Atwood’s Machine
mcup1
=
grams
Partner
Data
mcup2
(In the example below, mcup1 = 5.2 g and mcup2
Name
FNet in
FNet in
= 5.9 g; Substitute in your own masses of the cups.)
=
grams
Part A: Total Mass Held Constant
mass 1
mass 2
m1–m2
FNet
71+5.2 =
70+5.9 =
0.300 g
294 dynes
76.2 g
75.9 g
980*0.3 =
“g” = 980 cm/s2
“g” = 9.8 m/s2
Dynes

Newtons 
Plotting FNet vs acceleration
slope
Use LineFit: Plot FNet vs accel
Example ONLY
1.31 cm/s2
(
y-int
x-int
slope
(cm/s2)
)
(grams)
mpulley
Ffriction
Don’t Forget Questions from
Part B: Net Force Held Constant (m1 – m2 = _____ grams)
mass 1
(grams)
mass 2
(grams)
accel
(
)
1/a
(s2/cm)
m1 + m2
(
)
A
&
B !!!
Plotting 1/a vs m1 + m2
Start at origin for graph by hand
y-int
x-int
FNet  (m1 – m2) g
slope
FNet
mpulley
(from data)
(from graph/data
includes friction)
Ffriction
LineFit for comparison
y-int
Part C: Predicting the mass distribution that will
yield a given acceleration
eq. 4
(m1 – m2)g – Ff = mT a
Predicted
(calculated)
mtotal
(grams)
100.0
mass1
(grams)
mass2
(grams)
Slope
(cm/s2)
60.0
x-int
slope
Part D: Predicting the acceleration of the
freely spun pulley
Experimental
acceleration
(3 values for uncertainty)
Predicted acceleration
Note: Ff pulley = mpulley apulley
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
a = -____ ± ___
a = -______ cm/s2
For predicted Part C and Part D, please use mpulley from Part A and Ff from Part B;
Reasoning: unknowns obtained from slopes typically have less % uncertainty than unknowns obtained from intercepts.