Addition of Torques Prepared for Frank Reed Technical Physics I Indian Hills Community CollegeByChad Boal Roy SlaymakerLarry Fox Rebecca Hopkins Meagan 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) o A fiberglass meter stick is to be used.
Suspend this meter stick using string. o Hang 100 gram weight from the meter stick with a string a the 10 cm point on the meter stick. o 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) o Place a string at 65 cm to support the meter stick. o 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.
o Take found torque and calculate mass to be placed at the 15 cm mark in order to balance the meter stick. o Hang weights to meter stick at the 15 cm location until the meter stick acquires equilibrium to prove your calculations. Procedure: (Lab part C) o Suspend a meter stick with string placed at the 65 cm point. o Hang 100 grams of weight at the 45 cm mark, and 500 grams at the 90 cm mark on the meter stick. o Hang 200 grams of weight between 0 – 45 cm mark and move this weight until equilibrium is achieved. Record this measurement.
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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) m 1 = m 2 (L 1/ L 2) Therefore: m 1 = 100 (26. 4/13. 6) m 1 = 100 (1. 94111) m 1 = 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 = 90 cm – 65 cm (500 g) t = 12, 500 Then we calculated the left side torque: t = 65 cm – 40 cm (100 g) 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 = 50 cm (? ) 7025/50 = 140. 5 grams Calculate weight to be placed at 15 cm. = 140.
5 grams Data 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 = 40 cm (47 g) t = 1880 Right side torque: t = 90 cm – 60 cm (500 g) t = 15000 Subtracted the offset torque from the right side torque. = 15000 – 1880 = 13120 Right side torque total = 13120 Left side torque: t = 60 cm – 45 cm (100 g) t = 1500 Total left side torque = 1500 Then we subtracted left side torque from right side torque to find missing torque. 13120 – 1500 = 11620 Missing torque = 11620 t = f l therefore: 11620 = 200 g (? cm) This formula produced the placement from the support line (60 cm) which calculated to be 58. 1 cm from 60 cm. Which yielded a calculation that the 200 g weight would have to be place at 1.
9 cm. Calculated: 1. 9 cm Measured: 1. 25 cm Conclusion: 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.
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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..
