A chain of mass M and length L is held vertically with its lower end just touching a table. It is released from rest. What is the normal force exerted by the table on the chain when length x has fallen onto the table?

Answer and explanation

Correct answer: A. 3 (M / L) g x

  1. Step 1: Weight of chain already lying on table = (M / L) x g.
  2. Step 2: Rate of momentum destroyed by falling links = v (dm/dt) = v [ (M / L) v ] = (M / L) v^2.
  3. Step 3: Since links fall from rest through height x: v^2 = 2 g x.
  4. Step 4: Impact force = (M / L) (2 g x) = 2 (M / L) g x.
  5. Step 5: Total normal force N = Weight + Impact force = (M / L) g x + 2 (M / L) g x = 3 (M / L) g x.

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