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NAEST 2020: Rotational Equilibrium and Toppling of Box using

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Question: As the small ring magnets are stuck one by one, at some stage the plastic box topples on the floor. Take a small area dS of the floor in contact with the plastic box near the edge of the box about which it topples. Call this area A. Take a similar area dS near the opposite edge of the box. Call this area B. The normal force by A on the box is denoted by $N_A$ and that by B is denoted by $N_B$. The weight of the box together with all magnets and stand inside it is denoted by W.
Consider the situation when 4 extra magnets are stuck and the box stays in equilibrium. Choose the correct option.

  1. $N_A = N_B$
  2. $N_A = N_B = W / 2 $
  3. $N_A > N_B$
  4. $N_A > NB$, $N_A + N_B = W$

Question 2: Consider the situation when 5 extra magnets are stuck and the box topples. Just before toppling,
  1. $N_A < W$ and $N_B = 0$
  2. $N_A > W$ and $N_B = 0$
  3. $N_A = W$ and $N_B = 0$
  4. $N_A = W$ and $N_B > 0$

Solution 1: Let us consider the box, the pencil, and the magnets as a system. In the absence of ring magnets, the system is symmetric, and the normal force on the box is distributed uniformly over the base area. The addition of a magnet introduces asymmetry, increases instability, and affects normal force in two ways. Firstly, it increases the magnitude of the normal force to balance the increase in weight of the system. Secondly, it shifts the net normal force towards the periphery (in a direction parallel to the line formed by the ring magnet), to balance the torque. In this situation, the normal force is not uniformly distributed over the base area. It has more value towards the magnet and a lesser value away from the magnet. Thus, $N_A > N_B$ i.e., option (C) is correct.

Apart from the area A and area B, the normal force also act at other contacts. Thus, $N_A + N_B < W$.

Solution 2: Just before toppling, A is the only contact point and the net force on the system is zero. The normal force at all other points is zero. Thus, $N_A = W$ and $N_B = 0$. Option (C) is correct.

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