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Addition of Torques
Prepared for
Frank Reed
Technical Physics I
Indian Hills Community College
Chad Boal
Roy Slaymaker
Larry Fox
Rebecca Hopkins
Meagan Requena
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