I am having troubles with this physics question, I'm not really sure on how to set it up and everything. question: A mass M of 3.60×10-1 kg slides inside a hoop of radius R=1.20 m with negligible friction. When M is at the top, it has a speed of 5.33 m/s. Calculate size of the force with which the M pushes on the hoop when M is at an angle of 36.0 degrees. (see attachment). my answer: Centripetal a = v^2/r = (5.33 m/s)^2 / 1.20 m = 23.7 m/s^2 Normal force = y component of gravity (if we take y axis to be perpedicular to the tangent of the hoop at that point). = mgcos(36.0) = (3.60×10-1 kg)(9.81 N/kg)(cos36.0) =2.86 N If anyone could help me, I'm not sure what I'm suppose to do. 2.86 N is not the answer, I know. But I'm so confused. Thanks.