A wooden block of mass 2 kg rests on a soft horizontal floor. When an iron cylinder of mass 25 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 m s^(-2)`. What is the action of the block on the floor (a) before and (b) after the floor yields ? Take g = `10 m s^(-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