Explanation Given information : The maximum tension sustained by the cables AB and AC is 500 N. The maximum compression supported by the pole AO is 300 N. Show the free body diagram of the forces acting on the cables as in Figure (1). Determine the position vector for point AO . r A O = r O − r A = ( x O i + y O j + z O k ) − ( x A i + y A j + z A k ) = ( x O − x A ) i + ( y O − y A ) j + ( z O − z A ) k (I) Here, the position vector for point O is r O , the position vector for point A is r A , the distance of point O from the origin along x direction is x O , the distance of point O from the origin along y direction is y O , the distance of point O from the origin along z direction is z O , the distance of point A from the origin along x direction is x A , the distance of point A from the origin along y direction is y A , the distance of point A from the origin along z direction is z A , the vector component along x direction is i, vector component along y direction is j, and the vector component along z direction is k. Determine the modulus of position vector for point AO . | r A O | = ( x O − x A ) 2 + ( y O − y A ) 2 + ( z O − z A ) 2 (II) Determine the position vector for point AB . r A B = r B − r A = ( x B i + y B j + z B k ) − ( x A i + y A j + z A k ) = ( x B − x A ) i + ( y B − y A ) j + ( z B − z A ) k (III) Here, the position vector for point B is r B , the distance of point B from the origin along x direction is x B , the distance of point B from the origin along y direction is y B , and the distance of point B from the origin along z direction is z B . Determine the modulus of position vector for point AB . | r A B | = ( x B − x A ) 2 + ( y B − y A ) 2 + ( z B − z A ) 2 (IV) Determine the position vector for point AC . r A C = r C − r A = ( x C i + y C j + z C k ) − ( x A i + y A j + z A k ) = ( x C − x A ) i + ( y C − y A ) j + ( z C − z A ) k (V) Here, the position vector for point C is r C , the distance of point C from the origin along x direction is x C , the distance of point C from the origin along y direction is y C , and the distance of point C from the origin along z direction is z C . Determine the modulus of position vector for point AC . | r A C | = ( x C − x A ) 2 + ( y C − y A ) 2 + ( z C − z A ) 2 (VI) Force in Rope AO : F A O = F A O [ r A O | r A O | ] (VII) Here, the force in cable AO is F A O . Force in Rope AB : F A B = F A B [ r A B | r A B | ] (VIII) Here, the force in cable AB is F A B . Force in Rope AC : F A C = F A C [ r A C | r A C | ] (IX) Here, the force in cable AC is F A C . Write the Equation of equilibrium condition. Σ F = 0 F A O + F A B + F A C + W = 0 (X) Conclusion : Substitute (2 m) for x O , 0 m for x A , (–1.5 m) for y O , 0 m for y A , (6 m) for z O , and 0 m for z A in Equation (I). r A O = ( 2 − 0 ) i + ( − 1.5 − 0 ) j + ( 6 − 0 ) k = ( 2 i − 1.5 j + 6 k ) m Substitute (2 m) for x O , 0 m for x A , (–1.5 m) for y O , 0 m for y A , (6 m) for z O , and 0 m for z A in Equation (II). | r A O | = ( 2 − 0 ) 2 + ( − 1.5 − 0 ) 2 + ( 6 − 0 ) 2 = 42.25 = 6.5 m Substitute (–6 m) for x B , 0 m for x A , (3 m) for y B , 0 m for y A , (–6 m) for z B , and 0 m for z A in Equation (III). r A B = ( − 6 − 0 ) i + ( 3 − 0 ) j + ( − 6 − 0 ) k = ( − 6 i + 3 j − 6 k ) m Substitute (–6 m) for x B , 0 m for x A , (3 m) for y B , 0 m for y A , (–6 m) for z B , and 0 m for z A in Equation (IV). | r A B | = ( − 6 − 0 ) 2 + ( 3 − 0 ) 2 + ( − 6 − 0 ) 2 = 81 = 9 m Substitute (–2 m) for x C , 0 m for x A , (3 m) for y C , 0 m for y A , (–6 m) for z C , and 0 m for z A in Equation (V). r A C = ( − 2 − 0 ) i + ( 3 − 0 ) j + ( − 6 − 0 ) k = ( − 2 i + 3 j − 6 k ) m Substitute (–2 m) for x C , 0 m for x A , (3 m) for y C , 0 m for y A , (–6 m) for z C , and 0 m for z A in Equation (VI)