# Science

Essay by   •  September 6, 2010  •  774 Words (4 Pages)  •  1,680 Views

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Prepared for

Frank Reed

Technical Physics I

Indian Hills Community College

By

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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