Explanation Given data: Refer to the FIGURE P5.3 in the textbook for the required data. In an amusement park, the riders are suspended from a lower by a long cable. A second cable then lifts them until they reach the starting position indicated in the given figure. The lifting cable is then released, and the riders swing down the circular arc. Mass of the riders ( m ) is 270 kg. From the given figure, the angle made by the left side cable with the horizontal ( θ left ) is 15 ° . The angle made by the right side cable with the horizontal ( θ right ) is 26 ° . Formula used: Write the expression for equilibrium equation of tension force for the x -component as follows. ∑ F x = 0 (1) Here, ∑ F x is the sum of the forces in the x -component. Write the expression for equilibrium equation of tension force for the y -component as follows. ∑ F y = 0 (2) Here, ∑ F y is the sum of the forces in the y -component. Write the expression for force due to weight as follows. w = m g (3) Here, m is the mass of the loudspeaker, and g is the acceleration due to gravity, which is 9.8 m s 2 . Explanation: From the Newton’s first law, the sum of the forces acts on the riders towards each component ( x -component and y -component) is zero. Modify the expression in equation (1) for the forces in the x -component as follows. ( T left ) x + ( T right ) x = 0 (4) Here, ( T left ) x is the tension force T left (tension force in the left side cable) in the x -component, and ( T right ) x is the tension force T right (tension force in the right side cable) in the x -component. From the given figure, write the tension force ( T left ) x as follows. ( T left ) x = − T left cos ( θ left ) From the figure, write the tension force ( T right ) x as follows. ( T right ) x = T right cos ( θ right ) Substitute [ − T left cos ( θ left ) ] for ( T left ) x , and T right cos ( θ right ) for ( T right ) x in equation (4), [ − T left cos ( θ left ) ] + [ T right cos ( θ right ) ] = 0 T left cos ( θ left ) = T right cos ( θ right ) Write the expression for tension in the right side cable as follows. T right = T left [ cos ( θ left ) cos ( θ right ) ] Substitute 15 ° for θ left , and 26 ° for θ right , T right = T left [ cos ( 15 ° ) cos ( 26 ° ) ] = T left ( 0.9659 0.8987 ) T right = 1.0747 T left (5) Modify the expression in equation (2) for the forces in the y -component as follows. ( T left ) y + ( T right ) y − w = 0 (6) Here, ( T left ) y is the tension force T left in the y -component, ( T right ) y is the tension force T right in the y -component, and w is the weight of the riders

4th Edition

ISBN: 8220106755235

Author: Field

Publisher: PEARSON

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Question

Chapter 5, Problem 3P

To determine

To find: The tension in the cables just before the release of lifting cable.

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Chapter 5, Problem 2P

Chapter 5, Problem 4P

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