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my last asked question is answered by vikas askiitins expert and it is wrong plese give the correct answer and the answer is 10 N
dear aditya, i found no mistakes in my solution which i had posted ...similar question is in arihant DC pandey try to solve that with my method and and u will get exactly same ans.... may be the book from which u had taken this q has printed wrong ans .....if u get the solution then please send it to me...
at this point let me clear ur doubtu say answer is 10if it is so from eq 2N=9.8=>f(friction)=4.9 but 4.9 is not sufficient to hold the small blockat its placethus bw m and M there is normal reaction let it be N, then at m F-N=ma N=F-ma or =F(1 - m/(m+M) ) N=F( M/m+M)................2 if both blocks are at rest wrt each other then it can be said that friction is sufficient to balance the weight of block of mass m.. (friction force) f =mg f =u(N) (u is cofficient of friction and N is normal reaction) uN=mg u(F)(M/m+M) =mg F=mg(m+M)/uM on solving F=240Nthis is the minimum force required for which block remmains in equilibrium...if force if more then friction is more and then also block remains in equilibrium,so 240N is the value is minimum possible value of force... should be the answer
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