Tuesday, January 21, 2020
Science :: essays research papers
Addition of Torques Objective:      To ascertain equilibrium of the meter stick. Doing so by finding missing variables consisting of torque, length, weight and mass. Record all results and compare to calculated results. Procedure:      (Lab part A) †¢Ã‚     A fiberglass meter stick is to be used. Suspend this meter stick using string. †¢Ã‚     Hang 100 gram weight from the meter stick with a string a the 10 cm point on the meter stick. †¢Ã‚     Move the loop that suspends the meter stick left or right horizontally until the meter stick balances. (with the 100 g weight still attached at the 10 cm point) Procedure:      (Lab part B) †¢Ã‚     Place a string at 65 cm to support the meter stick. †¢Ã‚     Find the torque produced by the off centered string support by hanging weights on the shorter end of the meter stick to make it balance. †¢Ã‚     Take found torque and calculate mass to be placed at the 15 cm mark in order to balance the meter stick. †¢Ã‚     Hang weights to meter stick at the 15 cm location until the meter stick acquires equilibrium to prove your calculations. Procedure:      (Lab part C) †¢Ã‚     Suspend a meter stick with string placed at the 65 cm point. †¢Ã‚     Hang 100 grams of weight at the 45 cm mark, and 500 grams at the 90 cm mark on the meter stick. †¢Ã‚     Hang 200 grams of weight between 0 – 45 cm mark and move this weight until equilibrium is achieved. Record this measurement. Data Part A:           Mass of weight (m-2) = 100 grams Position string balanced = 36.4 cm Distance from center of meter stick to balance point. (L-1) = 13.6 cm Distance from balance point to suspended weight. (L-2) = 26.4 cm Mass 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 Part B:           Found natural torque (off set support string) = t = fl 85 grams placed at 100 cm balanced the off set support string at 65 cm. Therefore: t = 85 * (100 – 65) t = 2975 Total torque of right side of support string: t = 90cm – 65cm (500 g) t = 12,500 Then we calculated the left side torque: t = 65cm – 40cm (100g) t = 2500      Then we took the right torque and subtracted the left torque: 9525 – 2500 = 7025 (this is the missing force on the left side) Missing torque 7025 = 50cm ( ? ) 7025/50 = 140.5grams Calculate weight to be placed at 15cm. = 140.5 grams Data Part C:
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