Explanation Given information : The given tensile force value is 450 N. Show the free body diagram of the forces acting on the rope system as in Figure (1). Using Figure (1), 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 (I) Here, the position vector for point B is r B , the position vector for point A is r A , 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 , the distance of point B from the origin along z direction is z B , 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 AB . | r A B | = ( x B − x A ) 2 + ( y B − y A ) 2 + ( z B − z A ) 2 (II) 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 (III) 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 (IV) Determine the position vector for point AD . r A D = r D − r A = ( x D i + y D j + z D k ) − ( x A i + y A j + z A k ) = ( x D − x A ) i + ( y D − y A ) j + ( z D − z A ) k (V) Here, the position vector for point D is r D , the distance of point D from the origin along x direction is x D , the distance of point C from the origin along y direction is y D , and the distance of point D from the origin along z direction is z D . Determine the modulus of position vector for point AD . | r A D | = ( x D − x A ) 2 + ( y D − y A ) 2 + ( z D − z A ) 2 (VI) Force in Rope AB : F A B = F A B [ r A B | r A B | ] (VII) Here, the force in rope AB is F A B . Force in Rope AC : F A C = F A C [ r A C | r A C | ] (VIII) Here, the force in rope AC is F A C . Force in Rope AD : F A D = F A D [ r A D | r A D | ] (IX) Here, the force in rope AD is F A D . Write the Equation of equilibrium condition. Σ F = 0 F A B + F A C + F A D + F = 0 (X) Conclusion : Substitute (–1.5 m) for x B , 0 m for x A , (–2 m) for y B , 0 m for y A , (–6 m) for z B , and 0 m for z A in Equation (I). r A B = ( − 1.5 − 0 ) i + ( − 2 − 0 ) j + ( − 6 − 0 ) k = − 1.5 i − 2 j − 6 k Substitute (–1.5 m) for x B , 0 m for x A , (–2 m) for y B , 0 m for y A , (–6 m) for z B , and 0 m for z A in Equation (II). | r A B | = ( − 1.5 − 0 ) 2 + ( − 2 − 0 ) 2 + ( − 6 − 0 ) 2 = 42.25 = 6.5 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 (III). r A C = ( 2 − 0 ) i + ( − 3 − 0 ) j + ( − 6 − 0 ) k = 2 i − 3 j − 6 k 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 (IV)