# Laws of Motion-4

2305
Shaswata Roy ·

If 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
996
Swarna Kamal Dhyawala ·

m3g=m1g+m2g

• Akash Anand Not the right approach
• Ayush Lodha if we know the which mass is greater m1 or m2 ,sir we can then get the condition by constraint relation
996
Swarna Kamal Dhyawala ·

2T=m3g

• Akash Anand What is T???
229
Dwijaraj Paul Chowdhury ·

Tension 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)

158
Anik Chatterjee ·

m1=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..
661
Gaurav Gardi ·

m3=|m1-m2|/(m1+m2)

• Akash Anand Wrong ...try again :)
661
Gaurav Gardi ·

m3=2|m1-m2|/(m1+m2)

• Akash Anand Wrong ...try again :)
661
Gaurav Gardi ·

T=[|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.
1161
Akash Anand ·

Till now ..No one is even close to the right answer. Keep trying.

• Akash Anand This is a very good and very important question.
661
Gaurav Gardi ·

if 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)

661
Gaurav Gardi ·

if 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 different
• Gaurav Gardi there iss a calculation mistake here it is T=2m1m2 so m3=2T=4m1m1/(m1+m2)
229
Dwijaraj Paul Chowdhury ·

m3= (m1-m2)/(m1+m2)+ m2...considering. m1 is more than m2

1133
Sourish Ghosh ·

m3 = 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.
2305
Shaswata Roy ·

m3 = 4m1m2m1+m2