2.3 Newton’s Third Law — Worked Solutions
Work through Topic 2.3
These are the full worked solutions to the lesson questions. Try the questions first, then check each step and the reason for it.
Question 1
A car and truck collide. Which statement about their interaction forces is correct?
- The truck exerts the larger force
- They exert equal and opposite forces on each other
- They exert equal forces only if their masses match
Step 1. Identify the pair: force of truck on car and force of car on truck.
Why this step? Both forces come from the same collision interaction, with the source and receiving object exchanged.
Step 2. Newton’s third law gives equal magnitudes and opposite directions at each instant.
Why this step? The pair equality does not depend on equal masses, rest, or equal collision damage.
Step 3. The lighter object can still have larger acceleration because a = F/m.
Why this step? The force magnitude can match while division by different masses produces different responses, so damage is not evidence of unequal paired forces.
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Question 2
A book rests on a table. Are gravity and the normal force on the book a third-law pair?
- Yes, because they cancel
- No, because both act on the book
- Only if the book has zero acceleration
Step 1. Gravity on the book is Earth-on-book; its partner is book-on-Earth.
Why this step? A third-law partner exchanges the source and receiver within the same gravitational interaction.
Step 2. Normal force on the book is table-on-book; its partner is book-on-table.
Why this step? The table contact is a separate interaction from gravity, even when the two book forces balance.
Step 3. Gravity and normal force on the book are therefore not a third-law pair.
Why this step? Both selected forces act on the same object; equal and opposite equilibrium forces are not sufficient to make an interaction pair.
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Question 3
A heavy chain hangs at rest. Where is its tension greatest?
- Near the top
- Near the bottom
- It must be identical everywhere
Step 1. Consider the portion of chain hanging below a cut near the top. The tension at that cut supports its weight.
Why this step? The chain is at rest, so the force exerted across the cut must balance the gravitational force on the lower portion.
Step 2. A cut farther down has less chain below it and requires less supporting tension.
Why this step? The mass below the cut decreases with height down the chain, reducing the weight the cut must support.
Step 3. The tension is greatest near the top.
Why this step? Uniform tension assumes negligible string mass; that assumption fails for the explicitly heavy chain.
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Question 4
A 2 kg cart pushes a 5 kg cart with 10 N on a frictionless track. Find each acceleration due to that interaction alone and identify which cart feels each force.
Step 1. Take the 10 N force on the 5 kg cart as positive. Its interaction-only acceleration is 10/5 = +2 m/s².
Why this step? The stated force acts on the 5 kg cart; dividing by that cart’s own mass gives its response.
Step 2. The 2 kg cart receives the opposite force, −10 N, and acceleration −10/2 = −5 m/s².
Why this step? Newton’s third law reverses the force but does not equalize masses, so this cart has the larger acceleration magnitude.
Step 3. The interaction-only mass-weighted force sum is 5·2 + 2·(−5) = 0 N.
Why this step? This checks internal cancellation for both carts together; any separately applied external force would have to be added to its recipient’s equation.
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Question 5
Draw separate FBDs for a hanging mass and the ideal string supporting it. Explain why an arrow for mass-on-string doesn’t belong on the mass’s diagram.
Step 1. On the hanging mass, draw string-on-mass upward and Earth-on-mass downward.
Why this step? The mass FBD shows only forces received by the mass; the source labels distinguish the two interactions.
Step 2. On the ideal string, draw mass-on-string downward at the lower end and support-on-string upward at the upper end.
Why this step? The downward string force is the third-law partner of string-on-mass, so it acts on the string rather than on the mass.
Step 3. For a negligible-mass string with finite acceleration, the endpoint force magnitudes balance, giving the same tension at both ends.
Why this step? Applying ΣF = m_string a with m_string ≈ 0 leaves no finite net force on the string; a massive chain would need a different model.
Step 4. If the mass is also at rest, its own equation gives T = mg.
Why this step? Equal tension along an ideal string does not alone establish T = mg; that additional equality requires zero acceleration of the mass.
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Question 6
Two force sensors give F₁ = +12.0 N and F₂ = −11.7 N, each with uncertainty ±0.3 N. Is this clear evidence against Newton’s third law? Explain the comparison.
Step 1. Reverse the sign of the second reading to compare paired magnitudes: 12.0 N and 11.7 N differ by 0.3 N.
Why this step? The third law predicts opposite signed forces, so comparing F₁ with −F₂ tests equal magnitudes rather than incorrectly demanding the same sign.
Step 2. The first magnitude range is 11.7–12.3 N; the second is 11.4–12.0 N. They overlap from 11.7 to 12.0 N.
Why this step? Each reading allows ±0.3 N; at least one common magnitude is compatible with both measurements, so the mismatch is not resolved by these data.
Step 3. The predicted signed sum is zero; the measured sum is +0.3 N with a conservative combined interval of −0.3 to +0.9 N.
Why this step? Adding the stated uncertainty bounds gives ±0.6 N for the sum without assuming an unprovided statistical model; that interval includes zero.
Step 4. These readings are not clear evidence against the third law; compare calibrated, simultaneous readings and their uncertainties.
Why this step? The prediction remains within the measurement bounds, and time mismatch or sensor offset could also imitate unequal forces.