Atwood’s Machine Data             Partner                     Name                    

 

y = ½ a t2; e.g. 100 cm = ½a(3.08s)2 ;  a = 21.1 cm/s2

y = _____ cm  (this can change in Part A,B,C, but must be recorded each measurement if changing.)

FNet  in    Dynes   à       “g” = 980 cm/s2  

FNet  in    Newtons à       “g” = 9.8 m/s2  

 

Part A: Total Mass Held Constant m1 + m2 = ____ grams      Plotting FNet vs acceleration

mass 1

mass 2

m1–m2

FNet

time

accel

Use LineFit:  Plot FNet vs accel

98+50 =

148 g

102+50 =

152 g

0.004 kg

9.8*0.004 =

0.0392 N

s

m/s2

y-int

(           )

x-int

(m/s2)

slope

(kg)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

mpulley

Ffriction

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Part B: Net Force Held Constant (m1 – m2 = _____ grams)         Plotting 1/a vs m1 + m2

mass 1

(grams)

mass 2

(grams)

time

(s)

accel

(       )

1/a

(s2/m)

m1 + m2

(       )

Hand Graph:  Never force line through origin

y-int

x-int

slope (from data)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

FNet à (m1 – m2) g

FNet  (from graph/data includes friction)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

mpulley

Ffriction

 

 

 

 

 

 

 

 

 

 

 

 

 

 

LineFit (or Excel) for comparison

y-int

x-int

slope

 

 

 

 

 

 

 

 

 

 

Part C: Predicting the mass distribution that will yield a given acceleration

eq. 4

(m1 – m2)g – Ff = mT a

mtotal

(grams)

mass1

(grams)

mass2

(grams)

Slope (cm/s2)

 

Predicted

  (calculated)

200

 

 

60.0

Experimental

  (Actual mass)

 

 

 

 

Since you have a limited number of masses in your mass holder set,

actual m1 & m2 usually varies slightly from calculated m1 & m2

For predicted Part C, please use mpulley from Part A and Ff from Part B;

Reasoning:  unknowns obtained from slopes typically have less uncertainty than from intercepts.