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Two blocks sit side by side on a smooth floor and you push the first one. Both move, so the first block must push the second. How hard? Students often answer "F", or "the same as the push", and lose the mark. This post gives one rule that settles every version of the question, with worked numbers for two blocks, three blocks, opposing pushes and friction.
Where is this in the syllabus?
The JEE (Main) 2026 syllabus lists Newton's second and third laws under Laws of Motion, and the NEET (UG) 2026 syllabus has the same entries. The NCERT Class 11 Laws of Motion chapter (2026-27 reprint) covers both, with a section on solving problems in mechanics and Example 4.12 on identifying action-reaction pairs between a block, a cylinder and the floor. Its Exercise 4.15 asks the closely related string version: two bodies tied together and pulled from either end. Check the current year's syllabus.
What is the one rule?
Choose the block, or group of blocks, that lies ahead of the contact you care about. The only horizontal force on that group from outside is the contact force, so:
contact force = (mass ahead of the contact) × (common acceleration)
The blocks stay touching, so they share one acceleration, a = F/(total mass), on a smooth floor.
Worked example 1: two blocks, pushed from each side
Blocks A (2 kg) and B (3 kg) touch on a smooth floor. A horizontal force of 20 N pushes A towards B.
a = 20/(2 + 3) = 4 m/s².
Only the contact force N acts horizontally on B, so N = 3 × 4 = 12 N.
Check with A: 20 − N = 2 × 4, so N = 12 N. Both diagrams agree.
Now push B towards A with the same 20 N. The acceleration is still 4 m/s², but the block ahead is A, so N = 2 × 4 = 8 N. The contact force changed from 12 N to 8 N just by pushing from the other side.
Worked example 2: three blocks in a row
Blocks of 1 kg, 2 kg and 3 kg sit in a line, and 24 N pushes the 1 kg block.
a = 24/6 = 4 m/s².
Contact between the 1 kg and 2 kg blocks: mass ahead = 5 kg, so 5 × 4 = 20 N.
Contact between the 2 kg and 3 kg blocks: mass ahead = 3 kg, so 3 × 4 = 12 N.
The contact forces fall along the row, because each contact has less mass left to accelerate. Check the middle block: 20 − 12 = 8 N = 2 × 4. It works.
Worked example 3: pushes from both ends
A (4 kg) is pushed right with 30 N, and B (6 kg), touching A on its right, is pushed left with 10 N.
Net force 20 N right on 10 kg, so a = 2 m/s² to the right.
On B: N − 10 = 6 × 2, so N = 22 N.
On A: 30 − N = 4 × 2, so N = 22 N.
The contact force, 22 N, is more than the 10 N push on B, because it has to cancel that push and still accelerate B.
Does friction change the answer?
Worked example 4. Go back to Example 1 (20 N on A, 2 kg and 3 kg), but now both blocks slide on a floor with μ = 0.2. Take g = 10 m/s².
Total friction: 0.2 × 50 = 10 N, so a = (20 − 10)/5 = 2 m/s².
On B: N − 0.2 × 30 = 3 × 2, so N = 6 + 6 = 12 N.
The same 12 N as on the smooth floor. That is no accident: with the same μ for both blocks, N = m₂(a + μg) = Fm₂/(m₁ + m₂). Change the coefficients and it changes. With μ = 0.1 under A and 0.3 under B, a = (20 − 2 − 9)/5 = 1.8 m/s² and N = 3 × 1.8 + 9 = 14.4 N.
Why do students get this wrong?
The usual error is thinking the push "passes through" unchanged, or that the bigger or faster block pushes harder on the other. Brown (1989) found that students tend to hold a naive view of force as a property of objects, and argued that Newton's third law, which makes the relational nature of force explicit, needs more emphasis in teaching. In this problem the third law says A pushes B exactly as hard as B pushes A, whichever block is pushed and however heavy each one is. What differs between cases is the size of that shared force.
Which mistakes should you avoid?
Setting the contact force equal to F. That would leave nothing to accelerate the first block.
Drawing both third-law forces on one block. One goes on A's diagram, the other on B's.
Using the mass behind the contact. Look at the blocks the contact force has to drive.
Forgetting friction on the block ahead when the floor is rough.
For the diagram habit that makes these quick, see our guide to free-body diagrams and Newton's second law. If friction between stacked blocks is the problem, read friction block problems.
A practice routine
Solve Example 1 from both sides without notes and explain in one sentence why the answers differ.
Do three rows of three or four blocks, finding every contact force and checking each middle block.
Add friction with equal, then unequal, coefficients.
Finish with the string version: NCERT Exercise 4.15.
If the same step trips you up each time, the Fermi Problem Library has a step-by-step solver that explains the reasoning rather than printing an answer.
Next, find both contact forces when 30 N pushes a row of 3 kg, 2 kg and 1 kg blocks from the 3 kg end.
