In the arrangement shown in the figure mass `m_(A)=2kg, m_(B)=4kg and m_(C)=6kg`. Co-efficient of static friction between A and B `(mu_(1))=0.4`, between B and C `(mu_(2))=0.3` and between C and ground `(mu_(3))=0`. A constant force F = 32 N is applied horizontally on block B. Find friction force developed between A and B and that between B and C.
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Maximum frictional force, `f_("1max")=0.4xx2xx10=8N` `f_("2max")=0.3 xx 6xx10=18N` Case 1: Assuming all blocks moves together. For block A: `f_(1)=2a` For block B: `32-f_(1)-f_(2)=4a` For Block C : - `f_(2)=6a` solving above equations, `f_(1)=5.33N, f_(2)=16N` `f_(1) lt f_("1max") and f_(2) lt f_("2max")`. Hence assumption is correct .
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