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Rigid Body Equilibrium

Rigid Body Equilibrium. By Leo Takahashi, PSU Beaver Campus. PB = L ; PA = D ; PC = S. Weight of Sign = W. B. Weight of Pole = mg. A. steel cable. . Thin chain. Pole’s Center of gravity . . C. P. Frictionless Pivot . PB = L ; PA = D ; PC = S. B. Weight of Sign = W.

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Rigid Body Equilibrium

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  1. Rigid Body Equilibrium By Leo Takahashi, PSU Beaver Campus

  2. PB = L ; PA = D ; PC = S Weight of Sign = W B Weight of Pole = mg A steel cable  Thin chain Pole’s Center of gravity  C P Frictionless Pivot

  3. PB = L ;PA= D ; PC = S B Weight of Sign = W Weight of Pole = mg A   +x B C P FT A  W  +y C fx fy mg  P

  4. PB = L ; PA = D ; PC = S +x B Fx = 0 Fy = 0 FT A  W +y C P = 0  fx fy mg C = 0 A = 0 B = 0  P

  5. +x L FT  W D +y S fx fy mg   Fx = 0 P fx – mgCos - FTCos - WCos = 0 Fy = 0 fy – mgSin + FTSin - WSin = 0 P = 0 -(mgS)Sin + FTDSin - WLSin = 0

  6. 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. Fx = 0 fx – mgCos - FTCos - WCos = 0 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). Fy = 0 fy – mgSin + FTSin - WSin = 0 P = 0 -(mgS)Sin + FTDSin - WLSin = 0

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