Explanation Given information: The length of the boom is 2.4 m . The 3.6 kN load applied at A . Calculation: Assumption: Apply the sign convention for calculating the equations of equilibrium as shown below. For the horizontal forces equilibrium condition, take the force acting towards right side as positive ( → + ) and the force acting towards left side as negative ( ← − ) . For the vertical forces equilibrium condition, take the upward force as positive ( ↑ + ) and the downward force as negative ( ↓ − ) . For moment equilibrium condition, take the clockwise moment as negative and counterclockwise moment as positive. Sketch the Free Body Diagram as shown in Figure 1. Calculate the position vector ( r ) as shown below. The position of A is r A = ( 2.4 m ) k . The position of B is r B = ( 1.2 m ) k . The position of D is r D = − ( 0.8 m ) i + ( 0.6 m ) j . The position of E is r E = ( 0.8 m ) i + ( 1.2 m ) j . Calculate the position of cable AD as shown below. A D → = r D − r A Substitute − ( 0.8 m ) i + ( 0.6 m ) j for r D and ( 2.4 m ) k for r A . A D → = [ − ( 0.8 m ) i + ( 0.6 m ) j ] − [ ( 2.4 m ) k ] = − ( 0.8 m ) i + ( 0.6 m ) j − ( 2.4 m ) k Calculate the length of cable AD as shown below. A D = ( − 0.8 ) 2 + ( 0.6 ) 2 + ( − 2.4 ) 2 = 6.76 = 2.6 m Calculate the position of cable AE as shown below. A E = r E − r A Substitute ( 0.8 m ) i + ( 1.2 m ) j for r E and ( 2.4 m ) k for r A . A E = [ ( 0.8 m ) i + ( 1.2 m ) j ] − [ ( 2.4 m ) k ] = ( 0.8 m ) i + ( 1.2 m ) j − ( 2.4 m ) k Calculate the length of cable AE as shown below. A E = ( 0.8 ) 2 + ( 1.2 ) 2 + ( − 2.4 ) 2 = 7.84 = 2.8 m Calculate the tension in cable AD ( T A D ) as shown below. T A D = T A D A D → A D Substitute − ( 0.8 m ) i + ( 0.6 m ) j − ( 2.4 m ) k for A D → and 2.6 m for A D . T A D = T A D − 0.8 i + 0.6 j − 2.4 k 2.6 = T A D 2.6 ( − 0.8 i + 0.6 j − 2.4 k ) Calculate the tension in cable AE ( T A E ) as shown below. T A E = T A E A E → A E Substitute ( 0.8 m ) i + ( 1.2 m ) j − ( 2.4 m ) k for A E → and 2.8 m for A E . T A E = T A E 0.8 i + 1.2 j − 2.4 k 2.8 = T A E 2.8 ( 0.8 i + 1.2 j − 2.4 k ) Apply the Equations of Equilibrium as shown below. Apply the Equilibrium equation of moment about point C and equate to zero for equilibrium. ∑ M C = 0 r A × T A D + r A × T A E + r A × ( − 3 j ) = 0 Substitute ( 2.4 m ) k for r A , T A D 2.6 ( − 0.8 i + 0.6 j − 2.4 k ) for T A D , and T A E 2.8 ( 0.8 i + 1

Vector Mechanics for Engineers: Statics, 11th Edition

11th Edition

ISBN: 9780077687304

Author: Ferdinand P. Beer, E. Russell Johnston Jr., David Mazurek

Publisher: McGraw-Hill Education

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Question

Chapter 4.3, Problem 4.109P

To determine

Find the tension in each cable and the reaction at C.

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Chapter 4.3, Problem 4.108P

Chapter 4.3, Problem 4.110P

Chapter 4 Solutions

Vector Mechanics for Engineers: Statics, 11th Edition

Ch. 4.1 - Two crates, each of mass 350 kg, are placed as...Ch. 4.1 - A lever AB is hinged at C and attached to a...Ch. 4.1 - A light rod AD is supported by frictionless pegs...Ch. 4.1 - A tension of 20 N is maintained in a tape as it...Ch. 4.1 - A gardener uses a 60 N wheelbarrow to transport a...Ch. 4.1 - The gardener of Prob. 4.1 wishes to transport a...Ch. 4.1 - A 2100-lb tractor is used to lift 900 lb of grave....Ch. 4.1 - For the beam and loading shown, determine (a) the...Ch. 4.1 - A load of lumber of weight W = 25 kN is being...Ch. 4.1 - A load of lumber of weight W = 25 kN is being...

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