NAEST 2020: The Rotational Mechanics of Sliding Fingers Unde

Question: The person tries to slide both the fingers at the same time under the scale. When the finger on the left slides, the finger on the right remains at rest. In this situation, the frictional force on the scale by the finger on the left has magnitude $f_1$ and that by the finger on the right has magnitude $f_2$. Choose the correct option.

1. $f_1 = f_2$
2. $f_1 > f_2$
3. $f_1 < f_2$
4. $f_1 = f_2 = 0$

Question 2: The directions of the forces $f_1$ and $f_2$ are

1. both towards right
2. both towards left
3. towards left and towards right respectively
4. towards the right and towards left respectively

Question 3: Let $N_1$ be the normal force by the finger on the left on the scale and $N_2$ be that by the finger on the right. The mass of the scale is$M$. Choose the correct option

1. $N_1 = N_2$
2. $N_1 = N_2 = Mg / 2$
3. $N_1 > N2$
4. $N1 < N_2$

Solution 1: The scale remains stationary as the fingers slide below it. Thus, the net force on the scale is zero. The forces on the scale in the horizontal direction are frictional forces $f_1$ and $f_2$. These two forces shall be equal in magnitude but opposite in the direction to make the net force in the horizontal direction zero. Hence, option (A) is correct i.e., $f_1 = f_2$.

Solution 2: The sliding finger will apply a force on the scale in the direction of its motion i.e., $f_1$ will be towards the left. The frictional force by the other finger will be in the opposite direction i.e., towards the right. Hence, option (C) is correct.

Solution 3: The sliding finger (on the left) is far away from the centre of mass of the scale as compared to the stationary finger. The net torque on the scale about the centre of mass is zero. Thus, to balance the torque, $N_1 < N_2$ and option (D) is correct. Note that $N_1 + N_2 = Mg$.

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