Explanation Given information: The tension in cable A B is 259 N . The below figure represents the schematic diagram of the arrangement. Write the position of points A , B , C and D with respect to point O . A = ( 0 , 5.6 m , 0 ) B = ( − 4.2 m , 0 , 0 ) C = ( 2.4 m , 0 , 4.2 m ) D = ( 0 , 0 − 3.3 m ) Write the vector A B → . A B → = ( − 4.2 m-0 ) i ^ + ( 0 ft − 5 .6 m ) j ^ + ( 0 -0 ) k ^ = ( − 4.2 i ^ − 5.6 j ^ ) m …… (I) Write the vector A C → A C → = ( 2.4 m-0 ) i ^ + ( 0 − 5 .6 m ) j ^ + ( 4.2 m-0 ) k ^ = ( 2.4 i ^ − 5.6 j ^ + 4.2 k ^ ) m …… (II) Write the vector A D → A D → = ( 0 -0 ) i ^ + ( 0 − 5 .6 m ) j ^ + ( − 3.3 m-0 ) k ^ = ( − 5.6 j ^ − 3.3 k ^ ) m …… (III) Write the expression for modulus of the 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 axisis z from the 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 the 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 the cable A B is T A B → , the magnitude of tension in cable A B is T A B . Write the expression for tension T A C → . T A C → = T A C ⋅ A C ^ …… (IX) Here, the tension vector in the cable A C is T A C → , the magnitude of tension in cable A C is T A C . Write the expression for tension T A D → . T A D → = T A D ⋅ A D ^ …… (X) Here, the tension vector in the cable A D is T A D → , the magnitude of tension in cable A D is T A D . Write the equilibrium equation of forces at the point A . T A B → + T A C → + T A D → + P = 0 …… (XI) Here, the force acting at point A is P . Calculation: Substitute − 4.2 m for x , − 5.6 m for y , 0 for z and A B → for D → in equation (IV) | A B → | = ( − 4.2 m ) 2 + ( − 5.6 m ) 2 + ( 0 ) 2 = 17.64 m 2 + 31.36 m 2 + 0 = 7 m Substitute 2.4 m for x , − 5.6 m for y , 4.2 m for z and A C → for D → in equation (IV) | A C → | = ( 2.4 m ) 2 + ( − 5.6 m ) 2 + ( 4.2 m ) 2 = 5.76 m 2 + 31.36 m 2 + 17.64 m 2 = 7.4 m Substitute 0 for x , − 5.6 m for y , − 3.3 m for z and A D → for D → in equation (IV) | A D → | = ( 0 ) 2 + ( − 5.6 m ) 2 + ( − 3.3 m ) 2 = 0 + 31.36 m 2 + 10.89 m 2 = 6.5 m Substitute ( − 4

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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Chapter 2, Problem 112RP

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The magnitude of forceexerted by the balloon at point A.

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Chapter 2, Problem 111RP

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The magnitude of forceexerted by the balloon at point A.

Solution Summary: The author explains the magnitude of forceexerted by the balloon at point A. The tension in cable AB is 259.

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The magnitude of forceexerted by the balloon at point A.

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