Atwood Machine
Below is an Atwood machine which is simply a pulley with a mass at each end of a
string passed over the top of the pulley and the free body diagram for each mass.
There are a few
assumptions made to
simplify the study of an Atwood machine. First, the pulley is considered to be an ideal
pulley. An ideal pulley is frictionless and massless. Friction would result in accelerations
being smaller than the calculated values. The mass of the pulley would introduce a
small amount of rotational inertia which results in the pulley resisting changes in its state
of rotation. The string is assumed to be massless and not stretchable. This ensures that
the mass of the system consists only of the two masses at each end of the string and
the tension is the same throughout the length of the string. The reason for using an
Atwood machine is to provide a simple way to produce a small acceleration compared
to the acceleration due to gravity. The assumptions that have been discussed make the
following equations valid:
Fnet = m1a (1) Fnet = m2a (1)
Fnet = T - Fw1 (2) Fnet = Fw2 – T (2)
m1a = T - m1g (3) m2a = T – m2g (3)
(1) is Newton's 2nd law in its general form, (2) is the net force applied to each mass
from the free body diagram, and (3) results by equating (1) and (2). Adding both
equations (3) cancels the T and rearranging terms yield:
a = (m2 - m1)/(m2 + m1)*g
where the masses can be expressed in any mass units.
Atwood Machine Problems
1) Given the following:
(a) Draw a free body diagram for each mass.
m1 = 3.0 kg m2 = 4.5 kg
(b) Applying Newton's 2nd law, Fnet = ma, to each mass yields:
T - Fw1 = m1a Fw2 - T = m2a
T - m1g = m1a (1) m2g - T = m2a (2)
Whenever you have two equations such as (1) and (2) with the same two
unknowns you are in luck. Adding or subtracting them will eliminate one of the
unknowns.
T - m1g = m1a (1)
m2g - T = m2a (2)
m2g - m1g = m1a + m2a
Rearranging and solving for a:
a = (m2-m1)/(m2+m1)*g = (4.5 kg-3.0 kg)/(4.5 kg+3.0 kg)*9.80 m/s2 = 1.96 m/s2
(c) To find the tension, you can use either equation 1 or 2:
T - m 1g = m 1a
T = 3.0 kg*1.96 m/s2 + 3.0 kg*9.80 m/s2 = 35.3 N
(d) How would the answers above change if you were on the moon where
g ≅ 1.63 m/s2? Use both a qualitative and quantitative analysis to support your
answer.
The value of the two masses would remain the same because mass is
independent of location. The force diagrams would remain the same but Fw and T
would be smaller because of a smaller g.
Mathematically, the acceleration and the tension would be:
a = (m2-m1)/(m2+m1)*g = (4.5 kg-3.0 kg)/(4.5 kg+3.0 kg)*1.63 m/s2 = 0.33 m/s2
T - m1g = m1a = 3.0 kg*0.33 m/s2 + 3.0 kg*1.63 m/s2 = 5.88 N
2) In the figure shown below, a 50. N weight
is resting on a bathroom scale while
Fw2 changes in value.
Determine the scale reading when Fw2 is:
(a) 10. N
The same force diagram exists for (a) and (b).
Applying Newton's 2nd law to Fw2:
Fnet = ma = T - Fw2 but Fnet = 0 because the problem states that Fw1 and Fw2 are at
rest. Therefore: T = Fw2 = 10. N
Similarly, for Fw1:
Fnet = T + FN - Fw1 = 0
FN = Fw1 - T = 50. N - 10. N = 40. N
Fw1 will appear to weigh 40. N. The resultant of Fw1 and FN is equal to 40. N straight
down which is the force being exerted on the scale. Because of Newton's 3rd law, the
scale pushes up on Fw1 with a force of 40. N
(b) 40. N
Following the same logic as in (a), FN = Fw1 - T = 50. N - 40. N = 10. N.
(c) 70. N
Following the same logic as in (a) and (b) , FN = Fw1 - T = 50. N - 70. N = -20. N.
What???
How can FN = -20. N? The scale either pushes up on Fw1 or not at all. In this case
the logic must change. The net force on Fw1 is T - Fw1 and the tension and
acceleration can be determined as it was in problem 2.
A couple of notes before leaving this problem:
The force diagram for parts (a) and (b) would be the same except for the
relative sizes of T and FN.
The force diagram for part (c) would not have a F N but would look like the
diagram in problem 1 (a).
3) Given that m1 = 4.50 kg and m2 = 3.00 kg, determine the acceleration of the system
and the tension in the cord.
m1= 4.50 kg m2 = 3.00 kg
Applying Newton's 2nd law to m1, Fnet = ma, it follows that
T = m1a (1)
Applying Newton's 2nd law to m2:
Fw2 - T = m2a (2)
Substituting equation 1 into equation 2 and solving for a gives:
a = m2/(m1 + m2)*g = 3.00 kg/7.50 kg*9.80 m/s2 = 3.92 m/s2
To find the tension, you can use either equation1 or equation 2:
T = m1a = 4.50 kg*3.92 m/s2 = 17.6 N
T = Fw2 – T
T = 3.00 kg*9.80 m/s2 - 3.00 kg*3.92 m/s2 = 17.6 N
