Difference between revisions of "Forest UCM NLM AtwoodMachine"

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== Step 4: Define the Force vectors using the above coordinate system==
 
== Step 4: Define the Force vectors using the above coordinate system==
 +
 +
since the system is one dimensional I will omit the vector notation
 +
 +
:<math>T_1=</math>  Tension in the rope attached to mass <math>m_1</math>
 +
:<math>T_2=</math>  Tension in the rope attached to mass <math>m_2 = T_1</math>
 +
:<math>T_3=</math>  Tension in the rope attached to mass <math>m_3</math>
 +
:<math>F_g</math> = force of gravity on each mass <math>= m_1 g</math> or <math>m_2 g</math> or <math>m_3 g</math>
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 +
==Step 5: Use Newton's second law==
  
 
;for mass 1
 
;for mass 1
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:<math>a_r =</math> acceleration of <math>m_1</math> with respect to the lower pulley
 
:<math>a_r =</math> acceleration of <math>m_1</math> with respect to the lower pulley
  
with respect to the earth
+
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>
 
:<math>a_1 = a_r - a_3</math>  : <math>a_3 =</math> acceleration of lower pully as well as <math>m_3</math>
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similarly
 
similarly
  
:<math>a_2=-a_r-a_3</math>
+
:<math>a_2=-a_r-a_3</math> : if <math>m_1</math> is accelerating upwards then <math>m_2</math> is accelerating downwards
 +
 
 +
=== 3 equations and 3 unknowns===
 +
 
 +
:<math>T_1 - m_1 g = m_1 \left ( a_r - a_3 \right )</math>
 +
 
 +
 
 +
:<math>T_1 - m_2 g = m_2 \left ( -a_r - a_3 \right )</math>
 +
 
 +
 
 +
:<math>\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>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>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>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>T_1, a_r, a_3</math> are the unkowns)
 +
 
 +
<math>\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:
  
==Step 5: Use Newton's second law==
 
  
 +
:<math>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>
  
 
[[Forest_UCM_NLM#Atwoods_Machine]]
 
[[Forest_UCM_NLM#Atwoods_Machine]]

Latest revision as of 12:52, 22 August 2014

Simple Atwood's machine

TF UCM SAM 1.gif


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

Double Atwood's machine

TF UCM DAM 1.gif


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

[math]T_1=[/math] Tension in the rope attached to mass [math]m_1[/math]
[math]T_2=[/math] Tension in the rope attached to mass [math]m_2 = T_1[/math]
[math]T_3=[/math] Tension in the rope attached to mass [math]m_3[/math]
[math]F_g[/math] = force of gravity on each mass [math]= m_1 g[/math] or [math]m_2 g[/math] or [math]m_3 g[/math]

Step 5: Use Newton's second law

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

3 equations and 3 unknowns

[math]T_1 - m_1 g = m_1 \left ( a_r - a_3 \right )[/math]


[math]T_1 - m_2 g = m_2 \left ( -a_r - a_3 \right )[/math]


[math]\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]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]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]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]T_1, a_r, a_3[/math] are the unkowns)

[math]\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]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]

Forest_UCM_NLM#Atwoods_Machine