First slide

Location of centre on mass for multiple point masses

Question

A wooden log of mass 120kg is floating on still water perpendicular to the shore. A man of mass 80kg is standing at the centre of mass of the log and he is at a distance of 30m from the shore. When he walks through a distance of 10m towards the shore and halts, now his distance from the shore is

Easy

A

20m
B

24m
C

26m
D

28m

Solution

If r₁, r₂ and r_c be the distances of COM of the man, boat and COM of the system, then m₁r₁+m₂r₂ = (m₁+m₂).r_c $\large \Rightarrow {m_1}.\Delta {r_1} + {m_2}.\Delta {r_2} = ({m_1} + {m_2})\Delta {r_c} = ({m_1} + {m_2}) \times 0$ . If Δr₁= (10-x) and Δr₂=x
$\large \therefore {m_1}(10 - x) + {m_2}( - x) = 0 \Rightarrow x = \frac{{10 \times 80}}{{80 + 120}} = 4m$
$\large \therefore$ Distance of man from the shore = 30 - (10 - 4) = 24m

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