Explanation The truss is loaded as shown in the figure. There is load at P , E , G , I and L . The free body diagram of the given arrangement is given by Figure 1. Figure 1 The tangent of the angle can be written from the diagram as, tan α = 7.5 6 The angle is derived from the tangent value as, α = tan − 1 7.5 6 = 51.34 ° From the diagram the sum of the moments in the counter clockwise direction about F is given as, P ( 18 ft ) − 2.5 P ( 12 ft ) − P ( 6 ft ) + F E G ( 7.5 ft ) = 0 (I) From the diagram the sum of the moments in the counter clockwise direction about A is given as, ( F E F sin α ) ( 12 ft ) − P ( 12 ft ) + ( 2.5 P ) ( 6 ft ) = 0 (II) Conclusion: Solve for F E G by substituting 20 kips for P from (I). ( 20 kips ) ( 18 ft ) − 2

Connect 1 Semester Access Card for Vector Mechanics for Engineers: Statics and Dynamics

11th Edition

ISBN: 9781259639272

Author: Ferdinand P. Beer, E. Russell Johnston Jr., David Mazurek, Phillip J. Cornwell, Brian Self

Publisher: McGraw-Hill Education

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Chapter 6.2, Problem 6.52P

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The force in the members EG and EF.

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Chapter 6 Solutions

Connect 1 Semester Access Card for Vector Mechanics for Engineers: Statics and Dynamics

Ch. 6.1 - 6.1 through 6.8 Using the method of joints,...Ch. 6.1 - 6.1 through 6.8 Using the method of joints,...Ch. 6.1 - Prob. 6.3P Ch. 6.1 - 6.1 through 6.8 Using the method of joints,...Ch. 6.1 - Prob. 6.5P Ch. 6.1 - Using the method of joints, determine the force in...Ch. 6.1 - 6.1 through 6.8 Using the method of joints,...Ch. 6.1 - Prob. 6.8P Ch. 6.1 - 6.9 and 6.10 Determine the force in each member of...Ch. 6.1 - Prob. 6.10P

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