24 Replies
@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.Anyone ????
You just have to draw FBDs as usual
try to split the forces along and perpendicular to the incline for m
just write the Normal force tension force and mg n solve the eqns
isme pseudo kyu lgega?
use T.v
once th system is released m will exert some force thus pushing 2m behind
usi mein to dikkat aa rhi
that is after t=0
guess mera bhi constrained motion bahut weak hai
as soon as the system releases conditions change
we solve at t=0
Bhai angle change thodi hi ho rha hai
@hardcoreisdead we goood?
Do you want another ping on the physics helpers? @hardcoreisdead
i would tbh
Do you have the answer
There's a minor mistake in my solution
2T = 2ma + m(a - bcostheta) hoga instead of (bcostheta - a)
After that a = (2gsintheta)/(7-4costheta)
Mere and ashish sir ke equations dono se ara
but i want approach from ground frame, without pseudo force
pseudo use nhi kiya
Relative motion use kiya
From ground frame
oh , my bad . ill check out your solutuon in a few
you good?
how do i mark it solved ,cnat find this persons @
@iTeachChem
Gotcha
+fsolved
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