A 23 m-long log of irregular cross section is lying horizontally… Where is the logs center of gravity?
This is an equilibrium problem, meaning that if we sum the torques acting on the log the net torque equals zero. Only the weight and tension have the potential to produce torque, since the normal force supplied by the wall acts through the axis of rotation. Torque is given by the cross-product of the lever arm distance (distance from point of application of the force to axis of rotation) with the applied force. Because both the tension and the weight are perpendicular to this axis, the cross product is equivalent to the simple product of the distance and force. In the following equation, r represents the distance from the wall to the cable support, x the center of gravity, T the tension, and W the log’s weight: Torque = xW – rT = 0 Thus, x = rT/W = (19 m)*(5100 N)/(6000 N) = 16.2 m So the center of gravity of the log is 16.2 m from the wall.
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