MecMovies 4.0 : M5.7: Rigid Bar and Two Axial Members

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The figure shows the free body diagram of the rigid bar A B C D pinned at A and supported by vertical rods labeled 1 and 2 at B and C. The separation between A and B is 380 millimeters. The separation between B and C is 380 millimeters and the separation between C and D is 760 millimeters. The load at D is P which equals 36 kilo newtons directed vertically downwards. The free body diagram cuts through the vertical members 1 and 2. The reaction forces on the members 1 and 2 are F subscript 1 and F subscript 2 directed vertically upwards.

$\sum_{} M_{A} = F_{1} (380 mm) + F_{2} (760 mm) - P (1, 520 mm) = 0$

Begin the solution of this problem by considering equilibrium. Draw a free-body diagram of rigid bar $AD$ and write an equilibrium equation for $\sum_{} M_{A}$ .