The torque τ of a force has a magnitude that is given by the magnitude F of the force times the lever arm : = F . The lever arm is the perpendicular distance between the line of action of the force and the axis of rotation. A rigid body is in equilibrium if it has zero translational acceleration and zero angular acceleration , so the net external force and the net external torque acting on the body are zero. For forces acting only in the x , y plane, the conditions for equilibrium are . The center of gravity of a rigid object is the point where its entire weight can be considered to act when calculating the torque due to the weight. For a symmetrical body with uniformly distributed weight, the center of gravity is at the geometrical center of the body. When a number of objects whose weights W 1 , W 2 , . . . are distributed along the x axis at locations x 1 , x 2 , . . . , the center of gravity is located at x cg = ( W 1 x 1 + W 2 x 2 + . . . )/ ( W 1 + W 2 + . . . ). The center of mass
This is the end of the preview. Sign up
access the rest of the document.
This note was uploaded on 04/09/2008 for the course PHYS 101 taught by Professor Sharp during the Spring '08 term at Ohio State.