**Archery Mechanics: Analyzing Forces in the Bow**An archer aiming at a target is pulling with a 64-lb force on the bowstring.**Diagram Explanation:**The diagram shows an archer standing and holding a bow. The bow is drawn with a force of 64 lbs. The diagram includes measurements that relate to the positions and forces involved in the action:- **A** is the point where the archer holds the bowstring.- **B** is directly above A, 8 inches up.- **C** is 24 inches to the right of A, at the midpoint of the bowstring.- **D** is directly below point C, 32 inches down.- Points **J** represents a point along the bow where internal forces will be analyzed, 16 inches above and below point C.Assuming that the shape of the bow can be approximated by a parabola, determine the internal forces at point J.1. **Axial Force \( F \)** at point J: (Enter value and angle) 2. **Shear Force \( V \)** at point J: (Enter value and angle) 3. **Bending Moment \( M \)** at point J: (Enter value and angle)

Free Body diagram showing all forces and tension can be drawn as: The angle θ, can be calculated as: θ=tan-13224=53.13o Since the tension forces in AB and AD are in symmetry. ⇒T1=T2 Now, taking summation of all forces in x direction, we get ∑Fx=0-45+2T1Cos53.13o=0⇒2T1Cos53.13o=45⇒T1=452×0.6=37.5 lb Since the shape of the bow taking the shape of parabola, So, let the curve equation for portion CJB be y = ax2-------------(1) Differentiating the above equation,we get dydx=2ax----------------------------------------------------------(2) At point B, the coordinates are:-x =32", y = 8" Substituting the values of a and b in parabolic equation (1), we get 8=a×322⇒a=8322=1128in.The coordinates and slope at the point J can be calculated as: At J,xj=16"From equation1,yj=a162Substitute a = 1128in., we getyj=1128162=2" Therefore, at point J, coordinates are x =16" and y = 2". Now, slope at point J can be calculated as: θ=tan-1dydxBut from equation2, dydx=2ax⇒θ=tan-12axjSubstitute a =1128in. and xj=16 in⇒Slope at J =θJ=tan-12×1128×16So, θJ=tan-10.25⇒θJ=14.036o To determine the internal forces at point J, consider FBD of BJ: Taking equilibrium of all forces in x direction, we get ∑FX=0⇒-37.5×Cosθ+VCosθj-FCos90-θj=0⇒-37.5×Cos53.13o+VCos14.036o-FCos90-14.036o⇒-22.5+0.97014V-0.2425F=00.97014V-0.2425F=22.5--------------(3) Taking equilibrium of all forces in Y direction, we get ∑FY=0⇒-37.5×Sinθ+VSinθj+FSin90-θj=0⇒-37.5×Sin53.13o+VSin14.036o+FSin90-14.036o⇒-29.99+0.24254V+0.9701F=0or 0.24254V+0.9701F=29.99--------------(4) On solving equation, 3 and 4, we get V = 29.01 lb and F = 23.63 lb Now, taking moments about J, we get ∑MJ=0⇒37.5×Cos53.13o16+37.5×Sin53.13o8-2+M=0⇒360+180.39+M=0⇒M=-540.39 lb-in. Therefore, the bending moment at point J on the member CJ = 540.39 lb-in (clockwise direction).