need answer asap with work

10. Equilibrium of a Rigid Body Accessibility Mode Print EXPERIMENT Note: the center of gravity of a uniform stick is at its geometric center (50cm mark) PART 1: PIVOT POINT AT THE CENTER OF GRAVITY (c.E) X M. = 200kg My = 100kg 1. Set the pivot point at the center of the stick Xom =0 .5 (or 50cm) and choose the stick mass M, = 100kg- 2. Place M1= 200 kg mass at x1= 0.3 (30cm) 3. Find the position where M. = 100 kg will produce equilibrium (zero net torque) 4. Determine the distance X between the Ma and the pivot point