Explanation Given information : The tension developed by a cord is 400 N. Draw the free body diagram of the forces acting in the cord DEF as shown below in Figure 1. Draw the free body diagram of the forces acting in the cord ABCD as in Figure 2. Determine tension in each cord by applying the Equation of equilibrium. At joint D . Along the vertical direction : ∑ F y = 0 F D E sin θ D E − W = 0 F D E = W sin θ D E (I) Here, the force acting on cord DE is F D E , the angle of cord DE is θ D E , and weight of the lamp is W . Along the horizontal direction : ∑ F x = 0 F D E cos θ D E − F C D = 0 F C D = F D E cos θ D E (II) Here, the force acting on cord CD is F C D . At joint D . Along the horizontal direction : ∑ F x = 0 F C D − F C A cos θ C A − F C B cos θ C B = 0 (III) Here, the force acting on cord CA is F C A , the angle of cord CA is θ C A , the force acting on cord CB is F C B , and the angle of cord CB is θ C B . Along the vertical direction : ∑ F y = 0 F C A sin θ C A − F C B sin θ C B = 0 (IV) Determine the mass of the lamp. W = m g (V) Here, the greatest tensile force is F and the acceleration due to gravity is g . Conclusion : Substitute 30 ° for θ D E in Equation (I). F D E = W sin 30 ° = 2 W Substitute 30 ° for θ D E and 2 W for F D E in Equation (II). F C D = 2 W × cos 30 ° = 1.732 W Substitute 1.732 W for F C D , ( 3 5 ) for cos θ C A , and 45 ° for θ C B in Equation (IV)

Engineering Mechanics: Statics & Dynamics (14th Edition)

14th Edition

ISBN: 9780133915426

Author: Russell C. Hibbeler

Publisher: PEARSON

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Chapter 3.3, Problem 30P

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The maximum mass of the lamp.

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Chapter 3 Solutions

Engineering Mechanics: Statics & Dynamics (14th Edition)

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The maximum mass of the lamp.

Solution Summary: The author explains how to determine the maximum mass of a lamp by drawing the free body diagram of the forces acting in the cord DEF as shown in Figure 1.

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Chapter 3 Solutions