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Addition of TorquesObjective To ascertain equilibrium of the mensuration stick. Doing so by finding missing variables consisting of tortuousness, length, weight and mass. Record all told results and compare to calculated results.Procedure(Lab part A)A fiberglass bar stick is to be used. obviate this prison term stick using puff. sticktle 100 gram weight from the molarity stick with a pull in a the 10 cm point on the measuring stick stick. roleplay the loop that suspends the meter stick left or right horizontally until the meter stick balances. (with the 100 g weight still wedded at the 10 cm point)Procedure (Lab part B)Place a string at 65 cm to support the meter stick. chance upon the crookedness produced by the off centered string support by hanging weights on the shorter end of the meter stick to make it balance. affect found torque and calculate mass to be placed at the 15 cm mark in order to balance the meter stick.Hang weights to meter stick at the 15 cm muddle un til the meter stick acquires equilibrium to prove your calculations.Procedure(Lab part C)Suspend a meter stick with string placed at the 65 cm point.Hang 100 grams of weight at the 45 cm mark, and 500 grams at the 90 cm mark on the meter stick. Hang 200 grams of weight between 0 45 cm mark and move this weight until equilibrium is achieved. Record this measurement.Data map AMass of weight (m-2) = 100 gramsPosition string equilibrate = 36.4 cmDistance from center of meter stick to balance point. (L-1) = 13.6 cmDistance from balance point to suspended weight. (L-2) = 26.4 cmMass of meter stick. (at center gravity) m1 = m2 (L1/ L2)Therefore m1 = 100 (26.4/13.6) m1 = 100(1.94111) m1 = 194.1176 grams (mass of the meter stick)Data trigger B Found natural torque (off set support string) = t = fl85 grams placed at 100 cm balanced the off set support string at 65 cm.Therefore t = 85 * (100 65) t = 2975Total torque of right side of support stringt = 90cm 65cm (500 g)t = 12,500 Then we calculated the left side torquet = 65cm 40cm (100g)t = 2500 Then we took the right torque and subtracted the left torque9525 2500 = 7025 (this is the missing force on the left side)Missing torque 7025 = 50cm ( ? )7025/50 = 140.5gramsCalculate weight to be placed at 15cm. = 140.5 gramsData Part C