Calculate the position a 200 g mass would have to be if a 100 g mass were placed at the 5 cm mark. Use the center of mass of the meter stick you found above in your calculations. Be sure to make a free body diagram before doing your calculations. Use provided data from lab

1. The masses from step 1 and step 2. Meter stick mass m,: 0.0887 Kg Meter stick center of gravity x,: 0.50 m Meter stick weight w,: 0.8708(N) m.: 0.0213 Kg mgs: 0.0213 Kg 2. The fulcrum location from step 5. X 0671 m 3. The table from the top of page 3. Mass (Kg) X (X-%) T(N m) Direction of (weight+clamp) | (m) (m) torque 0.5+0.0213 = 0.750 0.079 -0.0552 0.0713 0.5+0.0213= 0.900 0.229 -0.0927 cw 0.0413 0.500 -0.171 0.149 CCw 0.0887 T, o 20:0952-0.0927+0.149 = 0.0011 Nm to 4. The fulcrum for part I, step 2. X 0.277 m 5. The table and total torque for part |l from step 4. Mass Weight Force X Direction of (Kg) (m) Torque (Weight + Clamp) 0.5+0.0213 = 0.69874 0.05 -0.227 0.159 CCw 0.0713 0.5+0.0213= 0.40474 0.19 0.0413 0.50 0.0887 0.86926 Total: 0.159+0.0352-0.194 = 0.0002 (N m) 6. the x2 location for part lll step 3 and the table and total torque for part ll| step 5. X?:0.824 m Total: 0. 159+0.0352-0.194 = 0.0002 (N m) 6. the x2 location for part III step 3 and the table and total torque for part III step 5. X2: 0.824 m Xf = 0.500 Mass Weight Force X (X-XF) T(N m) (weight+clamp) (m) (m) (Kg) 0. 100+0.0213 = 0.980 0.05 0.45 -0.674 0.1213 0.150+0.0213 =[ 1.47 0.824 -0.324 -0.552 Total: -0.674 + -0.552 = -1.23 (N m)