A wooden block of mass 1 kg rests on a soft horizontal floor. When an iron cylinder of mass 20kg is placed on top of the block, the floor yields steadily and booth go down with an acceleration of `0.1ms^(-2)`. What is the action of the block on the floor (a) before and (b) after the floor yields? (Take `g=10ms^(-2)`). Identify the action and reaction pairs in the problem.

Explanation

(a) The block is at rest on the floor. Its FBD shows two forces on it,
Weight = mg
`=1xx10`
`=10N`
and the normal force R of the floor on the block.
Net force = 0,
According to first law `rArr R=10N` directed upwards
(b) The system (block + cylinder) accelerates downward with `0.1ms^(-2)`. The FBD shows weight of hte system `(=(20+1)xx10=210N)` and normal force R' by the floor. FBD does not show internal forces between the block and the cylinder.
`210-R'=21xx0.1 " "` (Applying second law to the system)
`R'=207.9N`
Action of the system on the floor = 207.9 N vertically downwards
Action-reaction pairs
For (a) : (i) Action : Force of gravity (10N) on the block by the earth.
Reaction: Force of gravity on the earth by the block.
(ii) Action: Force on the floor by the block.
Reaction: Force on the block by the floor.
For (b): (i) Action: The force of gravity (210 N) on the system by the earth.
Reaction: The force of gravity (210 N) on the earth by the system.
(ii) Action: Force on the floor by the system.
Reaction: Force on the system by the floor.
Remember that an action-reaction pair consists of mutual forces which are always equal and opposite between two bodies. The forces on the same body connot constitute an action-reaction pair. For example, the force of gravity on the block in (a) or (b) and the normal reaction on the block by the floor are not action - reaction pairs. These forces are equal and opposite for (a) as the block is at rest. For (b), they are not same. Weight = 210 N, while R' = 207.9 N.