Q.A wooden block of mass $2\ \text{kg}$ rests on a soft horizontal floor. When an iron cylinder of mass $25\ \text{kg}$ is placed on top of the block, the floor yields steadily and the block and the cylinder together go down with an acceleration of $0.1\ \text{m s}^{-2}$. What is the action of the block on the floor

Newton's Third Law: the action of the block on the floor equals in magnitude, and is opposite to, the floor's reaction on the block.

1. Before the floor yields — the block+cylinder system is in equilibrium. Total weight $=(2+25)\times 10=270$ N.

   $N = 270\ \text{N upward (floor on block)} \implies \text{action of block on floor} = 270\ \text{N downward}$

2. After the floor yields — the system accelerates downward at $a=0.1\ \text{m/s}^2$:

   $(M+m)g - N' = (M+m)a \implies 270 - N' = 27\times 0.1 = 2.7 \implies N' = 267.3\ \text{N}$

   Action of block on floor $= 267.3$ N downward. …

The "action of the block on the floor" is the normal force that the block exerts downward on the floor. By Newton's third law, the floor exerts an equal and opposite normal force upward on the block. So finding the normal force on the block automatically gives the action of the block on the floor — they are a third-law pair.

The normal force is not always equal to the weight. It adjusts to match the net force required by the motion. Before the floor yields, everything is in equilibrium, so the normal force equals the total weight. After the floor yields, the system accelerates downward, so the normal force must be less than the weight — the net downward force is what causes the acceleration.

1. Identify the masses and the total weight.

   Mass of wooden block: $m_b = 2\ \text{kg}$

   Mass of iron cylinder: $m_c = 25\ \text{kg}$

   Total mass: $M = m_b + m_c = 27\ \text{kg}$

   Total weight: $W = Mg = 27 \times 10 = 270\ \text{N}$

2. Case (a): Before the floor yields.

   Before yielding, the floor is rigid enough to prevent any downward motion, so the system is in static equilibrium: net force is zero.

By Newton's third law, the action of the block on the floor is equal in magnitude and opposite in direction to $N$: the block pushes down on the floor with $270\ \text{N}$.

3. Case (b): After the floor yields.

   Now the floor gives way, and the whole system accelerates downward at $a = 0.1\ \text{m/s}^2$. Taking downward as positive:

Again, by Newton's third law, the action of the block on the floor is $267.3\ \text{N}$ downward. …

Concept: Newton's Second/Third Law for a system supported by a yielding floor

Step 1: Total weight of block + cylinder

Step 2 (before yielding): the system is in equilibrium

By Newton's third law, the block pushes down on the floor with 270 N.

Step 3 (after yielding): the system accelerates downward at $a = 0.1\ \text{m s}^{-2}$

Taking downward as positive:

By Newton's third law, the block now pushes on the floor with 267.3 N. …