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alaix
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Homework Statement
Note: My textbook and first language is french, I am sorry if I mistranslated some terms.
A friction-less rope goes around the fix pulley A. At one side of the rope, there is a mass M1. On the other side, there is a pulley of mass M2. Another friction-less rope goes around that second pulley. Mass m1 and m2 are at the extremeties. Show that the acceleration of mass m1 is given by:
[itex]\frac{3*m2*M2-m1*M1-m1*M2-m2*M1-4*m1*m2}{(m2+m1)*(M1+M2)+4*m1*m2}[/itex]
Homework Equations
[itex]\sum F = ma[/itex]
The Attempt at a Solution
[URL]http://imgur.com/1OzYU[/URL]
For the first part, where we take The pulley M2 and the mass M1:
M1*a = M1*g-T where T is the tension
-(M2+m1+m2)a = (M2+m1+m2)g - T
T = M1g-M1a
a = [itex]\frac{-(M1-M2-m1-m2)}{M1+M2+m1+m2} g[/itex]
Then I take the second part separately:
m2a = m2g -T2 where T2 is the tension in that second rope
-m1a = m1g - T2
T2 = m2g-m2a
a = [itex]\frac{m2-m1}{m2+m1} g[/itex]
Then I add both acceleration to have the total acceleration of m1
[itex]a_{total} = (\frac{m2-m1}{m2+m1} - \frac{M1-M2-m1-m2}{M1+M2+m1+m2}) g [/itex]
I tried to put both fractions on same denumerator both my answer =/= to the answer I should get...
What am I doing wrong?
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