Atwood Machines

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Atwood Machines

1   Fixed Pulley

Let’s begin our analysis by studying the fundamental Atwood machine – a fixed pulley with two masses.

basic

Question

Find the accelerations of the masses and the tension in the string.

We solve problems involving Atwood machines by using F=ma equations and an equation of conservation of string.

F=ma equations

T-m_1g=m_1a_1 

T-m_2g=m_2a_2 

Conservation of String

a_1+a_2=0 

Accelerations

a_1=\frac{m_2-m_1}{m_1+m_2}g 

a_2=\frac{m_1-m_2}{m_1+m_2}g 

Tension

T=\frac{2m_1m_2g}{m_1+m_2} 

We can use the idea of effective mass of a fixed pulley to simplify our analysis in fixed pulley combinations.

Effective Mass of a Fixed Pulley

M=\frac{2T}{g}=\frac{4m_1m_2}{m_1+m_2} 

1.1   Fixed pulley combination

multiple

Question

Find the accelerations of the masses and the tension in the string.

F=ma equations

T-m_1g=m_1a_1 

T/2-m_2g=m_2a_2 

T/2-m_3g=m_3a_3 

We can use the idea of effective acceleration of a fixed pulley to simplify finding the equation for conservation of string in fixed pulley combinations.

Effective Acceleration of Fixed Pulley

A=\frac{a_2+a_3}{2} 

Conservation of String

a_1+\frac{a_2+a_3}{2}=0 

Accelerations

a_1=\frac{4m_2m_3-m_1(m_2+m_3)}{4m_2m_3+m_1(m_2+m_3)}g 

a_2=\frac{-4m_2m_3+m_1(3m_3-m_2)}{4m_2m_3+m_1(m_2+m_3)}g 

a_3=\frac{-4m_2m_3+m_1(3m_2-m_3)}{4m_2m_3+m_1(m_2+m_3)}g 

Tension

T=\frac{8m_1m_2m_3g}{2m_2m_3+m_1(m_2+m_3)} 

2   Free Pulley

Now let’s analyze systems with free pulleys in which they are not attached to supports.

free

Question

Find the accelerations of the masses and the tension in the string.

F=ma equations

T-m_1g=m_1a_1 

2T-m_2g=m_2a_2 

Conservation of String

a_1+2a_2=0 

Accelerations

a_1=\frac{2m_2-4m_1}{4m_1+m_2}g 

a_2=\frac{2m_1-m_2}{4m_1+m_2}g 

Tension

T=\frac{3m_1m_2g}{4m_1+m_2} 

2.1   3  Strings

3

Question

Find the accelerations of the masses and the tension in the string.

F=ma equations

T-m_1g=m_1a_1 

3T-m_2g=m_2a_2 

Conservation of String

a_1+3a_2=0 

Accelerations

a_1=\frac{3m_2-9m_1}{9m_1+m_2}g 

a_2=\frac{3m_1-m_2}{9m_1+m_2}g 

Tension

T=\frac{4m_1m_2}{9m_1+m_2}g 

2.2   n Strings

We can generalize this to n  strings.

Question

Find the accelerations of the masses and the tension in the string.

F=ma equations

T-m_1g=m_1a_1 

nT-m_2g=m_2a_2 

Conservation of String

a_1+na_2=0 

Accelerations

a_1=\frac{nm_2-n^2m_1}{n^2m_1+m_2}g 

a_2=\frac{nm_1-m_2}{n^2m_1+m_2}g 

Tension

T=\frac{(n+1)m_1m_2}{n^2m_1+m_2}g 

2.3   Free Pulley In Between

between

Question

Find the accelerations of the masses and the tension in the string.

F=ma equations

T-m_1g=m_1a_1 

2T-m_2g=m_2a_2 

T-m_3g=m_3a_3 

Conservation of String

a_1+2a_2+a_3=0 

Accelerations

a_1=\frac{3m_2m_3-m_1(m_2+4m_3)}{m_2m_3+m_1(m_2+4m_3)}g 

a_2=\frac{-m_2m_3+m_1(4m_3-m_2)}{m_2m_3+m_1(m_2+4m_3)}g 

a_3=\frac{-m_2m_3+m_1(3m_2-4m_3)}{m_2m_3+m_1(m_2+4m_3)}g 

Tension

T=\frac{4m_1m_2m_3}{m_2m_3+m_1(m_2+4m_3)}g 

2.4   Massless Free Pulley

Equivalence of Free Pulleys

eq

The subject of massless free pulleys is an interesting topic in Atwood machines. We can use a key idea derived here to solve more interesting problems.

massless

Question

Find the acceleration of the free pulley.

Procedure

Assume the free pulley has mass m_2 . This is equivalent to the case with a free pulley derived above.

T-m_1g=m_1a_1 

2T-m_2g=m_2a_2 

a_1+2a_2=0 

We have m_2=0 . The force on the pulley must be zero so T=0 . We can use the first equation and third equation to solve for a_2 . Note that we cannot use the second equation.

a_2=\frac{g}{2} 

strange

Question

Find the accelerations of the masses and the tension in the string.

Accelerations

Note that the net force on the left massless pulley must be zero.

a_1=-g 

a_2=-g 

Tension

T=0 

The techniques developed here can be applied to any Atwood Machine problem.

For more information see: Introduction to Classical Mechanics (David Morin)