## Friday Physics 146

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

[Fig. 1]
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
 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.