Explanation Given: Tension in cable AB is 288 lb . Concept used: Write the expression for the position vector for point B with respect to A. d → A B = ( − 64 i ^ − 128 j ^ + 16 k ^ ) ft Here, d → A B is the position vector for point B 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 B with respect to A. | d A B | = ( − 64 ) 2 + ( − 128 ) 2 + ( 16 ) 2 = 20736 = 144 ft Here, | d A B | is the magnitude of position vector for pointB with respect to A. Write the expression for unit vector along line AB. n ^ A B = d → A B | d A B | ........ (1) Here, n ^ A B is the unit vector along line AB. Write the expression for representation of tension in cable AB in vector. T → A B = | T A B | ⋅ n ^ A B ....... (2) Here, T → A B is the representation of tension in vector and | T A B | is the magnitude of tension in cable AB. Write the expression for moment about point O. M O = d → O A × T → A B ....... (3) Here, M O is the moment about point O. Calculation: Substitute ( − 64 i ^ − 128 j ^ + 16 k ^ ) ft for d → A B and 144 ft for | d A B | in equation (1) n ^ A B = ( − 64 i ^ − 128 j ^ + 16 k ^ ) 144 = ( − 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 68P

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To replace force in cable AB at A with equivalent force couple system at point O.

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Chapter 3.3, Problem 67P

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