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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 (in N), [Take $$g = 10$$ m s$$^{-2}$$ and $$\mu_s = 0.2$$]
$$M = m_{\text{block}} + 3 \times m_{\text{cylinder}} = 10 + 3(20) = 70\ \text{kg}$$
$$Mg - R' = Ma$$
$$\implies R' = M(g - a)$$ $$\implies R' = 70(10 - 0.2) = 70 \times 9.8 = 686\ \text{N}$$
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