(ii) Consider block A only and write Newton's second law for it:
Ox:−T cos30∘+μkNA=0,Oy:T sin30∘+NA−mAg=0. Express the normal force from the second equation:
NA=mAg−T sin30∘,−T cos30∘+μkmAg−μkT sin30∘=0, T=cos30∘+μk sin30∘μkmAg=14.5 N.
Consider only block B and find its acceleration from Newton's second law:
Oy:NB−mBg−NA=0,Ox:mBa=−μkNB−μkNA+N. a=mBN−μk(NB+NA)= =mBN−μk[mBg+2(mAg−T sin30∘)]=0.36 m/s2.
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