Put a uniform meter scale horizontally on your extended index fingers with the left one at 0.00 cm and the right one at 90.00 cm. When you attempt to move both the fingers slowly towards the center, initially only the left finger slips with respect to the scale and the right finger does not. After some distance, the left finger stops and the right one starts slipping. Then the right finger stops at a distance $$x_{R}$$ from the center (50.00 cm) of the scale and the left one starts slipping again. This happens because of the difference in the frictional forces on the two fingers. If the coefficients of static and dynamic friction between the fingers and the scale are 0.40 and 0.32, respectively, the value of $$x_{R}$$ (in
cm) is ______.

Let the left and right fingers be at positions $$x_L$$ and $$x_R$$ (in cm) measured from the left end of the meter scale. Let the normal reactions on the two fingers be $$N_L$$ and $$N_R$$ and the weight of the uniform scale be $$W$$ acting at its centre 50 cm.

Vertical equilibrium of the scale gives $$N_L + N_R = W \qquad -(1)$$

Taking moments about the left finger, $$N_R\,(x_R - x_L) = W\,(50 - x_L) \qquad -(2)$$

The coefficients of friction are static : $$\mu_s = 0.40 = \dfrac{2}{5}$$,   dynamic : $$\mu_d = 0.32 = \dfrac{8}{25}$$.

Case 1: Left finger slides, right finger sticks
Left contact is kinetic, so the friction there is $$F_L = \mu_d N_L$$. Right contact is still static; to keep the scale nearly at rest in the horizontal direction we must have $$F_L = F_R \; \Longrightarrow \; \mu_d N_L = F_R \le \mu_s N_R$$

The right finger starts slipping when the limiting value is reached: $$\mu_d N_L = \mu_s N_R \qquad -(3)$$

Using $$(1)$$ and $$(3)$$, $$\dfrac{N_L}{N_R} = \dfrac{\mu_s}{\mu_d} = \dfrac{0.40}{0.32} = \dfrac{5}{4}$$

Hence $$N_L = \dfrac{5}{9}W ,\qquad N_R = \dfrac{4}{9}W$$

At this instant the right finger is still at its original position $$x_R = 90$$ cm, while the left finger has moved to a new position $$x_L$$. Substituting $$N_R = \dfrac{4}{9}W$$ in moment equation $$(2)$$:

$$\dfrac{4}{9}W\,(90 - x_L) = W\,(50 - x_L)$$

Simplifying, $$4(90 - x_L) = 9(50 - x_L)$$ $$360 - 4x_L = 450 - 9x_L$$ $$5x_L = 90 \;\Longrightarrow\; x_L = 18\text{ cm}$$

So the left finger becomes static at 18 cm and the right one now starts sliding.

Case 2: Right finger slides, left finger sticks
Now the right contact is kinetic $$F_R = \mu_d N_R$$ and the left is static. The reversal to left-finger sliding occurs when $$\mu_s N_L = \mu_d N_R \qquad -(4)$$

From $$(1)$$ and $$(4)$$, $$\dfrac{N_L}{N_R} = \dfrac{\mu_d}{\mu_s} = \dfrac{0.32}{0.40} = \dfrac{4}{5}$$

Therefore $$N_L = \dfrac{4}{9}W ,\qquad N_R = \dfrac{5}{9}W$$

During this interval the left finger has remained fixed at 18 cm, while the right finger has slid to a new position $$x_R$$. Insert $$N_R = \dfrac{5}{9}W$$ into moment equation $$(2)$$:

$$\dfrac{5}{9}W\,(x_R - 18) = W\,(50 - 18) = 32W$$

Cancel $$W$$ and solve: $$\dfrac{5}{9}(x_R - 18) = 32$$ $$x_R - 18 = 32 \times \dfrac{9}{5} = 57.6$$ $$x_R = 18 + 57.6 = 75.6\text{ cm}$$

The distance of this stopping point from the centre (50 cm) is $$x_R - 50 = 75.6 - 50 = 25.6\text{ cm}$$

Hence, the right finger stops at a distance $$\boxed{25.6\ \text{cm}}$$ from the centre of the scale.