Explanation Given: Tension in cable AD is 270 lb . Concept used: Write the expression for the position vector for point D with respect to A. d → A D = ( − 64 i ^ − 128 j ^ − 128 k ^ ) ft Here, d → A D is the position vector for point D with respect to A. Write the expression for the position vector for point A with respect to O. d → O A = ( 128 j ^ ) ft Here, d → O A is the position vector for point A with respect to O. Write the expression for the magnitude of position vector for point D with respect to A. | d A D | = ( − 64 ) 2 + ( − 128 ) 2 + ( − 128 ) 2 = 36864 = 192 ft Here, | d A D | is the magnitude of position vector for point D with respect to A. Write the expression for unit vector along line AD. n ^ A D = d → A D | d A D | ……. (1) Here, n ^ A D is the unit vector along line AD. Write the expression for representation of tension in cable AD in vector. T → A D = | T A D | ⋅ n ^ A D ......(2) Here, T → A D is the representation of tension in vector and | T A D | is the magnitude of tension in cable AB. Write the expression for moment about point O. M O = d → O A × T → A D ......(3) Here, M O is the moment about point O. Calculation: Substitute ( − 64 i ^ − 128 j ^ − 128 k ^ ) ft for d → A D and 192 ft for | d A D | in equation (1) n ^ A D = ( − 64 i ^ − 128 j ^ − 128 k ^ ) 192 = ( − 0

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ISBN: 9780073398167

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

Publisher: McGraw-Hill Education

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Chapter 3.3, Problem 69P

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By replacing force in cable AD at A with equivalent force couple system at point O.

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