Centre of Mass: Edge Cases and Subtle Traps (4)

easy • 3 min read
Tags Center Of Mass

Question

A man of mass 60 kg60 \text{ kg} stands at one end of a 40 kg40 \text{ kg} boat of length 4 m4 \text{ m} floating on still water. He walks to the other end. Ignoring water friction, by how much does the boat move (relative to the ground), and in which direction?

Solution — Step by Step

No external horizontal force acts on the man + boat system, so the centre of mass cannot move horizontally. This is the key to every “man on boat” problem.

Initially: man at x=0x = 0, boat’s centre at x=2x = 2 (length 4 m, man at left end). Take rightward positive.

 xcm=60(0)+40(2)60+40=80100=0.8 mx_{\text{cm}} = \tfrac{60(0) + 40(2)}{60 + 40} = \tfrac{80}{100} = 0.8 \text{ m}

After walking, the man is at the right end of the boat. If the boat shifts left by dd, then:

  • New boat centre: 2−d2 - d
  • New right end of boat: 4−d4 - d (this is where the man stands)
 0.8=60(4−d)+40(2−d)1000.8 = \tfrac{60(4 - d) + 40(2 - d)}{100} 80=60(4−d)+40(2−d)=240−60d+80−40d=320−100d80 = 60(4 - d) + 40(2 - d) = 240 - 60d + 80 - 40d = 320 - 100d 100d=240  ⟹  d=2.4 m100d = 240 \implies d = 2.4 \text{ m}

Final answer: Boat moves 2.4 m\mathbf{2.4 \text{ m}} in the direction opposite to the man’s motion (to the left).

Why This Works

In an isolated system, the COM stays put — no external horizontal force can move it. So when the man walks right, the boat must slide left to keep the COM fixed.

The shortcut formula: if the man (mass mm) walks displacement LL (relative to the boat), the boat moves mLm+M\tfrac{mL}{m+M} in the opposite direction. Here 60×4100=2.4\tfrac{60 \times 4}{100} = 2.4 m. Same answer in one line.

Alternative Method

Work in the COM frame. From the COM’s perspective, the boat and the man move toward each other so that m⋅xman=M⋅xboatm \cdot x_{\text{man}} = M \cdot x_{\text{boat}} at all times. Their relative motion is L=4L = 4 m, so xman+xboat=4x_{\text{man}} + x_{\text{boat}} = 4 and 60xman=40xboat60 x_{\text{man}} = 40 x_{\text{boat}}. Solving: xboat=2.4x_{\text{boat}} = 2.4 m.

Common Mistake

Treating L=4L = 4 m as the man’s displacement relative to the ground. It’s the displacement relative to the boat. The man’s actual ground-frame displacement is 4−2.4=1.64 - 2.4 = 1.6 m to the right.

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