Forest UCM NLM AtwoodMachine

Simple Atwood's machine

$\Rightarrow T = \frac{2m_1m_2}{m_1+m_2} g$

Double Atwood's machine

The problem

Determine the acceleration of each mass in the above picture.

Step 1: Identify the system

Each block is a separate system with two external forces; a gravitational force and the rope tension.

Step 2: Choose a suitable coordinate system

A coordinate system with one axis that defines the posive direction as up is one possible orientation.

Step 3: Draw the Free Body Diagram

200 px

Step 4: Define the Force vectors using the above coordinate system

since the system is one dimensional I will omit the vector notation

$T_1=$ Tension in the rope attached to mass $m_1$

$T_2=$ Tension in the rope attached to mass $m_2 = T_1$

$T_3=$ Tension in the rope attached to mass $m_3$

$F_g$ = force of gravity on each mass $= m_1 g$ or $m_2 g$ or $m_3 g$

Step 5: Use Newton's second law

for mass 1

$T_1 - m_1 g = m_1 a_1$

for mass 2

$T_2 - m_2 g = m_2 a_2$

for mass 3

$T_3 - m_3 g = m_3 a_3$

If we know the mass of all the objects in the system then we are left with three unkown Tensions and three unknown acceleratios. In total we currently have 6 unkowns and 3 equations.

Using Newton's third law we know that $T_1 = T_2$ reducing the unkowns to 5.

We need 2 more equations!

External Forces on Lower pulley

Consider the external forces acting on the MASSLESS lower pulley

$T_3-T1-T2 =T_3-T1-(T1) =(0)a$

$T_3=2T_1$

Now we have 4 unkwons and 3 equations

relative acceleration

let

$a_r =$ acceleration of $m_1$ with respect to the lower pulley

assuming that $a_1$ is moving upwards with respect to the earth

$a_1 = a_r - a_3$ : $a_3 =$ acceleration of lower pully as well as $m_3$

similarly

$a_2=-a_r-a_3$ : if $m_1$ is accelerating upwards then $m_2$ is accelerating downwards

3 equations and 3 unknowns

$T_1 - m_1 g = m_1 \left ( a_r - a_3 \right )$

$T_1 - m_2 g = m_2 \left ( -a_r - a_3 \right )$

$\left ( 2 T_1 \right ) - m_3 g = m_3 a_3$

Solutions

solving the above system of equations leads to the solutions

$a_1 = \frac{3m_2m_3 -m_1m_3 -4m_1m_2}{m_1m_3 + m_2 m_3 + 4 m_1 m_2}g$

$a_2 = \frac{3m_1m_3 -m_2m_3 -4m_1m_2}{m_1m_3 + m_2 m_3 + 4 m_1 m_2}g$

$a_3 = \frac{4m_1m_2 -m_2m_3 -m_1m_3}{m_1m_3 + m_2 m_3 + 4 m_1 m_2}g$

Matrix method solution ($T_1, a_r, a_3$ are the unkowns)

$\left( \begin{array}{ccc} 1 & -m_1 & m_1 \\ 1 & m_2 & m_2 \\ 2 & 0 & -m_3 \end{array} \right)\left( \begin{array}{ccc} T_1 \\ a_r \\ a_3 \end{array} \right) = \left( \begin{array}{ccc} m_1 g \\ m_2 g \\ m_3 g\end{array} \right)$

Cramer's Rule:

$a_3 = \frac{\left| \begin{array}{ccc} 1 & -m_1 & m_1g \\ 1 & m_2 & m_2g \\ 2 & 0 & m_3g \end{array} \right|}{\left| \begin{array}{ccc} 1 & -m_1 & m_1 \\ 1 & m_2 & m_2 \\ 2 & 0 & -m_3 \end{array} \right|}$

Forest_UCM_NLM#Atwoods_Machine