Pulleys and Atwood Machines

A pulley changes the direction of a string and, in some arrangements, changes the relation between the motions and forces of connected bodies. The central ideas are not new force laws. They are the same tension model and Newton’s second law developed earlier, combined with a geometric constraint from an inextensible string.

For an ideal pulley system, three modeling assumptions are especially important:

1. the string is massless, so an ideal continuous string can have one tension magnitude when no other tangential interaction changes it;
2. the string is inextensible, so its total length is fixed and the coordinates of the attached bodies are constrained;
3. the pulley is ideal, meaning its axle is frictionless and its Rotational Inertia is neglected unless stated otherwise.

The first assumption controls force transmission. The second controls kinematics. The third determines whether the tension magnitude can remain the same on the two sides of a pulley.

1 What an ideal fixed pulley does

A fixed pulley is attached to a support and does not translate. Its primary role is to redirect the string. For an ideal fixed pulley, the tension magnitude is the same on both sides of one continuous massless string:

(1)

This equality is not a general property of every real pulley. A pulley with appreciable rotational inertia or axle friction can require different tensions on its two sides. The ideal result is a model assumption justified by neglecting those effects.

A single fixed pulley also does not provide a force advantage. In a static arrangement where one end supports a load of Weight W, the other end must sustain the same ideal tension,

(2)

Its advantage is directional: a downward pull on one end can raise a load on the other.

2 The fixed length constraint

Consider a single inextensible string passing over a fixed pulley. Let x₁ and x₂ be downward coordinates measured from fixed points near the pulley. Apart from the constant length wrapped around the pulley,

(3)

where C is constant.

Because the total length L is fixed,

(4)

Differentiating with respect to time gives

(5)

and differentiating again gives

(6)

Thus the two ends have equal speed magnitudes and equal acceleration magnitudes, but they move in opposite directions when the same downward coordinate convention is used on both sides.

If instead the positive axis for each body is chosen along its actual direction of motion, both scalar acceleration equations may use the same positive symbol a. This is only a coordinate choice. The underlying vector accelerations still point in opposite directions.

3 The Atwood machine

The classic Atwood machine consists of two masses connected by one light inextensible string over an ideal fixed pulley. Let

(7)

Then m₂ accelerates downward while m₁ accelerates upward. Because the string is inextensible, both acceleration magnitudes are the same. Let that magnitude be a.

For m₁, choose upward positive:

(8)

For m₂, choose downward positive:

(9)

Adding the equations eliminates the internal tension:

(10)

Therefore

(11)

Substituting into either mass equation gives the tension

(12)

4 Physical checks on the Atwood result

The formula

(13)

contains several useful checks.

If m₁ = m₂, then

(14)

Equal masses balance, and the tension is simply

(15)

If m₁ → 0 while m₂ remains finite, then

(16)

and

(17)

The remaining mass approaches free fall because the opposite side contributes almost no inertia or tension.

For positive finite masses,

(18)

An ideal Atwood mass cannot accelerate faster than free fall.

5 Why adding the two equations works

For the two mass system, the two tension forces are internal. They are equal in magnitude and opposite in their effect on the system coordinate. Adding the two Newton equations removes the internal force and leaves only the net external driving force,

(19)

The total inertial mass being accelerated is

(20)

This gives the compact system interpretation

(21)

The tension is then recovered by isolating one body.

6 A block on a table connected to a hanging mass

A second standard arrangement places one block of mass m₁ on a frictionless horizontal table and connects it over an ideal fixed pulley to a hanging mass m₂.

Because one inextensible string connects the two bodies, their acceleration magnitudes are equal. For the block on the table,

(22)

For the hanging mass, taking downward positive,

(23)

Adding gives

(24)

so

(25)

The tension is

(26)

This system is not an Atwood machine in the strict two hanging mass sense, but it uses the same modeling pattern: string constraint, equal ideal tension, and one Newton equation for each body.

7 Movable pulleys change the kinematic ratio

A movable pulley translates with the load. This changes the string length relation.

Consider one end of a string fixed to the ceiling. The string descends around a movable pulley and then rises to a free end. Let y_p be the downward coordinate of the movable pulley and y_f the downward coordinate of the free end.

