Solve Friction Problem: Find Acceleration of M | N1/N2 Normal Forces

[sachin123]

[URL]http://img600.imageshack.us/i/22258756.jpg/[/URL]
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Find acceleration of M.
N1 is normal force between the two blocks and N2 is that between the ground and M.

Let M move with acceleration 'a' towards right and so,m moves with acceleration '2a' downward and 'a' to the right.
Here are my equations:
For M,
2T-N1n1-N2=Ma(horizontal)

and
Mg-T-N2n2=0(vertical)

For m,
mg-T-N2n2=m(2a)

and
N2=ma
Are all these correct?
I am not getting the correct answer.

Thank You.

 

[graphene]

Hello,

Check your second equation - vetical for M.
you have missed $$N_1$$. and are the signs for $$N_2n_2$$ and T alright?

 

[sachin123]

Hi.
Oh yes how dumb of me.
So the correct ones are:

For M,
2T-N1n1-N2=Ma(horizontal)

and
Mg-T+N2n2-N1=0(vertical)

For m,
mg-T-N2n2=m(2a)

and
N2=ma
Are all these correct now?

 

[graphene]

Still, in the 2nd eqn, T is in the direction of Mg. So, its got to have a +ve sign.

 

[sachin123]

Actually i am little muddled up with the vertical eq.
T pulls the big block M right?Mg brings it down.Why +ve signs for both then?
What are the vertical forces on M?
Mg,N1,friction between the two blocks right?
Where does the T come from anyway?
I don't know why I put T.

 

[graphene]

they pulley on the top-right of the big block, experiences T towards the right & towards the bottom.

 

[sachin123]

Oh and pulley is attached to M.That was clearer.
So finally,For M,
2T-N1n1-N2=Ma(horizontal)

and
Mg-T+N2n2-N1=0(vertical)

For m,
mg+T-N2n2=m(2a)and
N2=ma
Correct?

 

[graphene]

Still not
T should have a +ve sign (2nd eqn).
Mg, T and $$N_2n_2$$ point downwards while $$N_{1}$$ points upwards.

 

[sachin123]

Oh I'm sorry graphene,I understood but didn't put it carefully.

For M,
2T-N1n1-N2=Ma(horizontal)

and
Mg+T+N2n2-N1=0(vertical)

For m,
mg-T-N2n2=m(2a)


and
N2=ma