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)
• 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.
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 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:
m= 10km2 x 1000m x 1000m = 107m2 107m2 x 15= 1.5 x 1.8m3 = 1.5 x 1011kg
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