Consider a (uniform) rod of mass *m* and length *l* leaning against a wall at an angle *θ* from the horizontal; both the wall and floor are made of the same material, with the same coefficient of static friction *μ*. What is the minimum value of *μ* for the rod to remain static, without slipping down?

Drawing a diagram, we see that there are five forces:

*N*, the normal force exerted by the wall on the upper end of the rod_{w}*N*, the normal force exerted by the floor on the lower end of the rod_{f}*f*, the friction force exerted by the wall on the upper end of the rod_{w}*f*, the friction force exerted by the floor on the lower end of the rod_{f}*W*=*mg*, the weight of the rod, acting on the center of the rod

[Fig. 1]

To have equilibrium we need forces and torques to cancel.

Cancellation of forces in the horizontal direction gives us *N _{w}*=

*f*, and cancellation in the vertical gives us

_{f}*f*+

_{w}*N*=

_{f}*mg*.

Now, for torque, we consider the torque about the lower end. The torque exerted by the weight is thus , while the forces exerted by the wall produce torques in the opposite direction of

and .

Equating, we have

,

and multiplying by , we get

and combining with the vertical equation,

.

Now, by the definition of coefficient of static friction, we have

and . Thus, we have the minimum value when these ratios, and are equal; thus, the relation between them gives us

.

Solving for positive

*μ*,

,

(Here, we use that for 0<

*θ*<π/2, the secant of

*θ*is positive, and so .)

Note that

*μ*goes to zero as

*θ*approaches π/2, and increases with decreasing angle. Thus, leaning the rod in a near-vertical position requires less friction, and so is easier, than leaning in a shallower angle; as one should expect from practical experience.