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@Gyro Gearloose
Note for OP
+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.Where am I wrong
ig u have not taken a3 for those m3 and m4 in constraint relation eqn where u have just written a1=2a2, acc to me it should be a3+2a2+a1=0 (i took all acc in downward dirn.)
3 and 4 would not have an acceleration w.r.t each other, so consider the pulley-m3-m4 to be a system of mass 2 kg. Now, a1+a_system+2a2=0, sue to symmetry between 1 and system, a1=a(system).
isnt a3=a4 so after applying virtual work method they just get cancelled
ohh , will try again like this
They won't as they dont have any acceleration w.r.t each other. In the ground frame tho, both have same accelerations in the same direction
hmm makes sense
also for m3 and m4 we use reduced mass formula toh get M=2 right ?
Not reduced mass formula, rather you consider the system as one rigid body with net mass being mass of pulley+mass of m3+mass of m4=2
when do we reduced mass formula then
...
You use reduced mass system when mass sort of acts in "parallel" sort of condition. For instance, a binary star system or spring joining 2 masses.
how to approach for it's (A) part?
how did you approach for (A) @hardcoreisdead
we good?
+solved @SirLancelotDuLac
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