A common technique in statics problems is to assume that static friction is saturated everywhere. Consider, for example, the following problem:
A thin, uniform bar of mass $m$ lies on a uniform floor, with coefficient of static friction $\mu$. What is the minimum force required to make the object start moving, if the force can only be applied horizontally? Assume the normal pressure on the floor remains uniform.
The minimum force required comes from pushing perpendicular to the bar at its edge. If we assume the friction has the maximum magnitude everywhere, that is $df = \mu g dm$ for every infinitesimal segment of mass $dm$, and it points either directly along or against the applied force, we then have enough information to solve the problem. The friction will oppose the applied force on the segment of length $l$ closest to the applied force, while the friction will align with the applied force for the remaining $L-l$ segment, where $L$ is the total length of the bar. Balancing forces yields
$$F = \mu m g \left(\frac{l}{L} - \frac{L-l}{L} \right)$$
While balancing torques about where the applied force acts yields
$$\mu m g \frac{l}{L} \left(\frac{l}{2} \right) = \mu m g \frac{L-l}{L} \left(\frac{L+l}{2} \right)$$
Solving the set of equations yields $l=L/\sqrt{2}$ and $\boxed{F=(\sqrt{2}-1)\mu mg}$.
My question is two-fold:
- In the above example, in order for the bar to be right at the threshold between motion and being static, friction must be maximal in at least one location along the rod. Intuitively, why must it be maximal at all locations along the rod?
- Is there an example of a statics problem/situation in which when at the threshold between motion and being static, friction does not need to be maximal along all locations? (other than trivial situations such as one foot sliding and the other foot being static—such a situation has essentially independent sources of friction).
