A lot of work

Suppose that the left-hand finger is a distance $u_L$ from the centre of mass of the rod, while the right-hand finger is a distance $u_R$ from the centre of mass, and that the left-hand finger is sliding on the rod. Then it is clear that the fingers exert normal reactions $N_L$ and $N_R$ respectively, where $N_L \; = \; \frac{u_R}{u_L + u_R}mg \hspace{2cm} N_R \; = \; \frac{u_L}{u_L + u_R}mg$ Since the motion is assumed to be slow, the rod does not move horizontally, and so the horizontal forces $P_L,P_R$ exerted by the two fingers must be equal. While the left-hand finger is sliding, we must have $P_L = \mu_KN_L$ , and $P_R$ cannot exceed $\mu_S N_R$ . This means that the motion, with the left-hand finger sliding, can take place so long as $\mu_K u_R \le \mu_S u_L$ , and hence so long as $u_L \ge ku_R$ . Once the point $u_L = ku_R$ is reached, the left-hand finger will stop sliding, and the right-hand finger will start sliding until the point $u_R = ku_L$ is reached, at which point the right-hand finger will stop sliding, the left-hand finger will start sliding again, and so on until (eventually) the fingers meet in the middle.

The work done by the left-hand finger as it slides from $u_L = u_0$ to $u_L = u_1$ , while the right-hand finger is stationary at $u_R = v$ , is simply $W(u_0,u_1,v) \; = \; -\int_{u_0}^{u_1} \mu_K N_L \,du \; = \; -\int_{u_0}^{u_1} \mu_K mg \frac{v}{v+u}\,du \; = \; \mu_K mgv\ln\left(\tfrac{v+u_0}{v+u_1}\right)$ This is of course the same as the work done by the right-hand finger as it slides from $u_R = u_0$ to $u_R = u_1$ , with the left-hand finger being stationary with $u_L = v$ . We need to add up the work done over an infinite number of stages:

• the first stage has the left-hand finger moving, with the right-hand finger stationary, with $u_0 = \tfrac12\ell$ , $u_1 = \tfrac12k\ell$ and $v = \tfrac12\ell$ ,
• the second stage has the right-hand finger moving, with the left-hand finger stationary, with $u_0 = \tfrac12\ell$ , $u_1 = \tfrac12k^2\ell$ and $v = \tfrac12k\ell$ ,
• the third stage has the left-hand finger moving, with the right-hand finger stationary, with $u_0 = \tfrac12k\ell$ , $u_1 = \tfrac12k^3\ell$ and $v = \tfrac12k^2\ell$ ,
• and so on... In general the $(n+1)$ st stage has one finger moving from $u_0 = \tfrac12k^{n-1}\ell$ to $u_1 = \tfrac12k^{n+1}\ell$ , with the other finger stationary with $v = \tfrac12k^n\ell$ .

Thus the work done during the first stage is $W(\tfrac12\ell,\tfrac12k\ell,\tfrac12\ell) \; = \; \tfrac12 mg\ell\mu_K \ln\big(\tfrac{2}{1+k}\big)$ while the work done during the $(n+1)$ st stage is $W(\tfrac12k^{n-1}\ell,\tfrac12k^{n+1}\ell,\tfrac12k^n\ell) \; = \; \tfrac12mg\ell \mu_K k^n \ln k^{-1}$ for all $n \ge 1$ . Thus the total work done is $\tfrac12mg\ell\mu_K \left[\ln\big(\tfrac{2}{1+k}\big) + \sum_{n=1}^\infty k^n \ln k^{-1}\right] \; = \; \tfrac12mg\ell\mu_K \left[\ln\big(\tfrac{2}{1+k}\big) - \tfrac{k}{1-k} \ln k\right]$