Difference between revisions of "Forest UCM NLM AtwoodMachine"

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Using Newton's first law we know that T_1 = T_2
+
Using Newton's third law we know that <math>T_1 = T_2</math> reducing the unkowns to 5.
  
;We need three more equations!
+
;We need 2 more equations!
  
 
===External Forces on Lower pulley===
 
===External Forces on Lower pulley===
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T_3-T1-T2 = (0)a
+
:<math>T_3-T1-T2 =T_3-T1-(T1) =(0)a</math>
 +
::<math>T_3=2T_1</math>
  
 
==Step 5: Use Newton's second law==
 
==Step 5: Use Newton's second law==

Revision as of 11:47, 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

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]

Step 5: Use Newton's second law

Forest_UCM_NLM#Atwoods_Machine