The cable used to provide temporary support of the retaining wall extends from p

Question

The cable used to provide temporary support of the retaining wall extends from point B through a metal loop attached to the wall at A to point C. The rope exerts forces FAB and FAC on the loop at A with magnitudes equal to 200 lb each (the sense of each force is oriented out of A to the ground anchors at B and C). Obstruction due to underground electric lines requires that the x coordinate location of point C must be three feet as shown in the figure, as opposed to 1.5 ft for point B. Determine the magnitude of the resultant of the two forces acting on the metal loop, and the required y coordinate of point C if the resultant force must not have an x component (i.e., FR,x = 0) to achieve maximum support for the wall. The y-axis runs perpendicular to the plane of the wall, whereas the x- and z-axes run parallel to the plane of the wall.

Rab= (1.5, 2, -5) ft and Rac=(-3,y,-5)ft

Explanation / Answer

The Forces in Vector forms are Fab = 200 * ( 1.5 i + 2 j - 5 k)/sqrt(1.5^2 + 2^2 + 5^2) Fac = 200 * ( - 3 i + y j - 5 k)/sqrt(3^2 + y^2 + 5^2) Unit vector in x - direction is x = 1 i + 0 j + 0 k for resultant force in x direction to be zero, Dot product of ( Fac , X) + Dot product of (Fab,x ) = 0 1 * 200 * -3/sqrt(3^2 + y^2 + 5^2) + 0 + 0 + 1 * 200 * 1.5/sqrt(1.5^2 + 2^2 + 5^2) = 0 2 * sqrt( 1.5^2 + 2^2 + 5^2) = sqrt( 3^2 + y^2 + 5^2) y = sqrt ( 4 * ( 1.5^2 + 2^2 + 5^2) - 3^2 - 5^2) = 9.539 ft Now y = 9.539 ft To get resultant force we add Fab and Fac Fab + Fac = 200 *(( 1.5 i + 2 j - 5 k)/sqrt(1.5^2 + 2^2 + 5^2) + ( - 3 i + 9.539 j - 5 k)/sqrt(3^2 + 9.539^2 + 5^2) Fab + Fac = 200*[( 1.5/5.59 - 3/11.18) i + (2/5.59 + 9.539/11.18) j - (5/5.59 + 5/11.18)] Fab + Fac = 200 * ( 0 i + 1.211 j - 1.342 k) so magnitude of resultant = 200 * sqrt(1.211^2 + 1.342^2) = 361.52 N

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