1Oh and the 2nd FBD is of M.
PLease see page 98 vol1 friction chapter,28 problem.
I am not getting the answer.
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.
I have a put a sincere effort in doing this,so please help me.
Thank You.
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8 Answers
21just one FORCE is missing in the FBD of M
which is the reaction of friction force which is acting in downward direction :)
it opposes the downward motion of m and tries to accelerate M in downward direction:)
1hi shubhodip.I have included that in the vertical eq.(N2n2?)
In fact I missed the nomal force by ground on M.
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?What do you say?
21i just saw ur FBD there is no force in downward direction other than Mg.
u just carefully list up all forces answer will follow
