2305If T is the tension in the string joining m3 and T' be the tension in the string joining m1 to m2.
Without loss of generality we can assume m1>m2.
a be the acc of m3 upwards and a' be the acc. of m1 downwards.
Then,
m1g - T' = m1a + m1a'.....(1)
T' - m2g= m2a' - m2a....(2)
T - m3g = m3a.....(3)
Substituting T = 2T' and a=0 we get the required result.
Akash Anand Excellent work ..keep it up 
996m3g=m1g+m2g
Akash Anand Not the right approachAyush Lodha if we know the which mass is greater m1 or m2 ,sir we can then get the condition by constraint relation
229Tension in the string should be. m2g(1+m1-m2) by (m1+m2)
i.e m3g= to the expression above
and the acc. In the blocks m1 and m2 should be---
(g(m1-m2))/(m1+m2)
Dwijaraj Paul Chowdhury m3g=2 times the exp. for TAkash Anand Not Good
158m1=m2=m32......??
(if u consider pulley to be massless...)
@Dwijaraj...can u pls explain how u got it...i mean the equations??
Dwijaraj Paul Chowdhury For m3 to be at rest...m3g=2T
and force balancing equatiins are
m1g-T=m1a------(i)
T-m2g=m2a------(ii)
Solving these....a=g*(m1-m2)/m1+m2)
then putting 'a' in equation (i) or (ii)...find T
As already said...m3g should be =2T
Akash Anand WRONG..661T=[|m1-m2|+(m1+m2)]*m2g/(m1+m2)
also T=[|m1-m2|-(m1+m2)]*m1g/(m1+m2)
so m3=2*[|m1-m2|+(m1+m2)]/(m1+m2)
or we can also say m3=T+T
=[|m1-m2|+(m1+m2)]*m2g/(m1+m2)+
[|m1-m2|-(m1+m2)]*m1g/(m1+m2)
SO m3=|m1-m2|-(m2-m1)
now if m1>m2
m3=2(m1-m2)
if m1<m2
m3=0,which is not possible
so m3=2(m1-m2)
Akash Anand I did not get the logic behind first two equations. Its logically wrong I guess.1161Till now ..No one is even close to the right answer. Keep trying.
Akash Anand This is a very good and very important question.661if T is the tension in string connecting m1 and m2 then
2T=m3g
also T-m1g=m1a....(i)
m2g-T=m2a....(ii)
solving (i) and (ii) we get
|a|=|m1-m2|*g/(m1+m2)
so putting this in eq(i) we get
T=m1(m2-m1)*g/(m1+m2) + m1g(as we have assumed a for m1 in upward direction)
T=m1m2g/(m1+m2)
so m3=2T=2m1m2/(m1+m2)
661if T is the tension in string connecting m1 and m2 then
2T=m3g
also T-m1g=m1a....(i)
m2g-T=m2a....(ii)
solving (i) and (ii) we get
|a|=|m1-m2|*g/(m1+m2)
so putting this in eq(i) we get
T=m1(m2-m1)*g/(m1+m2) + m1g(as we have assumed a for m1 in upward direction)
T=m1m2g/(m1+m2)
so m3=2T=2m1m2/(m1+m2)
Akash Anand A bit more correction is needed. Right answer is bit differentGaurav Gardi there iss a calculation mistake here
it is T=2m1m2
so m3=2T=4m1m1/(m1+m2)229m3= (m1-m2)/(m1+m2)+ m2...considering. m1 is more than m2
Dwijaraj Paul Chowdhury Sorry m3= 2times the exp.Akash Anand Give the final answer
1133m3 = 4m1m2/m1+m2
Akash Anand Good Work Sourish
Why you are not active in the forum these days?Sourish Ghosh I'll be active again from now on. Needed a break for school exams.