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Physics
Dynamics: Atwood Machine
Science and Mathematics
Education Research Group
Supported by UBC Teaching and Learning Enhancement Fund 2012-2014
Department of
Curr iculum and Pedagogy
F A C U L T Y O F E D U C A T I O N a place of mind
The Atwood Machine
The Atwood Machine
The Atwood Machine is a pulley system consisting of two weights
connected by string. We will assume no friction and that both the
string and pulley are massless.
If the masses of the two weights are
different, the weights will accelerate
uniformly by a. Our axis is defined
such that positive a indicates that m1
accelerates downwards, while m2
accelerates upwards.
1m
2m
a a
The Atwood Machine I
Let W1 be the weight of m1, and W2 be the weight of m2. Let the
tension of the string be T. Assume that m1 > m2.
Which of the following is the correct free body
diagram (force diagram) of m1?
1m
2m
a a
A.T
1W
B.2W
1W
C.T
g
D.2W
g
E.2W
1W
T
Answer: A
Justification: There are only two forces acting on m1, the force of
tension due to the string pulling it up and the force of its own weight
pulling it down.
C and D are incorrect because g itself is not a force. B and E are
incorrect because W2 is not a force acting directly on m1. Any force
due to weight must point in the direction of g, which is downwards.
Solution
A.T
gmW 11
The Atwood Machine II
According to Newton’s second law,
which of the following expressions is true?
gmT
gmamT
gmTam
amTgm
amgmT
1
11
11
11
11
E.
D.
C.
B.
A.
T
gmW 11
1m
2m
a a
amF 11
Answer: B
Justification: From the force diagram and taking downwards as
positive acceleration for a,
Therefore, by Newton’s second law,
Solution
T
gmW 11
TgmF 11
(Add forces in the same direction
as a, subtract forces in the
opposite direction of a)
am1
amTgm
amF
11
11
The Atwood Machine III
Which of the following is the correct expression
obtained by applying Newton’s second law for m2?
gmT
gmamT
gmTam
amTgm
amgmT
2
22
22
22
22
E.
D.
C.
B.
A.
1m
2m
a aPress for hint T
gmW 22
Answer: A
Justification: This question is solved in the same way we as the
previous question, for m1. It is a good idea to start by drawing the free
body diagram for m2.
Notice that the magnitude of T is the same for both m1 and m2.
However, in the case of m2, it is acting in the same direction as the
acceleration rather than opposing it.
Solution
amgmTF 222
(Add forces in the same direction
as a, subtract forces in the
opposite direction of a)
T
gmW 22
am2
The Atwood Machine IV
What is the acceleration of the two weights?
21
21
21
21
2
21
1
21
21
E.
D.
C.
B.
)(A.
mm
mmga
mm
mmga
m
mmga
m
mmga
mmga
1m
2m
a a
amgmT
amTgm
22
11
Answer: E
Justification: We have already found two expressions for the
acceleration, although they contain the unknown quantity T.
Since T is equal along all points on the string, it is the same in both
equations. Adding the two equations will eliminate T:
Solution
amgmT
amTgm
22
11
21
21
2121
2121
)()(
mm
mmga
mmammg
amamgmgm
The Atwood Machine V
Now that we have found an equation for the
acceleration, we can analyze some how the
Atwood Machine behaves.
Under what conditions will a be negative?
21
21
21
21
21
E.
D.
C.
B.
A.
mm
mm
mm
mm
mm
1m
2m
a a
21
21
mm
mmga
Answer: E
Justification: Recall that positive a implies that m1 accelerates
downwards, while m2 accelerates upwards. If a is negative, this
means that m2 will accelerates downwards.
Solution
021
21
mm
mmga
1m
2m
a a
We know intuitively that m2 will accelerate
downwards if m2 > m1. This is also reflected in
the formula we derived.
021 mmsince
The Atwood Machine VI
How should the masses of the weights be
chosen such that the neither weight
accelerates?
21
21
21
21
21
E.
D.
C.
