# A problem of strings, pulleys and weights

## Homework Statement

The fig has three masses m1 , m2 , m3 .. all of which are unequal.
Find relation between masses so that m3 remains stationary.

( pulleys , strings are frictionless and massless)

## The Attempt at a Solution

I thought if mass on both side are equal then the system remains stationary.
So i said

m3 = m1 + m2

But the answers say a different story...

#### Attachments

• sum.bmp
89.5 KB · Views: 513

## Answers and Replies

is the answer only in terms of masses? or is there any other quantity (such as acceleration)

The answer is only in masses (that is the relation between them)

anyone please ..atleast give me a hint?

I guess the concept of reduced mass is to be used here.
reduced mass of m1 and m2 is (m1 x m2)/(m1 + m2) which is equal to m3
Is this the correct answer?
Iam not sure of this.

No it isnt right.
The correct answer is

4/m3 = 1/m1 + 1/m2

I still dont konw how to get it. And i dont think this is a problem of reduced mass.(neglect gravitational force between the masses m1 m2 m3)

I still didnt get a hint(though I feel I have followed the rules right).
I still tried out the sum but reached nowhere!

I got the correct answer.
Please ignore my previous post as I did not think in detail before writing.
I will give you a hint.
As it is given the masses are unequal, the masses attached to the lower pulley (m2 and m1) won't be at rest. So, they have some common acceleration. You first calculate the tension in the string connecting these two masses. Use this equation to find the tension in the string connecting the mass m3.

Now, for m3 to remain at rest, this tension should balance its weight.
equate its weight and tension.
You will get the answer.

A.Q.

aha! Thanks A.q ! I got it!