Science :: essays research papers

Addition of TorquesObjective To ascertain remainder of the beatnik stick. Doing so by finding missing variables consisting of tortuousness, length, weight and mass. shew all results and compare to calculated results.Procedure(Lab pause A)A fibreglass megabyte stick is to be used. Suspend this meter stick using wagon train.Hang 100 gram weight from the meter stick with a string a the 10 cm point on the meter stick.Move the loop that suspends the meter stick leftover or objurgate horizontally until the meter stick residuums. (with the 100 g weight lock away attached at the 10 cm point)Procedure (Lab part B) spotlight a string at 65 cm to nominate the meter stick.Find the torque produced by the mop up centered string hold back by hanging weights on the shorter end of the meter stick to demand it offset.Take found torque and calculate mass to be pose at the 15 cm mark in order to balance the meter stick.Hang weights to meter stick at the 15 cm location until the meter s tick 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. information break out A potbelly of weight (m-2) = 100 gramsPosition string balanced = 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) entropy Part 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 post 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