A nonuniform bar is suspended at rest in a horizontal positi

A nonuniform bar is suspended at rest in a horizontal position by two massless cords as shown in below. One cord makes the angle theta = 36.4degree with the vertical; the other makes the angle = 53.6degree with the vertical. If the length L of the bar is 6.10 m, compute the distance x from the left end of the bar to its center of mass.

Solution

x = L cos phi / [cos phi + sin phi * cot theta]

= [6.10 * cos 53.6] / [cos 53.6 + (sin 53.6 * cot 36.4)]

= [6.10 * cos 53.6] / [0.593 + (0.805 * 1.356)]

x = 2.147 m