PhysicsDynamics: Atwood Machine

Science and Mathematics Education Research Group

Supported by UBC Teaching and Learning Enhancement Fund 2012-2014

Department of Curriculum and Pedagogy

FACULTY OF EDUCATIONa 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.

1m2m

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?

1m2m

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

1m2m

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.

1m2m

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

1m2m

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

1m2m

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

1m2m

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

1m2m

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

1m2m

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?

1m2m

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

1m2m

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?

1m2m

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

mgm

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)

1m2m

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).