Explanation Given data: Refer to the Figure Q5.30 in the textbook for set of two block on frictionless surface. Consider the block on left side as first block and the block that lies on the right side of the surface as second block. From the Figure, the mass of the first block ( m 1 ) is 10 kg. Angle made by the first block with the horizontal line ( θ 1 ) is ( 23 ° ) . Angle made by the second block with the horizontal line ( θ 2 ) is ( 40 ° ) . Formula used: Write the expression for tension force for the given system as follows. T = m g sin θ (1) Here, m is the mass of the block, g is the acceleration due to gravity, which is 9.8 m s 2 , and θ is the angle made by the block with the horizontal line. Explanation: To maintain the stationary condition for the two-block system, the two blocks must be in the static equilibrium condition. In other words, the tension force of the two blocks must be equal. Therefore, write the mathematic expression to attain the stationary condition of two blocks as follows. T 1 = T 2 (2) Here, T 1 is the tension force of first block, and T 2 is the tension force of second block. Modify the expression in equation (1) to find the tension force of first block as follows. T 1 = m 1 g sin ( θ 1 ) Modify the expression in equation (1) to find the tension force of second block as follows. T 2 = m 2 g sin ( θ 2 ) Calculation of mass of second block: Substitute [ m 1 g sin ( θ 1 ) ] for T 1 , and [ m 2 g sin ( θ 2 ) ] for T 2 in equation (2) as follows

College Physics: A Strategic Approach, Books a la Carte Edition (4th Edition)

4th Edition

ISBN: 9780134700502

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

Publisher: PEARSON

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Question

Chapter 5, Problem 30MCQ

To determine

To find: The mass of the second block in order to maintain the stationary condition.

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Chapter 5, Problem 29MCQ

Chapter 5, Problem 31MCQ

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Chapter 5 Solutions

College Physics: A Strategic Approach, Books a la Carte Edition (4th Edition)

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