Explanation Given data: Refer to the Figure P5.84 in the textbook for the blocks and pulley system. Mass of the right block ( M ) is 100 kg. When the right block is released from the rest, the right block takes 6 s to reach the floor. From the Figure, the right block is initially at the height of 1.0 m from the floor. Formula used: Write the expression for equilibrium equation of force for the y -component as follows. ∑ F y = 0 (1) Here, ∑ F y is the sum of the forces in the y -component. Write the expression for force due to the weight of the object as follows. w = m g (2) Here, m is the mass of the object, and g is the acceleration due to gravity, which is 9.8 m s 2 . Consider the initial position of the right block as reference position and write the expression of kinematic equation for the given system to find the acceleration of the block as follows. y f = y i + ( v y ) i ( t f − t i ) + 1 2 a M ( t f − t i ) 2 (3) Here, y i is the distance between reference position of the right block and initial position of the block, which is 0 m, y f is the distance between reference position of the right block and final position of the block, which is ( − 1 ) m (the negative sign is due to the movement of the right block towards the downward direction), ( v y ) i is the initial speed of right block, which is 0 m s , t i is the initial time, which is 0 s, t f is the final time (time taken by the block to reach the floor), which is 6 s, and a M is the acceleration of the right block. Explanation: Consider the rope as massless and the pulley as both massless and frictionless. Draw the free-body diagram for the given system as shown in Figure 1. In the free-body diagram, T R on m is the tension of the rope on the left block of the pulley ( m ) , T R on M is the tension of the rope on the right block of the pulley ( M ) , w → m is the weight of the left block, and w → M is the weight of the right block. Calculation of acceleration of right block: Modify the expression in equation (3) to find the acceleration of the right block as follows. a M = 2 { y f − y i − [ ( v y ) i ( t f − t i ) ] } ( t f − t i ) 2 Substitute 0 m for y i , ( − 1 ) m for y f , 0 m s for ( v y ) i , 0 s for t i , and 6 s for t f , a M = 2 { ( − 1 m ) − 0 m − [ ( 0 m s ) ( 6 s − 0 s ) ] } ( 6 s − 0 s ) 2 = 2 ( − 1 m − 0 m ) 36 s 2 = − 0.0555 m s 2 ≃ − 0.0556 m s 2 From the free-body diagram of the block m , write the expression for sum of the forces in the y -component by modifying the expression in equation (1) as follows. T R on m + ( − w m ) = m a m (4) From the given data, the acceleration of both the blocks is equal and opposite in the direction

4th Edition

ISBN: 8220106755235

Author: Field

Publisher: PEARSON

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Question

Chapter 5, Problem 84GP

To determine

To find: The mass of the block on the left of the pulley.

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Chapter 5, Problem 83GP

Chapter 5, Problem 85MSPP

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