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# Force Table Lab and Scientific Method

By Roberto Harris,2014-04-13 10:48
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Force Table Lab and Scientific Method

Force Table Lab and Scientific Method

Technical Physics PHY-100

Instructor: Kevin Kimball

Author: Binshan Zheng

26/10/10

Proposal

Using given formulas, this lab has the following goals:

1. Observe the static system of vectors

2. Determining the Equilibrium Vector of the vectors model

List of Equipment

;Force Table ;Weight

Procedure #1

Observe the static system of force vectors:

Hypothesis:

a. In a static system of forces, any one force vector is equal and opposite

to the resultant of all remaining force vectors.

b. In a static system of forces, the sums of the components equal zero.

1. Set up force table with 4 vectors in equilibrium

a. Set 3 angles selected at random (V, V, and V). 123

b. Place weights on each hanger. Vary the weight totals per hanger.

V=150g, 65? 1

V=350g, 170? 2

V=160g, 291? 3

c. Set the fourth angle (V) by applying tension and moving it back and forth 4

until the ring “floats” around the central post, and lock it in position;

remove tension and add weights until the ring is restored to floating

aspect.

V=230g, 345? 4

2. Analyze/ organize the data

a. Convert grams to kilograms, kilograms to Newtons

b. Graph the vectors on a Cartesian system

c. Using right-angle trigonometry, determine A and O components of

vectors V, V, V, and V 1234

d. Record results

e. Compare results to hypothesis

Calculations

-3V= m*a = (150*10)*9.8 = 1.47N, 65? 11

-3V= m*a = (350*10)*9.8 = 3.43N, 170? 22

-3V= m*a = (160*10)*9.8 = 1.57N, 291? 33

-3V= m*a = (230*10)*9.8 = 2.25N, 345? 44

A= cos65*1.47 = 0.62N A= cos69*1.57 = 0.56N 13

O= sin65*1.47 = 1.33N O= sin69*1.57 = 1.46N 13

A= cos10*3.43 = 3.38N A= cos15*2.25 = 2.18N 24

O= sin10*3.43 = 0.60N O= sin15*2.25 = 0.58N 24

Direction Magnitude Direction Magnitude

A ? +0.62 O ? +1.33 11

A ? -3.38 O ? +0.60 22

A ? +0.56 O ? -1.46 33

A ? +2.18 O ? -0.58 44

A ? -0.02 O ? -0.11 NETNET

A none 0 O none 0 HYPHYP

Conclusion

It was hard to tell if the post was in the exact center of the ring so the slight variation

may have caused some small errors. This lab proved that the hypothesis is correct, in a

static system of forces, the sums of the components equal zero.

Procedure #2

Determining the Equilibrium Vector of the vectors model:

1. Construct a model

a. On a Cartesian graph, construct 3 force vectors (V, V, and V) that are not 123

in equilibrium

V=1.0N, 45? 1

V=1.0N, 150? 2

V=0.5N, 190? 3

b. Using right-angle trigonometry, determine resultant of V, V, and V 123

2. Predict Equilibrium Vector (V) eq

V=1.3N, 300? eq

3. Test

a. Set up force table angles according to model and predictions

b. Set weights on each hanger (convert Newtons to kilograms, kilograms to

grams)

? the hanger weight 50g

?m1 = 102g 50g = 52g

m2 = 102g 50g = 52g

m3 = 51g 50g = 1g

meq = 133g 50g = 83g

4. Record results and compare to hypothesis

Calculations

V= 1.0N, 45? 1

V= 1.0N, 150? 2

V= 0.5N, 190? 3

A= cos45*1 = 0.707N A= cos10*0.5 = 0.49N 13O= sin45*1 = 0.707N O= sin10*0.5 = 0.08N 13

A= cos30*1 = 0.87N 2

O= sin30*1 = 0.50N 2

Direction Magnitude Direction Magnitude

? +0.707 A 1 ? +0.707 O1A ? -0.87 2 O ? +0.50 2A ? -0.49 3O ? -0.08 3

A ? -0.653 RO ? +1.127 R A ? +0.653 eqO ? -1.127 eq

tanΘ=1.127/0.653=1.726

-1tan(1.127/0.653)=59.9? ? 60?

V=1.127/sin60? = 1.301N R

?VR=1.3N,120 ?

?The angle of the equilibrium vector is the opposite of the resultant

?Veq=1.3N,300 ?

TEST:

?F=ma

?m=F/a

m = 1/9.8 = 0.102kg = 102g m = 0.5/9.8 = 0.051kg = 51g 13

m = 1/9.8 = 0.102kg = 102g m = 1.3/9.8 = 0.133kg = 133g 2eq

Conclusion

The test was very accurate overall but there was still room for error in a couple places. Firstly the force table is almost impossible to get completely centered so that causes some minor error.

Another problem was we couldn’t find the exact weight so it was a couple grams off.

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