A steel block of $10\, {kg}$ rests on a horizontal floor as shown. When three iron cylinders are placed on it as shown, the block and cylinders go down with an acceleration $0.2\, {m} / {s}^{2}$. The normal reaction ${R}$ by the floor if mass of the iron cylinders are equal and of $20\, {kg}$ each, is .....$N.$ [Take ${g}=10\, {m} / {s}^{2}$ and $\mu_{{s}}=0.2$ ]
JEE MAIN 2021, Difficult
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Writing force equation in vertical direction
${Mg}-{N}={Ma}$
$\Rightarrow 70 {g}-{N}=70 \times 0.2$
$\Rightarrow {N}=70[\,{g}-0.2]=70 \times 9.8$
$\therefore {N}=686\, \text { Newton }$
Note : Since there is no compressive normal from the sides, hence friction will not act.
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