A flexible chain weighing 42.0 N hangs between two hooks located at the same height (Fig. P12.19). At each hook, the tangent to the chain makes an angle = 41.5° with the horizontal. (a) Find the magnitude of the force each hook exerts on the chain. (b) Find the tension in the chain at its midpoint. So the hook exerts both a X and Y force. Fx=Cos(41.5)(42)=31.5 Fy=Sin(41.5)(42)27.4 using pythagorean theorem you get total force=42N, or you could use cos^2+sin^2=1 to get the same answer..so 21N on each hook?