Some Examples of Static Equilibrium — Question 5

Section 12-4 · Physics pages 1840–1841

Question 5 A shallow suspension requires a large tension even for a modest load A weight 500 N hangs at the midpoint of two equal straight cable segments whose level supports are 6 m apart. Each segment has tension limit 1000 N. Neglect cable weight; let the midpoint sag be h. h 500 N Schematic; labels, not drawing lengths, specify the data. (a) Find the minimum segment angle above horizontal. (b) Find the minimum sag compatible with the tension limit.
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Question 5 - Solution A shallow suspension requires a large tension even for a modest load h 500 N (a) Derivation and calculation For equal cable tensions, 2T sin θ = W = 500 N. The tension cap requires sin θ ≥ 500 2(1000) = 1 4, θmin = 14.5◦. Using the 3 m half-span, hmin = 3 tan θmin = 3 √ 15 = .775 m . (b) Result and interpretation Geometry gives sin θ = h/ √ 32 + h2, so T(h) = 500 √ 9 + h2 2h . As h →0+, T ∼750 h −→∞. A perfectly horizontal ideal cable cannot supply a finite upward resultant with finite tension. Physical check At minimum sag each cable supplies 250 N vertically, while its total tension is 1000 N.

Question 5

A shallow suspension requires a large tension even for a modest load

A weight 500N500\,\mathrm N hangs at the midpoint of two equal straight cable segments whose level supports are 6m6\,\mathrm m apart. Each segment has tension limit 1000N1000\,\mathrm N. Neglect cable weight; let the midpoint sag be hh.

Diagram shown in the original SVG worksheet.

  1. Find the minimum segment angle above horizontal.

  2. Find the minimum sag compatible with the tension limit.

Question 5 - Solution

A shallow suspension requires a large tension even for a modest load

Diagram shown in the original SVG worksheet.

(a) Derivation and calculation

For equal cable tensions, 2Tsin⁡θ=W=500N.2T\sin\theta=W=500\ \mathrm N. The tension cap requires sin⁡θ≥5002(1000)=14,θmin=14.5∘.\sin\theta\ge\frac{500}{2(1000)}=\frac14, \qquad\theta_{\min}=14.5^\circ. Using the 3 m half-span, hmin=3tan⁡θmin=315=.775m.\boxed{h_{\min}=3\tan\theta_{\min}=\frac3{\sqrt{15}}=.775\ \mathrm m}.

(b) Result and interpretation

Geometry gives sin⁡θ=h/32+h2\sin\theta=h/\sqrt{3^2+h^2}, so T(h)=5009+h22h.\boxed{T(h)=\frac{500\sqrt{9+h^2}}{2h}}. As h→0+h\to0^+, T∼750h→∞.T\sim\frac{750}{h}\longrightarrow\infty. A perfectly horizontal ideal cable cannot supply a finite upward resultant with finite tension.

Physical check

At minimum sag each cable supplies 250 N vertically, while its total tension is 1000 N.