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

Explanation

(a) The block is at rest on the floor. Its free-body diagram shows two forces on the block, the force of gravitational attraction by the earth equal to `2 ×x 10 = 20 N` , and the normal force R of the floor on the block. By the First Law, the net force on the block must be zero i.e., R = 20 N. Using third law the action of the block (i.e. the force exerted on the floor by the block) is equal to 20 N and directed vertically downwards.
(b) The system (block + cylinder) accelerates downwards with `0.1 m s^(-2)` . The free-body diagram of the system shows two forces on the system : the force of gravity due to the earth (270 N), and the normal force R′ by the floor. Note, the free-body diagram of the system does not show the internal forces between the block and the cylinder. Applying the second law to the system,
`270 - R. = 27 xx 0.1 N`
`i.e., R. = 267.3N`

By the third law, the action of the system on the floor is equal to 267.3 N vertically downward