Why Don't Action-Reaction Forces Cancel Each Other Out?
Two-Block Contact Force: How a Push Splits Between Touching Blocks
One push, one acceleration. The touching blocks move as one system: a = F ÷ (m1 + m2). Only block 1 pushes block 2, so the contact force is block 2's share: Fc = m2a. With no friction and F = 0, Fc = 0, yet a moving pair keeps its speed.
Fc is a fixed fraction of F, F × m2 ÷ (m1 + m2), the slope of the graph. Make m2 large and m1 small: Fc nears F but never reaches it. The orange arrows are a Newton's third law pair, equal and opposite, one on each block.
Friction slows the pair but leaves the split alone while it slides: each block's μmg comes out of its own share. At rest the pair holds until F passes μ(m1 + m2)g; this sim lets block 1's grip take the push first, up to μm1g, and block 2 feels only the rest.
Why don’t action–reaction forces cancel each other out?
In short: Newton’s third law pairs are equal in size and opposite in direction, but they act on different objects. Forces can only cancel when they act on the same object, so a pair never adds up to zero on any single block. Whether a block accelerates depends only on the net force on that block.
Common doubts
- “If the forces are equal and opposite, why does anything move?” Because each force acts on a different block. Block 2 feels only the contact force Fc, so Fc = m₂a. Block 1 feels the push F forward and Fc backward, so F − Fc = m₁a.
- “Is the contact force equal to the push F?” No. Fc = F · m₂ ÷ (m₁ + m₂), only a fraction of F, because the push also has to accelerate block 1. Fc approaches F only when m₂ is much larger than m₁.
- “Does block 1 push harder on block 2 than block 2 pushes back?” No. The two forces are always equal, even while the blocks speed up.
Change F, m₁ and m₂ in the simulation above, press RUN, and compare the two orange arrows where the blocks touch.
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