Science
Essay by 24 • September 6, 2010 • 774 Words (4 Pages) • 1,883 Views
Addition of Torques
Prepared for
Frank Reed
Technical Physics I
Indian Hills Community College
By
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
Then
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