A uniform gantry AB AB\,AB has length 9 m and mass 2.4 kg. A load of weight 4W N is attached at AAA, and a load of weight W W\,W N is attached at BBB. The gantry is held horizontally in equilibrium by a single vertical support cable attached at a point z z\,z m from AAA.
Show that z=9W+10.8g5W+2.4gz=\dfrac{9W+10.8g}{5W+2.4g}z=5W+2.4g9W+10.8g.