A 7.6m bean is supported by two posts. Post A is on the left end of the beam. Post B is 2m to the right from post A. Them Beam weighs 1200 N. What forces must be exerted at A and B to maintain equilibrium? In what direction must each force act?

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anonymous · Answer

First, let's draw a FBD. This is always FIRST and foremost. |dw:1333845867547:dw|where w is the weight distribution of the beam, R is the resultant force of the distributed force, and x is the location from the left side of R. Take the reaction forces at A and B to be upwards. 

Second, we need to find the location of the reaction force. This is a simple case. The location of a reaction force coresponding to a square distributed force is half the length of the distribution, in this case, it is distributed over the length of the beam. $$x = {l \over 2} = {7.6 \over 2}$$The magnitude of R is simply the weight. $$R = 1200 ~\rm N$$

Finally, let's sum forces and torques. $$\sum F_Y = 0 \rightarrow F_A + F_B - R = 0$$$$\sum \tau_A = 0 \rightarrow F_B \cdot 2 - R \cdot x = 0$$We can solve the torque equation for $F_B$ and then solve the force equation for $F_A$. Note that $F_B$ will be directed upwards (positive) and $F_A$ will be downwards (negative).