Forest UCM NLM AtwoodMachine

Simple Atwood's machine

[math]\Rightarrow T = \frac{2m_1m_2}{m_1+m_2} g[/math]

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

for mass 1

[math]T_1 - m_1 g = m_1 a_1[/math]

for mass 2

[math]T_2 - m_2 g = m_2 a_2[/math]

for mass 3

[math]T_3 - m_3 g = m_3 a_3[/math]

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 [math]T_1 = T_2[/math] 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

[math]T_3-T1-T2 =T_3-T1-(T1) =(0)a[/math]

[math]T_3=2T_1[/math]

Now we have 4 unkwons and 3 equations

relative acceleration

let

[math]a_r =[/math] acceleration of [math]m_1[/math] with respect to the lower pulley

assuming that [math]a_1[/math] is moving upwards with respect to the earth

[math]a_1 = a_r - a_3[/math] : [math]a_3 =[/math] acceleration of lower pully as well as [math]m_3[/math]

similarly

[math]a_2=-a_r-a_3[/math] : if [math]m_1[/math] is accelerating upwards then [math]m_2[/math] is accelerating downwards

Step 5: Use Newton's second law

Forest_UCM_NLM#Atwoods_Machine