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:
- Nw, the normal force exerted by the wall on the upper end of the rod
- Nf, the normal force exerted by the floor on the lower end of the rod
- fw, the friction force exerted by the wall on the upper end of the rod
- ff, the friction force exerted by the floor on the lower end of the rod
- W=mg, the weight of the rod, acting on the center of the rod
To have equilibrium we need forces and torques to cancel.
Cancellation of forces in the horizontal direction gives us Nw=ff, and cancellation in the vertical gives us fw+Nf=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
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.