Forest UCM NLM AtwoodMachine
Simple Atwood's machine
[math]\displaystyle{ \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
File:TF UCM FBD DoubleAtwoodMach.png
Step 4: Define the Force vectors using the above coordinate system
since the system is one dimensional I will omit the vector notation
- [math]\displaystyle{ T_1= }[/math] Tension in the rope attached to mass [math]\displaystyle{ m_1 }[/math]
- [math]\displaystyle{ T_2= }[/math] Tension in the rope attached to mass [math]\displaystyle{ m_2 = T_1 }[/math]
- [math]\displaystyle{ T_3= }[/math] Tension in the rope attached to mass [math]\displaystyle{ m_3 }[/math]
- [math]\displaystyle{ F_g }[/math] = force of gravity on each mass [math]\displaystyle{ = m_1 g }[/math] or [math]\displaystyle{ m_2 g }[/math] or [math]\displaystyle{ m_3 g }[/math]
Step 5: Use Newton's second law
- for mass 1
- [math]\displaystyle{ T_1 - m_1 g = m_1 a_1 }[/math]
- for mass 2
- [math]\displaystyle{ T_2 - m_2 g = m_2 a_2 }[/math]
- for mass 3
- [math]\displaystyle{ 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]\displaystyle{ 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]\displaystyle{ T_3-T1-T2 =T_3-T1-(T1) =(0)a }[/math]
- [math]\displaystyle{ T_3=2T_1 }[/math]
Now we have 4 unkwons and 3 equations
relative acceleration
let
- [math]\displaystyle{ a_r = }[/math] acceleration of [math]\displaystyle{ m_1 }[/math] with respect to the lower pulley
assuming that [math]\displaystyle{ a_1 }[/math] is moving upwards with respect to the earth
- [math]\displaystyle{ a_1 = a_r - a_3 }[/math] : [math]\displaystyle{ a_3 = }[/math] acceleration of lower pully as well as [math]\displaystyle{ m_3 }[/math]
similarly
- [math]\displaystyle{ a_2=-a_r-a_3 }[/math] : if [math]\displaystyle{ m_1 }[/math] is accelerating upwards then [math]\displaystyle{ m_2 }[/math] is accelerating downwards
3 equations and 3 unknowns
- [math]\displaystyle{ T_1 - m_1 g = m_1 \left ( a_r - a_3 \right ) }[/math]
- [math]\displaystyle{ T_1 - m_2 g = m_2 \left ( -a_r - a_3 \right ) }[/math]
- [math]\displaystyle{ \left ( 2 T_1 \right ) - m_3 g = m_3 a_3 }[/math]
Solutions
solving the above system of equations leads to the solutions
- [math]\displaystyle{ 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 }[/math]
- [math]\displaystyle{ 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 }[/math]
- [math]\displaystyle{ 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 }[/math]
- Matrix method solution ([math]\displaystyle{ T_1, a_r, a_3 }[/math] are the unkowns)
[math]\displaystyle{ \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) }[/math]
Cramer's Rule:
- [math]\displaystyle{ 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|} }[/math]