Rigid Body Equilibrium This is an example of how to set up the free-body diagram for a rigid object, identify the forces, and use the conditions net force equal zero and net torque about any axis equals zero to create the equations that can then be solved for three unknown. By Leo Takahashi, PSU Beaver Campus

PB = L ; PA = D ; PC = S Weight of Sign = W Weight of Pole = mg steel cable  Thin chain Pole’s Center of gravity  This is the problem: a sign of weight W is suspended from a non-uniform rigid pole of length L and mass m that is held in equilibrium by a cable and a frictionless pivot C P Frictionless Pivot

FT W fx fy mg   B A C P  PB = L ; PA = D ; PC = S B Weight of Sign = W Weight of Pole = mg A   +x B C P FT A  W  +y The pole is modeled as a line; the vector placement is animated so that the students can try to do each one before it is shown on the slide. The action-at-a distance force vector is placed at the center of gravity of the pole; the contact forces are located and then their vectors are added, with the pivot force being shown as a pair of perpendicular components because both its magnitude and direction are unknown; the xy coordinate system is chosen and the pivot force components are then x and y components. C fx fy  mg P

Fx = 0 Fy = 0 P = 0 C = 0 A = 0 B = 0 PB = L ; PA = D ; PC = S B +x Fx = 0 Fy = 0 FT A  W  +y C P = 0 fx fy  mg C = 0 A = 0 B = 0 The sum of the torques is zero about any axis; the P axis is chosen so that the torque equation will not include the unknown pivot force components. P

+x +y Fx = 0 Fy = 0 P = 0 FT W fx fy mg L FT  W D  +y S fx fy  mg P Fx = 0 fx – mgCos - FTCos - WCos = 0 After getting to this slide showing only the FBD, give the students a chance to write the three equations showing (one at a time by clicking) them here. Fy = 0 fy – mgSin + FTSin - WSin = 0 -(mgS)Sin + FTDSin - WLSin = 0 P = 0

Fx = 0 Fy = 0 P = 0 fx – mgCos - FTCos - WCos = 0 fy – mgSin + FTSin - WSin = 0 P = 0 Fx = 0 -(mgS)Sin + FTDSin - WLSin = 0 fx – mgCos - FTCos - WCos = 0 If m, g, W, S, D, L, , and  are known, this set of equations can be solved for the three unknowns, fx , fy , and FT. The magnitude of the pivot force can be calculated from its components, and the angle the pivot force makes with the pole (+x axis) is P= Tan-1(fy/fx).