I'm trying to figure this code out in jupterlab The nodes of the following three-member truss have x-y coordinates (in feet) of A = [0,0], B = [20,2], and C = [10,10]. If the applied forces at the nodes are \( F_B = 1000 \) lbs and \( F_C = 1000 \) lbs in the directions shown, determine all the reactions and the member forces.**Diagram Explanation:**The diagram shows a triangular truss structure with nodes labeled A, B, and C. - Node A is located at the origin [0,0]. It is represented with a triangular support, indicating it is a fixed or pinned support.- Node B has a coordinate of [20,2]. It is shown with a roller support, indicating it can move horizontally.- Node C is at [10,10]. The truss members are represented by lines connecting the nodes:- A line connecting A and B.- A line connecting A and C.- A line connecting B and C.Applied forces are shown at nodes B and C:- \( F_B = 1000 \) lbs acts horizontally to the right at node B.- \( F_C = 1000 \) lbs acts vertically downward at node C.

Draw free body diagram Consider the moment equilibrium about point O 100010+10002-RB×20=0RB=600 lb Consider the summation of forces in the vertical direction RAy-RBy=1000RAy=400 lb Consider the summation of forces in the horizontal direction -RAx+1000=0(RA)=-1000 lb Draw Free body diagram 1.) At joint 'C' Consider the summation of forces in the horizontal direction FAC cos 45o-FBC cos 38.66o=0 0.71 FAC+0.62 FBC=0 ...(1) Consider the summation of forces in the vertical direction FAC sin 45o + FBC sin 38.66o=10000.71 FAC+0.62 FBC=1000 ...(2) by solving the equation 1 and 2, FAC=784.71 lb, FBC=714.29 lb 2.) At joint 'B' Consider the summation of forces in the horizontal direction FBC cos 38.66o-FAB cos 5.71o+1000=0714.29 cos 38.66o-FAB cos 5.71o+1000=0FAB=1565.53 lb