 In this problem we have a rigid frame A, B, C, D which has been supported at point C and which is restrained by two known rigid rods, so cables, at point A and B. We have to find what are the reaction forces in the rods A and B, giving a vertical load B applied at D. The rods are identical and they have a jammed modulus E and cross-sectional area A and length L. So the question is what is F A and F B? In order to do this as usually we can start drawing the free body diagram of the problem. We have here the applied force B and the reaction forces F A, F B, R X and R Y. Remember this has been supported so here the displacement in the vertical and horizontal direction is constrained. But now we can apply the equilibrium equations in order to find the values of these reactions in terms of the force B. First we have that the sum of forces in the horizontal direction is equal to zero. Then we have that and this is our first equation. We have also that the sum of forces in the vertical direction is equal to zero and from here we directly obtain that R Y is equal to minus B and this is our second equation. Finally we have that the sum of moments for instance at point C is equal to zero. Then we have that and if we rearrange this equation we have that and this is our third equation. Then we have three equations but one, two, three, four unknowns. Then we need to find another additional equation in order to solve this problem. As you can see here we have three equations but four unknowns. As you can see here we have three equations but one, two, three, four unknowns. Then because there is one redundancy in the problem we need to find one displacement compatibility equation. Then in order to find the displacement compatibility equation that we need we're going to start considering a rotational deformation of this rigid frame ABCD which is rotating over respect to point C. Then we have something like this. Then the deformation of the cable A delta A is equal to the length of this segment A A prime and the deformation of the cable B delta B of this segment B B prime. If we look at the geometry this angle is theta. Then this is equal to 2B times the tangent of this angle. And also we have here that this length B B prime is equal to B this length times the tangent. We can approximate the tangent to be more or less equal to the angle. Then we find that delta A is equal to 2 times times B times the angle and delta B is equal to B times the angle. So from here we find that delta A is equal to 2 times delta B and this is the fourth equation that we were looking for. Now if we use the force displacement relationship we know that for an axial loaded member delta is equal to P L divided by A E. So if we apply this equation into equation 4 we find that F A times L divided by A E is equal to 2 times F B times L divided by A E. So remember that the area, the modulus of elasticity and the initial length of those cables are identical. So from here we have that F A is equal to 2 times F B. Now we can substitute this result into equation 3. Remember that equation 3 says that 2 times F A plus F B is equal to 2 B. So if we plug this result here we have that 2 times 2 F B plus F B is equal to 2 B. Therefore we find the first result which says that F B is equal to 2 over 5 times B. Then now if we know F B we can use this equation in order to find that F A is equal to 2 times F B. So this is equal to 4 over 5 B.