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 , Is FBD shows two forces on it,
Weight ` = mg = 1 xx 10 = 10 N`
and the normal force R of the floor on the block .
Net force = 0 , according to first law `rArr R = 10 N " directed upwards "`
(b) The system (block + cylinder ) accelerates downward with `0*1 "ms^(-2)` . The FBD shows weight of the system `( = (20 +1)xx 10 = 210 N) ` and normal force R' by the floor. FBD does not show internal forces between the block and the cylinder .
` 210 - R' = 21 xx 0*1 ` (Applying second law of the system )
` R' = 207 . 9 N`
Action of the system on the floor = `207 .9 N ` vertically downwards
Action - reaction pairs
For (a) : (i) Action : Force of gravity (10 N) 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 consits of mutual forces which are always equal and oposite between two bodies . The forces on the same body cannot 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 . 210 N, while R' = 207.9 N