To determine The mass of the box A to prevent any motion from occurring. Explanation Given: The mass of the box B, m B = 2.0 kg The coefficient of static friction, μ s = 0.30 Formula: Boxes A and B are connected by a rope passing over a pulley. The box A lies on a table and the box B hangs freely. The box A has its weight m A g acting vertically downwards and the normal force F N acts vertically upwards due to Newton’s third law . Since the box is connected to the box B , it also experiences the tension force F T along the length of the rope. The box B hangs freely. Its weight m B g acts vertically downwards and since, the same rope passes over the pulley connecting the boxes A and B , the same tension force F T acts on the box B along the length of the rope. The forces act as shown in the diagram given below. Figure.1 The boxes will not move if the force of static friction acting between the box A and the table is atleast equal to the tension acting on the box. Draw the free body diagram of the system. For box B As per the condition applied, the box B neither moves up nor down, If there is no motion; m B g = F T ......(1) Draw the free body diagram for the box A . For vertical equilibrium, m A g = F N ......(2) For horizontal equilibrium, F f r = F T

7th Edition

ISBN: 9780321929013

Author: Douglas C. Giancoli

Publisher: PEARSON

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Chapter 4, Problem 46P

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The mass of the box A to prevent any motion from occurring.

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Chapter 4, Problem 45P

Chapter 4, Problem 47P

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