We talked about stability, and we saw two examples. The example of a beam with several rollers. It was a case where obviously we could see that the structure is unstable, because we had the reactions which were all vertical and therefore the equation Sigma fx could not be satisfied. In the second example, we had the frame with support reactions which all where concurrent, all the three reactions happen to pass through the same corner B. Therefore, in that case, as long as there is an external load which creates a moment about that point, the reactions cannot contribute to pick up this load. So now, we're going to look at an example where it is not that obvious whether the flexor is stable or unstable. We consider a case where we have a structure comprised of 2-digit members, member AB and BCD which are connected with a hinge at B. Now, if we count the number of unknowns, let's see first of all the unknown reactions. In this case, we have a reaction Ay, a reaction Ax, a reaction Cy and Cx. So we have four reactions, three equations of equilibrium. However, since we have two rigid bodies, we must satisfy equilibrium of both individual members. That means, we must consider the two members, member AB with reactions Ax and Ay and at B, of course, we must show the forces that the right member access on the left member, Bx and By. Similarly, we could draw the free body diagram of the right member. It is not necessary. However, we could consider the overall equilibrium equations for the entire structure, then one of its members, then automatically the equilibrium of the right member would be also satisfied. So let us try to calculate this 1, 2, 3, 4, 5, 6 unknowns, using three equations of equilibrium here and three overall equation. So we have six equations with six unknowns. Let's take sum of the moments about point A for the entire structure. So we have that Cy times 4 minus 100 times 6 should be equal to 0. From here, we can solve for Cy and Cy turns out to be 150 Newton, so this is 150. From the equation, sum of the forces in the vertical direction must be equal to zero. We get that Ay plus Cy minus 100 must be equal to 0. So from here, we get that Ay is equal to minus 50, so this number is minus 50. Of course, we must also satisfy Sigma fx, Ax plus Cx is equal to 0. So these are the three overall equilibrium equations. We go now and consider the left member. We take the sum of the moments about point B, not for the entire structure but just for the member AB. Taking moments along this point, we get minus Ay times the distance of 2 should be 0, which gives 100 equal to 0, which clearly is not the case. So the equation, sum of the moments of member AB about point B is not satisfied. So this is an unstable structure because we cannot satisfy one of the equilibrium equations.