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/s². 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/s² and μₛ = 0.2]

Solution

Physics sir · 4 steps

Take the block and all three cylinders as one 70 kg body sinking with a: the floor pushes less than 700 N.

1. One system. Everything moves together, so take the block and the three cylinders as one body. Its mass is 10 + 3 × 20 = 70 kg, and its weight is 700 N.
2. Newton's second law. The weight Mg acts down and R' acts up. The system accelerates downward, so the downward force is larger and the net force points down.

   Mg - R' = Ma ⇒ R' = M(g - a) = 70 × (10 - 0.2)

3. Friction is a distractor. μₛ governs horizontal sliding. Nothing here is pushed sideways, so friction does not enter the vertical balance.
4. So. R' = 70 × 9.8 = 686 N, which is option (a).

The idea to keep. For a body accelerating vertically, the normal force is M(g - a) when it goes down and M(g + a) when it goes up. Check the direction of a first.

Answer from the NTA final answer key. Solution by PrepWiser.