1. Setup and idea

A uniform meter scale has its center of mass at $50\ \text{cm}$.

Initially:

• Left finger at $x_L=0$
• Right finger at $x_R=90$

When both fingers are pushed inward slowly, the finger with smaller normal reaction slips first because the maximum static friction $f_s^{\max}=\mu_s N$ is smaller there.

Also, when one finger slips, friction on that finger is kinetic: $f_k=\mu_k N$ while the other finger remains stuck (static friction adjusts as needed).

The switching occurs when the slipping finger can no longer sustain the required friction and the other one begins to slip.

2. Normal reactions when fingers are at positions $x_L$ and $x_R$

Let the weight of the scale be $W$ acting at $50\ \text{cm}$.

From vertical equilibrium: $N_L+N_R=W$

Taking moments about the left finger: $N_R(x_R-x_L)=W(50-x_L)$ $\Rightarrow N_R=\frac{W(50-x_L)}{x_R-x_L}$

Similarly, $N_L=\frac{W(x_R-50)}{x_R-x_L}$

3. Why the left finger slips first

Initially, $x_L=0, x_R=90$.

So, $N_L=\frac{W(90-50)}{90-0}=\frac{40W}{90}=\frac{4W}{9}$ $N_R=\frac{W(50-0)}{90}=\frac{5W}{9}$

Since $N_L<N_R$, the left finger has smaller frictional capacity and slips first.

This matches the statement in the question.

4. Condition for switching from left slipping to right slipping

While the left finger slips and the right finger is stuck:

• Left friction is kinetic: $f_L=\mu_k N_L$
• Right friction is static and equals the same horizontal force needed for equilibrium, so at the switching point: $\mu_k N_L=\mu_s N_R$

At this instant, the right finger is just about to slip.

During this phase, the right finger stays fixed at $x_R=90$, and the left finger moves to some position $x_L$.

Using the reaction formulas: $N_L=\frac{W(90-50)}{90-x_L}=\frac{40W}{90-x_L}$ $N_R=\frac{W(50-x_L)}{90-x_L}$

Now apply switching condition: $\mu_k N_L=\mu_s N_R$ $0.32\cdot \frac{40W}{90-x_L}=0.40\cdot \frac{W(50-x_L)}{90-x_L}$

Cancel $W$ and $(90-x_L)$: $0.32\cdot 40=0.40(50-x_L)$ $12.8=20-0.4x_L$ $0.4x_L=7.2$ $x_L=18\ \text{cm}$

So the left finger stops at $18\ \text{cm}$, and then the right finger begins to slip.

5. Second phase: right finger slips, left finger fixed at $18$ cm

Now:

• Left finger remains fixed at $x_L=18$
• Right finger moves inward to some position $x_R$

At the next switching point, the right finger stops and the left finger starts slipping again.

So at that instant:

• Right finger friction is kinetic: $f_R=\mu_k N_R$
• Left finger is at limiting static friction: $f_L=\mu_s N_L$

Thus, $\mu_k N_R=\mu_s N_L$

With $x_L=18$, the reactions are $N_R=\frac{W(50-18)}{x_R-18}=\frac{32W}{x_R-18}$ $N_L=\frac{W(x_R-50)}{x_R-18}$

Apply the condition: $0.32\cdot \frac{32W}{x_R-18}=0.40\cdot \frac{W(x_R-50)}{x_R-18}$

Cancel $W$ and $(x_R-18)$: $0.32\cdot 32=0.40(x_R-50)$ $10.24=0.40(x_R-50)$ $x_R-50=25.6$ $x_R=75.6\ \text{cm}$

The question asks for the distance from the center $50\ \text{cm}$: $x_R=75.6-50=25.6\ \text{cm}$

6. Final answer

$\boxed{25.60}$

This matches the stored correct answer.