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 xx 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 to 20 N and directed the block ) is equal to 20 N and directed vertically downwards .
(b) The system (block + cylinder) accelerated downwards with 0.1 `ms^(-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 , brgt 270 - R' = `27 xx 0.1 N`
ie. R' = 267.3 N
By the third law , the action of the system on the floor is equal to 267.3 N vertically downward.
Action-reaction pairs
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) .
Fig (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 ,