Explanation Given information: The tension in cable A C is 920 lb . The below figure represent the schematic diagram of the arrangement. Figure-(1) Write the position of points A , B , C and D with respect to point O . A = ( 0 , 90 ft , 0 ) B = ( − 45 ft , 0 , 30 ft ) C = ( 30 ft , 0 , 65 ft ) D = ( 20 ft , 0 − 60 ft ) Write the vector A B → . A B → = ( − 45 ft-0 ft ) i ^ + ( 0 ft − 90 ft ) j ^ + ( 30 ft-0 ft ) k ^ = ( − 45 i ^ − 90 j ^ + 30 k ^ ) ft …… (I) Write the vector A C → A C → = ( 30 ft-0 ft ) i ^ + ( 0 ft − 90 ft ) j ^ + ( 65 ft-0 ft ) k ^ = ( 30 i ^ − 90 j ^ + 65 k ^ ) ft …… (II) Write the vector A D → A D → = ( 20 ft-0 ft ) i ^ + ( 0 ft − 90 ft ) j ^ + ( − 60 ft-0 ft ) k ^ = ( 20 i ^ − 90 j ^ − 60 k ^ ) ft …… (III) Write the expression for modulus of vector. | D → | = ( x ) 2 + ( y ) 2 + ( z ) 2 …… (IV) Here, horizontal distance along x axis is x , horizontal distance along y axis is y and horizontal distance along z axis is z from point O and the modulus of the vector is | D → | . Write the expression for unit vector A B ^ . A B ^ = A B → | A B → | …… (V) Here, the unit vector of A B → is A B ^ and the magnitude of A B → is | A B → | . Write the expression for unit vector A C ^ . A C ^ = A C → | A C → | …… (VI) Here, the unit vector of A C → is A C ^ and the magnitude of A C → is | A C → | . Write expression for unit vector A D ^ . A D ^ = A D → | A D → | …… (VII) Here, the unit vector of A D → is A D ^ and the magnitude of A D → is | A D → | . Write the equation for T A B → . T A B → = T A B ⋅ A B ^ …… (VIII) Here, the tension vector in cable A B is T A B → and the magnitude of tension in cable A B is T A B . Write the equation for tension T A C → . T A C → = T A C ⋅ A C ^ …… (IX) Here, the tension vector in cable A C is T A C → , the magnitude of tension in cable A C is T A C . Write the equation for tension T A D → . T A D → = T A D ⋅ A D ^ …… (X) Here, the tension vector in cable A D is T A D → and the magnitude of tension in cable A D is T A D . Write the equilibrium equation of forces at point A . T A B → + T A C → + T A D → + P = 0 …… (XI) Calculation: Substitute − 45 ft for x , − 90 ft for y and 30 ft for z in equation (IV) | A B → | = ( − 45 ft ) 2 + ( − 90 ft ) 2 + ( 30 ft ) 2 = 2025 ft 2 + 8100 ft 2 + 900 ft 2 = 105 ft Substitute 30 ft for x , − 90 ft for y and 65 ft for z in equation (IV) | A B → | = ( 30 ft ) 2 + ( − 90 ft ) 2 + ( 65 ft ) 2 = 900 ft 2 + 8100 ft 2 + 4225 ft 2 = 115 ft Substitute 20 ft for x , − 90 ft for y and 60 ft for z in equation (IV) | A B → | = ( 20 ft ) 2 + ( − 90 ft ) 2 + ( 60 ft ) 2 = 400 ft 2 + 8100 ft 2 + 3600 ft 2 = 110 ft Substitute ( − 45 i ^ − 90 j ^ + 30 k ^ ) ft for A B → and 105 ft for | A B → | in equation (V) A B ^ = ( − 45 i ^ − 90 j ^ + 30 k ^ ) ft 105 ft = − 0.4285 i ^ − 0.857 j ^ + 0.285 k ^ Substitute ( 30 i ^ − 90 j ^ + 65 k ^ ) ft for A C → and 115 ft for | A C → | in equation (VI) A C ^ = ( 30 i ^ − 90 j ^ + 65 k ^ ) ft 115 ft = 0

2nd Edition

ISBN: 9780073398167

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

Publisher: McGraw-Hill Education

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