B.
A.
mm
mm
mm
mm
mm
1m
2m
a a
21
21
mm
mmga
Answer: C
Justification: If neither weight accelerates, then a = 0. This occurs
when m1 = m2. No matter where the two weights are positioned, the
weights will not move.
Solution
0
0
11
21
21
mmg
mm
mmga
1m2m
21 mm since
The Atwood Machine VII
Suppose that m1 >> m2 (one of the weights
is much heavier than the other).
How can the acceleration be approximated
in this case?
0E.
D.
2C.
B.
)(A.
1
21
a
ga
ga
gma
gmma
1m
2m
a a
21
21
mm
mmga
Answer: D
Justification: If m1 >> m2, then we can make the following
approximations:
Solution
121
121
mmm
mmm
1m
gm
mga
1
1
Compared to m1, we can approximate that
m2 = 0. We can treat m1 as if it were in
freefall.
g
The Atwood Machine VIII
Depending on the values chosen for the
mass for m1 and m2, we can achieve a
variety of values for a.
What are all the possible values of a that
can be obtained by varying the masses?
ga
gag
ga
a
a
E.
D.
0C.
0B.
A.
1m
2m
a a
21
21
mm
mmga
Answer: D
Justification: The largest possible value for a is obtained when
m2 = 0.
The smallest value of a is obtained when m1 = 0.
Solution
1m
gm
mga
0
0
1
1
g
gm
mga
2
2
0
0
No matter how large we choose m1, the weights
cannot accelerate upwards or downwards faster
than g. What would happen if this were not true?
The Atwood Machine IX
What is the tension on the string?
1m
2m
a a
Press for hint
21
21
mm
mmga
amgmT
amTgm
22
11
gmgmT
mm
mmgT
mm
mmgT
mm
mmgT
mm
mgT
21
21
21
21
21
21
21
21
1
E.
2D.
C.
2B.
A.
Answer: B
Justification: We can substitute the solution for a into either
one of the force equations:
For example, using the top equation gives:
Solution
amgmT
amTgm
22
11
21
21
21
211211
21
2111
2
)()(
mm
gmm
mm
mmgmmmgm
mm
mmgmgmT
1m
2m
a a
Answer: B
Justification: Many answers can be ruled out by doing a few quick
calculations.
Answer A is incorrect because it does not have units of force.
Answer C and D are incorrect because it is possible to obtain
negative values for tension.
Suppose we let m1 = 0. We should expect our formula to return T = 0,
since m2 is in freefall. This is not true for answer E.
Solution Continued
The Atwood Machine X
Suppose m1 is fixed, although we are free to
choose any value for m2.
How should m2 be chosen (in terms of m1) such
that T = m1g?
1m
2m
a a
12
12
12
12
2
2
1E.
2D.
C.
B.
0A.
mm
mm
mm
mm
m
21
212mm
mmgT
Answer: C
Justification: If the tension of the string is equal to the weight of m1,
then m1 is not accelerating. Recall that there is no acceleration when
m1 = m2. From the formula:
Solution
1m2m
gmgm
mg
m
gm
mm
mmgT
21
2
21
21
2
2
2
mmm 21
The Atwood Machine XI
What is the force on the pulley required to
hold it up along the ceiling?
(Recall that we are assuming a massless
pulley and string)
1m
2m
a a
gmmF
gmgmF
Tmm
mmgF
Tmm
mmgF
Tmm
mmgF
)(2E.
D.
24C.
2B.
5.0A.
21pulley
12pulley
21
21pulley
21
21pulley
21
21pulley
Answer: C
Justification: The pulley must be held up by twice the tension on
the string.
Solution
21
21pulley 42
mm
mmgTF
TT
T2
Notice that in the formula above, F = 0 if either
m1 or m2 is zero. Since the masses will be in
freefall, no force is required to hold the
massless pulley up.
Also notice that Fpulley = m1g + m2g, the sum of
the weight of the two masses, only when the
weights are not accelerating (m1 = m2).