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UPSC 2021 Maths Optional Paper 1 Q6a — Step-by-Step Solution

20 marks · Section B

Common catenary · Dynamics & Statics · asked 4× in 13 yrs · Read the full method →

Question

A heavy string of non-uniform density hangs from two points. $T_1,T_2,T_3$ are tensions at intermediate points $A,B,C$ where inclinations to the horizontal are in AP with common difference $\beta$. Weights of parts $AB,BC$ are $\omega_1,\omega_2$. Prove: (i) Harmonic mean of $T_1,T_2,T_3=\dfrac{3T_2}{1+2\cos\beta}$. (ii) $T_1/T_3=\omega_1/\omega_2$.

Technique

Horizontal tension is constant on any catenary; $T=T_0\sec\psi$. Vertical tension change = weight of segment.

Solution

Setup. At any point on a catenary (uniform or not), the horizontal component of tension is constant ($T_0$, the tension at the lowest point). The tension at a point making angle $\psi$ with horizontal: $T\cos\psi=T_0$, so $T=T_0\sec\psi$.

Let $\psi_1,\psi_2,\psi_3$ be the inclinations at $A,B,C$. Given: AP with common difference $\beta$, so $\psi_1=\psi_2-\beta$, $\psi_3=\psi_2+\beta$.

Step 1 — Express tensions via $T_0$

$T_1=T_0\sec\psi_1=T_0\sec(\psi_2-\beta)$. $T_2=T_0\sec\psi_2$. $T_3=T_0\sec(\psi_2+\beta)$.

Step 2 — Harmonic mean

$\text{HM}(T_1,T_2,T_3)=\dfrac{3}{1/T_1+1/T_2+1/T_3}=\dfrac{3}{\cos\psi_1/T_0+\cos\psi_2/T_0+\cos\psi_3/T_0}=\dfrac{3T_0}{\cos\psi_1+\cos\psi_2+\cos\psi_3}$.

$\cos\psi_1+\cos\psi_3=\cos(\psi_2-\beta)+\cos(\psi_2+\beta)=2\cos\psi_2\cos\beta$.

So $\cos\psi_1+\cos\psi_2+\cos\psi_3=2\cos\psi_2\cos\beta+\cos\psi_2=\cos\psi_2(1+2\cos\beta)$.

$\text{HM}=\dfrac{3T_0}{\cos\psi_2(1+2\cos\beta)}=\dfrac{3T_0\sec\psi_2}{1+2\cos\beta}=\dfrac{3T_2}{1+2\cos\beta}$ ✓.

Step 3 — Ratio $T_1/T_3$ in terms of weights

For a catenary segment, the vertical component of tension changes by the weight of the segment.

At $A$ (angle $\psi_1$): vertical tension $T_1\sin\psi_1=T_0\tan\psi_1$. At $B$ (angle $\psi_2$): $T_2\sin\psi_2=T_0\tan\psi_2$.

Weight of segment $AB$: $\omega_1$. Vertical equilibrium gives the change in vertical tension equals the weight: $T_0(\tan\psi_2-\tan\psi_1)=\omega_1$.

Similarly $T_0(\tan\psi_3-\tan\psi_2)=\omega_2$.

So $\omega_1/\omega_2=(\tan\psi_2-\tan\psi_1)/(\tan\psi_3-\tan\psi_2)$.

Step 4 — Compute $\tan\psi_2-\tan\psi_1$ and $\tan\psi_3-\tan\psi_2$

$\tan\psi_2-\tan\psi_1=\tan\psi_2-\tan(\psi_2-\beta)=\dfrac{\sin\psi_2\cos(\psi_2-\beta)-\cos\psi_2\sin(\psi_2-\beta)}{\cos\psi_2\cos(\psi_2-\beta)}=\dfrac{\sin\beta}{\cos\psi_2\cos(\psi_2-\beta)}$.

Similarly $\tan\psi_3-\tan\psi_2=\dfrac{\sin\beta}{\cos\psi_2\cos\psi_3}=\dfrac{\sin\beta}{\cos\psi_2\cos(\psi_2+\beta)}$.

Ratio: $\dfrac{\omega_1}{\omega_2}=\dfrac{\sin\beta/[\cos\psi_2\cos(\psi_2-\beta)]}{\sin\beta/[\cos\psi_2\cos(\psi_2+\beta)]}=\dfrac{\cos(\psi_2+\beta)}{\cos(\psi_2-\beta)}=\dfrac{\cos\psi_3}{\cos\psi_1}=\dfrac{T_0/T_3}{T_0/T_1}=\dfrac{T_1}{T_3}$ ✓.

Answer