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*_{w}, the normal force exerted by the wall on the upper end of the rod
*N*_{f}, the normal force exerted by the floor on the lower end of the rod
*f*_{w}, the friction force exerted by the wall on the upper end of the rod
*f*_{f}, 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 *N*_{w}=*f*_{f}, and cancellation in the vertical gives us *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.

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Tags: Equilibrium, Friday Physics, physics, Static Friction, Torque

This entry was posted on December 10, 2010 at 2:54 am and is filed under Math/Science. You can follow any responses to this entry through the RSS 2.0 feed.
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