Problem 2B-13**: Analyzing the high wire act

When a performer walks on a high wire, he usually carries a long flexible pole. We want to analyze the balancing effect of this pole. The situation is simplified as in Fig. 1. The center of mass of the performer is depicted as point A so that the mass of the performer, m, can be considered as a mass point located at A. A massless needle of length h is attached to point A and stands on point O on a cross section of the high wire. Two massless and rigid rods of length d each are attached to point A, each forming an angle θ with the needle. At the end points of two rods, B and C, a weight of mass M is attached. We want to find the condition that the system is stable, that is, it will restoring itself if the needle is tilted around point O.






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Solution:

The left hand side of the needle is depicted in Fig. 2. The right hand side is identical to the left hand side. The length of BO is denoted as x and the angle ABO as φ. The triangle ABO is completely determined by the given parameters θ, h and d. Thus x and φ can be expressed by those three given parameters. Applying the rule of cosine on the triangle AOB with regard to angle θ, we get

                x2 = d2 + h2 - 2d·h·cosθ.                                 (1)

sinφ can be calculated by applying the rule of sine on the triangle ABO as

                sinφ/h = sinθ/x     →     sinφ = (h/x)·sinθ.                                 (2)

cosφ can be calculated by applying the rule of cosine on the triangle ABO with regard to angle φ as

                cosφ = (d2 + x2 - h2)/(2·d·x) = (d - h·cosθ)/x.                                 (3)

Furthermore, the angle BOP is denoted as α and can be expressed as

                α = φ + θ.                                 (4)

When the needle is tilted by an angle δ as shown in Fig. 3, we have

                ∠BOP = α + δ,     and     ∠COP = α - δ.

The torque around point O generated by the weight at point B is denoted as τB, and we have

                |τB| = xMg·sin(α + δ).

The torque around point O generated by the weight at point C is denoted as τC, and we have

                |τC| = xMg·sin(α - δ).

The torque around point O generated by the weight at point A is denoted as τA, and we have

                |τA| = h·mg·sinδ.

Torques τB, τC and τA all lie on a line that passes through point O and is perpendicular to the plane of Fig. 3. Torque τB comes out of the plane, and torques τC and τA go into the plane. Thus as long as

                |τB| - |τC| - |τA| › 0,

the system is self-restoring since the sum of torques around point O will rotate the needle counter-clockwise back to its vertical position. We can write

                |τB| - |τC| - |τA| = xMg·{ sin(α + δ) - sin(α - δ)} - h·mg· sinδ
                                       = xMg·{(sinα·cosδ + cosα·sinδ) - (sinα·cosδ - cosα·sinδ)} - h·mg· sinδ
                                       = 2xMg·cosα·sinδ - h·mg·sinδ
                                       = (2xM·cosα - hm)·g·sinδ.

Therefore, for the needle to be self-restoring we must have

                2xM·cosα - hm › 0,

or

                cosα › hm/(2xM).

From Eq. (4) we get

                cos(θ + φ) = (cosθ·cosφ - sinθ·sinφ) › hm/(2xM).

Substituting Eq. (2) and Eq. (3) for sinφ and cosφ respectively, we have

                {(d - h·cosθ)/x)·cosθ - (h·sinθ/x)·sinθ › (hm/2xM).

This leads to

                (d·cosθ - d·cos2 - d·sin2)/x › (hm/2M)/x.

Thus we have

                d·cosθ - h › hm/2M,

and finally

                cosθ › (h/d)·{1 + (m/2M)}.