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A force `F` is applied on block `A` and `B` and between the blocks `B` and `C` respectively are (Assume frictionless surface) A. `(F)/(7),(2F)/(7)`B. `(6F)/(7),(4F)/(7)`C. `F,(F)/(7)`D. `(4F)/(7),(6F)/(7)` |
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Answer» Correct Answer - B `a=(F)/(m+2m+4m)=(F)/(7m)` Let normal force between `A` and `B` is `N_(1)` , Then `N_(1)=(2m+4m)a=(6f)/(7)` and between `B` and `C` is `N_(2)` , then `N_(2)=4ma=(4F)/(7)` . |
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