Atwood's Machine
Purpose:
To demonstrate that for a system of fixed total mass, the ratio of the acceleration to the difference in masses is a
constant that is equal to the acceleration due to gravity.
To determine the acceleration due to gravity in Sydney Academy and to estimate the friction present in the pulley.
Discussion:
A pulley serves merely to redirect the one-dimensional motion of the string and attached masses, without
affecting the tension in the string or otherwise interfering with the motion.
The Atwood's machine system is driven by a net external force Fext equal in magnitude to the difference between
the two weights. If m1 is the larger mass and m2 is the smaller mass, then Fext = (m1 − m2)g.
The total inertia of the system is given by the total mass (m1 + m2).
The real pulleys used in the lab have some friction, which will be analyzed in this experiment. The pulleys also
have a mass, so they have a small amount of 'rotational inertia' that will be assumed to be insignificant in this
experiment.
The equation of motion for the system is
Fnet = mtotal  anet or
Fext −Ff = (m1 + m2)a
where a is the one-dimensional acceleration of the masses, and Ff is assumed to be a constant friction due to the
pulley. For a specific value of Fext, we expect the acceleration to be a constant. Rearranging the equation of
motion,
Fext − Ff = (m1 + m2)a or (m1 – m2)g −Ff = (m1 + m2)a
To get this equation in the form of y = mx + b, with a as the dependent variable (y) and the difference in the
masses (m1 – m2) as the dependent variable (x):
(m1+m2)a = (m1 – m2)g+ Ff
In this lab, the total mass (m1 + m2) will stay constant at 1.000 kg. Therefore, the formula can be expressed
as:
a = (m1 – m2)g+ Ff
A graph of a versus (m1 – m2) should be a straight line with slope g and intercept Ff.
For given values of m1 and m2, the time t it takes m1 to fall from a known height h to the floor will be measured. If
m1 is released with v0 = 0 and a constant acceleration is assumed, then the equations of uniformly accelerated
motion predict h = ½ a t2. Solving for the acceleration, a = 2h/t2.
Note: Use m, kg, N, and s in your calculations throughout this experiment.
Materials:
Procedure:
Equipment
Dependent Variables
Measuring Devices
Technology
ring stand
pulley
light string
set of known masses
2 mass hangers
Pasco Motion Sensor
digital balance ± 0.01g
Pasco SW 500 interface, Data
Studio software, and computer
Xplorer GLX
Set up
The masses of the known masses and the mass hangers were verified.
The pulley was secured to the ring stand so that it hung over the edge of the table at least 1.0 m above the floor.
A light string was put around the pulley and attached to the two mass hangers.
A thin barrier (poster board) was placed between the two suspended masses to avoid collisions.
The motion sensor was placed beneath the lighter mass that was to move upward.
Trials and Data Collection
In the first trial, the two masses (including the mass hanger) used were 450g and 550g.
The heavier 550g side was held near the pulley and released with as little initial velocity as possible.
The time interval to fall exactly 1.000 m was determined from the position-time graph.
The acceleration was determined from the line of best fit on the velocity-time graph.
Trials were repeated after transferring masses from one hanger to the other, keeping the total mass at 1.00 kg.
Data Analysis
All relevant data was recorded and a graph of acceleration versus mass difference was constructed.
The value of the acceleration due to gravity and the force of friction were determined
The percent error for the calculated value of g compared to the expected value (−9.81 m/s2) was determined.
Results
Accelerations of the larger mass for varying differences in mass. Each mass includes the mass of the mass
hanger. The height the larger mass (Mass 1) fell was 1.000 m for all trials.
Time Interval to
Mass 1
Mass 2
Mass Difference
Acceleration
Trial
fall 1.000m
(kg)
(kg)
(kg)
(m/s2)
(s)
1
Traditional
Table 1
2
3
4
5
Technological
1
2
3
4
5
Sample Calculations
Table 2
Calculations of values for graph of (m1 + m2) a = (m1 – m2) g – Ff .
x
Trial
Mass Difference
m1 – m2 (kg)
y
Total Mass
m1 + m2 (kg)
Acceleration
a (m/s2)
Traditional
1
2
3
4
5
Technological
1
2
3
4
5
Acceleration due to Gravity (g) = slope from graph =
_______________________
Force of Friction (Ff) = y-intercept from graph =
_______________________
Total Mass and Acceleration
(m1 + m2)a (kg m/s2)