Two variable string segments change length when the movable pulley moves. Therefore

(27)

Differentiating,

(28)

and

(29)

The free end therefore moves twice as far and twice as fast as the movable pulley, in the opposite direction when both coordinates are measured downward.

8 Force support from a movable pulley

The same ideal string passes around the movable pulley, so each supporting segment has tension magnitude T. The pulley and attached load therefore receive two upward tension forces. Neglecting the mass of the pulley itself, the upward support is

(30)

For static equilibrium of a load with weight W,

(31)

Hence

(32)

If a person pulls the free end with force magnitude F = T, the ideal equilibrium force needed is

(33)

The price for this force advantage is displacement: the free end must move twice as far as the load.

More generally, in a simple ideal arrangement where a moving load is supported by n segments of the same rope, static force balance often gives

(34)

so the ideal force advantage is associated with the number of supporting rope segments. The exact kinematic relation must still be derived from the actual rope geometry rather than memorized from the number n alone.

9 Constraint equations should be derived from geometry

Pulley problems become unreliable when acceleration relations are guessed. The safe procedure is to write the total length of each inextensible string explicitly.

For example, if one string contains variable straight segments s₁, s₂, and s₃, then

(35)

Because L is constant,

(36)

and

(37)

These derivatives produce the velocity and acceleration constraints. Fixed portions of rope, fixed pulley wrap lengths, and constant offsets are absorbed into C and disappear upon differentiation.

10 When tensions are not equal

The equality of tension on the two sides of a pulley is an idealization. If the pulley has appreciable rotational inertia, the two string tensions generally differ because a net torque is required to angularly accelerate the pulley. If axle friction is present, the tensions can also differ.

Those cases require rotational dynamics. They are not part of the ideal translational model developed here. The important diagnostic is:

(38)

Different strings should still be given different tension symbols unless the equations prove otherwise.

11 A systematic procedure for pulley problems

1. Identify each distinct string and each pulley that moves.
2. State the idealizations: massless string, inextensible string, ideal pulley, and any frictionless surfaces.
3. Choose coordinates for every moving body and pulley.
4. Write one fixed length equation for each inextensible string.
5. Differentiate the length equation to obtain the velocity or acceleration constraints.
6. Draw a separate free body diagram for each body or useful combined system.
7. Use one tension symbol along a single ideal string passing over ideal pulleys; use different symbols for different strings.
8. Apply Newton’s second law with a consistent sign convention.
9. Solve the force equations together with the kinematic constraints.
10. Check limiting cases, directions, and whether the assumed string remains taut.

12 Worked example 1: a basic Atwood machine

Two masses

(39)

are connected by a massless inextensible string over an ideal fixed pulley. Find the acceleration magnitude and the tension.

The acceleration is

(40)

Substituting,

(41)

so

(42)

The 5.00 kg mass accelerates downward and the 3.00 kg mass accelerates upward.

Using

(43)

we obtain

(44)

so

(45)

13 Worked example 2: a table and hanging mass

A 6.00 kg block rests on a frictionless table and is connected over an ideal pulley to a hanging 2.00 kg mass. Find the acceleration magnitude and tension.

The system acceleration is

(46)

Therefore

(47)

which gives

(48)

For the block on the table,

(49)

so

(50)

14 Worked example 3: an ideal movable pulley

A load has weight

(51)

and is supported by a single ideal movable pulley. One end of the rope is fixed to the ceiling and a person pulls the free end. Find the required force for static equilibrium and the free end displacement needed to raise the load by 0.250 m.

Two rope segments support the movable pulley, so

(52)

Thus

(53)

The person’s pull equals the rope tension in the ideal model:

(54)

For the string length,

(55)

If the pulley and load move upward by 0.250 m, then with downward positive

(56)

Therefore

(57)

The person must pull the free end downward by

(58)

The ideal force is halved while the input displacement is doubled.

15 Worked example 4: a hanging mass driving a movable pulley

A 4.00 kg mass m₁ hangs from the free end of an ideal rope. The same rope passes around a massless movable pulley that supports a 6.00 kg load m₂. The other end of the rope is fixed to the ceiling. Determine the acceleration of each mass and the rope tension.

