[The legend]

Hello everyone!

This isnt exactly homework but I wanted to try out the sum on my own so I felt right posting this here.

In the figure in attachment there are three masses m1 , m2 , m3 all of which are unequal.
Then what must be the relation between the masses to keep m3 stationary? (neglect friction and string and pulleys are massless)

So my attempt was that for it to remain stationary (mass on both sides should be equal)m3 = m2 + m1. But thats not what the answer is.
So can anyone give me a hint?

Attached File(s)

•  sum.bmp (89.49K)
  Number of downloads: 29

[swansont]

Do you have any more information about m1 and m2, or are you free to chose the values? (Your answer is correct for m1=m2, but not otherwise, and we know m1≠m2) If you don't, you will have a somewhat long expression when all is said and done.

You need to look at the tensions in the ropes supporting the pulleys.

[Mr Skeptic]

Remember you only need to keep M3 stationary. A hint: what is the tension in the string holding M3? What is the tension in the string holding M1 and M2?

[The legend]

Right i get the hint!

I made these equations

m3 - 2T = 0

m1 - T = m1 * -a

m2 - T = m2 * a

then substituting values of T and equating

I got
4/m3 = 1/m2 + 1/m1

which is the right answer.

Thanks for your help!!