I need help with 1 2 3

A VIRTUAL EXPERIMENT Non-uniform Body - "Loaded" Meter Stick In this experiment, you will apply the Second Condition of Eguilibrium to the problem of predicting the weight of a loaded meter stick. You will determine weight by three different methods. The center of gravity of a loaded meter stick is located by balancing the meter stick on a knife- edge as described in the Discussion section. The distance Ffrom the zero end (the end of the stick which corresponds to zero on the l-meter scale along its edge) to the location of the center of gravity (CG) is found to eb: i=0.74-5m 1. In this virtual experiment we now attach a mass hanger to the string at the zero end of the stick and place about 0.45 kilograms of mass on the hanger. Note that this generates a force = mg applied to this point. To balance the meter stick, with the mass at the end, we nd that knife edge (the distances 1'1 below) is 0.45 m. This means that $1 + I2 = 0.745111, 50 that 12 = 0.3 m center of gravity of "loaded" meter stick 2. Imagine an axis of rotation that passes through the center of gravity of the loaded meter stick and is perpendicular to the meter stick. Does the weight (W) of the loaded meter stick contribute a torque about this axis of rotation? Explain (hint: the lever arm of W is zero.. .). 3. With this axis of rotation (through the CG), use the 2\"(1 Condition for Equilibrium (balanced torques) to solve for the support forge, FN, exerted on the meter stick by the knife-edge