A wooden block of mass $2\mathrm{kg}$ rests on a soft horizontal floor. When an iron cylinder of mass $25\mathrm{kg}$ is placed on top of the block, the floor yields steadily and the block and the cylinder together go down with an acceleration of $0.1\mathrm{m}{\mathrm{s}}^{-2}$. What is the action of the block on the floor (a) before and (b) after the floor yields ? Take $g=10\mathrm{m}\mathrm{s}{}^{-2}$. Identify the action-reaction pairs in the problem.

For (a):
(i) the force of gravity (20 N) on the block by the earth (say, action); the force of gravity on the earth by the block (reaction) equal to 20 N directed upwards (not shown in the figure).
(ii) the force on the floor by the block (action); the force on the block by the floor (reaction).
For (b):
(i) the force of gravity (270 N) on the system by the earth (say, action); the force of gravity on the earth by the system (reaction), equal to 270 N, directed upwards (not shown in the figure).
(ii) the force on the floor by the system (action); the force on the system by the floor (reaction). In addition, for (b), the force on the block by the cylinder and the force on the cylinder by the block also constitute an action-reaction pair.