The resultant normal of all infinitesimal reactions MUST pass thru part B to neutralize the clockwise torque produced by friction.
The trend of normal reaction is shown..this trend may be linearly increasing, but i cannot already say.. clearly the reaction MUST be more on part B, else the resultant WILL NOT pass thru B.
but i did it another way i considered them as two blocks for A part we get mgsin(o)/2-N1-uN2=ma/2
ReplyDeletefor B
mgsin(o)/2+N1-uN3=ma/2
where u is cofficient of friction
N1 is normal reaction between A and B
N2 between A and surface
N3 between B and surface
from comparing we will get N3> N2
and thus N<N*
bhaiya plz tell is this way it is correct to do that ques
If you let them 2 blocks and coordinate axis along the incline and perpendicular to it.
ReplyDeleteas mass of 2 blocks is same so, you will directly get N2=mgcos(o)/2
N3=mgcos(o)/2
therefore N2=N3
and N1 is the action reaction pair.
got by balancing force perpendicular to incline.
bhaiya can we find how normal force is varying with distance???
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