Addition stick left or right horizontally until the

Addition of TorquesPrepared for Frank ReedTechnical Physics IIndian Hills Community CollegeByChad BoalRoy SlaymakerLarry FoxRebecca HopkinsMeagan RequenaObjective: 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)A fiberglass meter stick is to be used. Suspend this meter stick using string.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 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)Place a string at 65 cm to support the meter stick.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.Take 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 location until the meter stick acquires equilibrium to prove your calculations.

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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 Part A:Mass 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)Data 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 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 = 140.5gramsCalculate weight to be placed at 15cm. = 140.

5 gramsData Part C:Calculated off set torque:We found it took 47 grams at 100 cm to balance off set meter stick.Natural off set torque = t = 40cm(47g) t = 1880Right side torque: t = 90cm 60 cm (500 g)t = 15000Subtracted the offset torque from the right side torque. = 15000 1880 = 13120Right side torque total = 13120Left side torque: t = 60 cm 45 cm (100 g)t = 1500Total left side torque = 1500Then we subtracted left side torque from right side torque to find missing torque.

13120 1500 = 11620Missing torque = 11620t = f l therefore: 11620 = 200g( ? cm )This formula produced the placement from the support line(60cm) which calculated to be 58.1 cm from 60 cm. Which yielded a calculation that the 200g weight would have to be place at 1.

9 cm.Calculated: 1.9 cmMeasured: 1.25 cmConclusion:Our team believes that all of our objectives were met. We differentiated from the actual lab process, primarily because there were a few easier ways to find hidden variables.

For example to find hidden torque on the meter stick when the string was off set from the center we just hung weight on the end of the meter stick to balance it. Then we took the weight times the difference of the support string and 100 cm to find the off set torque. Other than that we followed the books write up on the procedures. Our errors were trying to calculate the hidden torque of the unbalanced meter stick. We finally just placed weight as I mentioned before, to help calculate this. Also we were pleasantly surprised to find how close our calculations actually were. Words/ Pages : 774 / 24

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