Recent content by Erdi

Okai, i Will try to solve that little later today.

Yes a couple People on here have mentioned LaTex. So i think i Will look into it.

Hey! yeah i got 2/15 that is equal to 0.1333kg. Its my confusing writing on here that is the problem. I DO NOT mean function or force when i wrote a(m1). I don't know how to write a small the subscript. But i mean acceleration for m1 that is zero!

I messed up on the solve for M part, So i got M = 0.1333kg ANy chance you solved for M with the values and got the same?

I could replace this with another expression. I see what you mean So i solved for 2*a(m1) = -(a(M)+a(m2)), and so i found expression for a(M,m2) 2*a(m1) =-(g-(T2/M) + g-(T2/m2)) Replacement: 2*a(m1) =-(2g-(T1/2) * ((1/M) + (1/m2)), T1 = m1g - m1a 2*a(m1) = -(2*g - (m1/2) * (g-a(m1)) *...

So M will be traveling upwards, if I am correct.

I can quickly go through my math here, firstly i got: m1: m1g -T1 = m1a m2: m2g-T2 = m2a These are the downward "forces" (m*g) minus the upwardforces tension(T) M: Mg - T2 = Ma Pulley B: 2*T2 - T1, from here i can see out that T2 equals T1/2 Then i got to the acceleration relative to each...

Thanks guys! I think i found an expression for T2 and i came to an answer for M.

Okay thanks for your help tho!

Back to stage 1 again. That is m2g isn't it??

I know i edited it, guess it didnt go throug..

Well i think the formula is correct, so what you mean is: T - Mg = M*(-a)

T1 + mg = ma?

Thats also got to T1 - Mg = ma

If that's not T - mg = ma or just T - mg , then i really don't understand at all