Example
Question:
A wooden block of mass \(2\,\mathrm{kg}\) rests on a soft horizontal floor. When an iron cylinder of mass \(25\,\mathrm{kg}\) is placed on top of the block, the floor yields steadily and the block and cylinder together go down with an acceleration of \(0.1\,\mathrm{m\,s^{-2}}\). What is the action of the block on the floor
(a) before, and
(b) after the floor yields?
(\(\mathrm{Take}\ g = 10\,\mathrm{m\,s^{-2}}\)). Identify the action-reaction pairs in the problem.
Solution:
(a) The block is at rest on the floor. Its free-body diagram shows two forces:
- The force of gravitational attraction by the earth: \(2 \times 10 = 20\,\mathrm{N}\)
- The normal force \(R\) of the floor on the block.
By Newton’s First Law, the net force must be zero: \(R = 20\,\mathrm{N}\).
By Newton’s Third Law, the action (force by block on floor) is \(20\,\mathrm{N}\) downward.
(b) The system (block + cylinder) accelerates downward at \(0.1\,\mathrm{m\,s^{-2}}\):
Free-body diagram of the system shows:
- Gravity: \(270\,\mathrm{N}\)
- Normal force \(R'\) by the floor.
Applying Newton’s Second Law:
\[
270 - R' = 27 \times 0.1 \implies R' = 267.3\,\mathrm{N}
\]
By Newton’s Third Law, the system’s action on the floor is \(267.3\,\mathrm{N}\) downward.
Action-reaction pairs:
- For (a):
- (i) Gravity on block by earth (action, \(20\,\mathrm{N}\)); gravity on earth by block (reaction, \(20\,\mathrm{N}\) upward).
- (ii) Force on floor by block (action); force on block by floor (reaction).
- For (b):
- (i) Gravity on system by earth (action, \(270\,\mathrm{N}\)); gravity on earth by system (reaction, \(270\,\mathrm{N}\) upward).
- (ii) Force on floor by system (action); force on system by floor (reaction).
- (iii) Force on block by cylinder and on cylinder by block is also an action-reaction pair.
Remember: Action-reaction pairs are always equal and opposite between two bodies. Two forces acting on the same body, though equal and opposite, are not action-reaction pairs.
