering Physics with Pearson eText -- Access Card Package (4th Edition) (What's New in Astronomy & Physics)

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