Difference between revisions of "Forest UCM NLM AtwoodMachine"

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solving the above system of equations leads to the 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}</math>
+
:<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}</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}</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>
  
  
 
[[Forest_UCM_NLM#Atwoods_Machine]]
 
[[Forest_UCM_NLM#Atwoods_Machine]]

Revision as of 12:42, 22 August 2014

Simple Atwood's machine

TF UCM SAM 1.gif


T=2m1m2m1+m2g

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

T1= Tension in the rope attached to mass m1
T2= Tension in the rope attached to mass m2=T1
T3= Tension in the rope attached to mass m3
Fg = force of gravity on each mass =m1g or m2g or m3g

Step 5: Use Newton's second law

for mass 1
T1m1g=m1a1
for mass 2
T2m2g=m2a2
for mass 3
T3m3g=m3a3


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 T1=T2 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


T3T1T2=T3T1(T1)=(0)a
T3=2T1


Now we have 4 unkwons and 3 equations

relative acceleration

let

ar= acceleration of m1 with respect to the lower pulley

assuming that a1 is moving upwards with respect to the earth

a1=ara3 : a3= acceleration of lower pully as well as m3


similarly

a2=ara3 : if m1 is accelerating upwards then m2 is accelerating downwards

3 equations and 3 unknowns

T1m1g=m1(ara3)


T1m2g=m2(ara3)


(2T1)m3g=m3a3


Solutions

solving the above system of equations leads to the solutions

a1=3m2m3m1m34m1m2m1m3+m2m34m1m2g
a2=3m1m3m2m34m1m2m1m3+m2m34m1m2g
a3=4m1m2m2m3m1m3m1m3+m2m34m1m2g


Forest_UCM_NLM#Atwoods_Machine