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

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

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

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.

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.