2013: Problem 4

Topic: Dynamics
Concepts: Statics, Torque from weight


Solution:

We choose the hinge as the pivot point and balance torques on one of the legs. The weight mgmg of the leg acts at its center while the normal force N=mgN=mg acts at the bottom to keep the leg at rest. Friction ff points inward to prevent the sign from collapsing. These are the only forces that contribute to the torque since the force at the hinge has no moment arm. We have

mg(l2sinθ2)+f(lcosθ2)=N(lsinθ2)=mg(lsinθ2)mg\left(\frac{l}{2}\sin\frac{\theta}{2}\right)+f\left(l\cos\frac{\theta}{2}\right)=N\left(l\sin\frac{\theta}{2}\right)=mg\left(l\sin\frac{\theta}{2}\right)

Solving for ff,

flcosθ2=mgl2sinθ2fl\cos\frac{\theta}{2}=mg\frac{l}{2}\sin\frac{\theta}{2}
f=mg2tanθ2f=\frac{mg}{2}\tan\frac{\theta}{2}

Since static friction fμNf \leq \mu N, we have

mg2tanθ2μN=μmg\frac{mg}{2}\tan\frac{\theta}{2} \leq \mu N=\mu mg
tanθ22μ\tan\frac{\theta}{2} \leq 2\mu

so the answer is E.

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