Let a₁ be positive downward for m₁, and let a₂ be positive upward for the movable pulley and load. From the length constraint,

(59)

For the hanging mass,

(60)

For the movable pulley and attached load, two rope segments pull upward:

(61)

Using a₁ = 2a₂ in the first equation,

(62)

Substituting into the second equation gives

(63)

Therefore

(64)

With m₁ = 4.00 kg and m₂ = 6.00 kg,

(65)

The free mass acceleration is twice as large:

(66)

Finally,

(67)

so

(68)

This example shows why a movable pulley problem cannot be solved by assuming all connected bodies have the same acceleration magnitude.

16 Practice problems

1. Two masses are connected by one inextensible string over a fixed pulley. If the left mass moves downward 0.180 m, how far and in what direction does the right mass move?
2. An ideal Atwood machine has m₁ = 2.00 kg and m₂ = 6.00 kg. Find the acceleration magnitude and tension.
3. An Atwood machine has masses m and 3m. Express the acceleration magnitude and tension in terms of g and m.
4. A 5.00 kg block on a frictionless table is connected over an ideal pulley to a hanging 3.00 kg mass. Find the acceleration and tension.
5. A 10.0 kg block on a frictionless table is connected to a hanging 2.50 kg mass. Starting from rest, how far does the hanging mass move in 1.50 s?
6. A 250 N load hangs motionless from one end of a single rope over an ideal fixed pulley. What force must be applied to the other end to maintain equilibrium?
7. A 400 N load is supported by one ideal movable pulley with two supporting segments of the same rope. Find the rope tension and the ideal force required at the free end for equilibrium.
8. In the system of Problem 7, the load rises 0.300 m. How far must the free end move, and in what direction?
9. A 20.0 kg load is attached to a massless movable pulley. A person pulls the free end downward with a constant force of 120 N. Neglect pulley mass. Find the load’s vertical acceleration and direction.
10. A 5.00 kg mass hangs from the free end of a rope that passes around a massless movable pulley supporting an 8.00 kg load; the other rope end is fixed. Let the 5.00 kg mass accelerate downward and the load upward. Find both acceleration magnitudes and the tension.
11. In a real Atwood machine, the pulley has rotational inertia. If the heavier mass descends and the pulley angularly accelerates, should the tension on the heavier side be equal to, greater than, or less than the tension on the lighter side? Explain qualitatively.
12. For each statement, identify the assumption responsible: (a) the two ends of one rope over a fixed pulley have equal speed magnitudes; (b) the tension magnitude is the same on both sides of one ideal pulley; (c) a movable pulley supported by two segments receives upward force 2T.

17 Answer check

1. The right mass moves 0.180 m upward. For a fixed pulley, the two end displacements have equal magnitudes and opposite directions.

2. (69)

   (70)

3. (71)

4. (72)

   (73)

5. (74)

   From rest,

   (75)

6. 250 N. A single ideal fixed pulley changes direction but not the tension magnitude.
7. 2T = 400 N, so

   (76)

8. The free end moves 0.600 m downward.
9. The upward rope force is 2T = 240 N. The weight is (20.0)(9.81) = 196.2 N, so

   (77)

   upward.

10. The constraint is a₁ = 2a₂. Solving

    (78)

    (79)

    gives

    (80)

    (81)

    (82)

11. The heavier side tension must be greater than the lighter side tension if the pulley is angularly accelerating in the direction driven by the heavier mass. Their difference supplies the net torque on the pulley.
12. (a) Inextensible string. (b) Ideal pulley together with the massless string model. (c) Geometry plus equal tension magnitude in the two supporting segments of the same ideal rope.

18 Summary

Pulley systems combine force modeling with geometric constraints. For an ideal fixed pulley and one inextensible string,

(83)

so the two ends have equal speed and acceleration magnitudes in opposite directions.

For an ideal Atwood machine,

(84)

and

(85)

A movable pulley changes the constraint. For the simplest one movable pulley arrangement,

(86)

which gives

(87)

The movable pulley is supported by two tension forces, so static equilibrium gives 2T = W. The force advantage and displacement ratio are therefore two aspects of the same rope geometry.

The next article, M02-07, introduces friction and shows how contact forces modify these connected body problems.

References