I have a general question about solving these types of problems. I had two different problems for calculating the tensions and compressions of a pin-jointed truss. both problems were of a trapezoidal shape with 5 pins in total. in one problem, a weight was applied to the center truss on the center of the bridge. on another problem, the weight was in the middle of the left beam. here is where i get confused. both problems involve solving the normal forces on the far left and far right pin which is easy enough knowing the dimensions of the truss bridge and the weight by taking a net torque equaling zero equation. my problem is when calculating the sum forces on the far left and far right pins. in the first problem, there is only the normal force and two beams creating equilibrium. in the second problem ( do to issues with sheer stress???) there is an extra force included in the free body diagram to find the results. i dont get why the first calculation involves just three forces in the equilbrium on the far pin, while when in the second problem the weight between two pins creates a need to add a partial force. if you could explain this, it would be much appreciated. i should mention that in the second problem, the first thing asked is whether the weight is equally distributed between the two pins (since the weght of the object is equidistant form both).