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. Draw the free body diagram of the assembly as in Figure (1). Refer 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

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 4.3, Problem 4.109P

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The tension in each cable and the reaction at C.

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Chapter 4.3, Problem 4.108P

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Chapter 4 Solutions

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

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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The tension in each cable and the reaction at C .

The tension in each cable and the reaction at C.

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Chapter 4 Solutions