• # question_answer Two blocks are placed on a wedge with coefficients of friction being different for two blocks. Choose the correct option (friction is not sufficient to prevent the motion) A)  if${{m}_{1}}<{{m}_{2}}$, then normal reaction between blocks will be non-zero B)  if${{m}_{1}}={{m}_{2}}$then normal reaction between the blocks will be zero C)  if${{\mu }_{1}}={{\mu }_{2}}$, then normal reaction between the blocks will be zero D)  None of the above

Idea The normal reaction between the blocks will be non-zero only if the initial acceleration of block ${{m}_{2}}$ is more than block${{m}_{1}}$. Then, only the block ${{m}_{2}}$ will push block ${{m}_{1}}$ and normal reaction will develop and it is possible only when ${{\mu }_{2}}<{{\mu }_{1}}$ The normal reaction between the blocks will not depend on the masses ${{m}_{1}}$and${{m}_{2}}$. If ${{\mu }_{1}}={{\mu }_{2}}=\mu$ $\Rightarrow$ ${{a}_{1}}=(g\sin \theta -\mu g\cos \theta )$ $\Rightarrow$${{a}_{2}}=(g\sin \theta -\mu g\cos \theta )$ $\Rightarrow$${{a}_{1}}={{a}_{2}}$ So, there is no relative motion between the blocks, so normal reaction between them is zero. TEST Edge Both are identical planes, if m does not slip then M will? M will also not slip because in this case motion of block will depend only on (l not on mass of the block.