UPSC 2026 Maths Optional Paper 1 Q6b-ii — Step-by-Step Solution

10 marks · Section B

Common catenary · Dynamics & Statics · asked 5× in 14 yrs · Read the full method →

Question

A uniform chain of length $l$ has one end fixed at a height $h$ above a rough table and rests in a vertical plane so that a portion of it lies in a straight line on the table. Prove that if the chain is on the point of slipping, the length on the table is $(l + \mu h) - \sqrt{(\mu^2 + 1)h^2 + 2\mu l h}$ , where $\mu$ is the coefficient of friction.

Technique

Common catenary + limiting friction. The hanging portion of a perfectly flexible heavy chain is a common catenary; because the chain continues along the table beyond the point of contact, it must meet the table tangentially, so that point is the vertex of the catenary. Then $y^{2}=s^{2}+c^{2}$ supplies the geometry and $T_{0}=wc=\mu wx$ supplies the statics.

Solution

Step 1 — Set up the geometry and the figure (in words).

Let $w$ be the weight per unit length of the chain, so its total weight is $wl$ .

Draw the vertical plane of the chain. Mark the table as a horizontal line. Let

$A$ be the fixed end, at height $h$ vertically above the table;
$B$ be the point at which the chain first touches the table (the point of separation between the hanging arc and the flat part);
$C$ be the free end, lying on the table, with $BC$ straight and horizontal.

Let $x=BC$ be the length lying on the table. Then the hanging arc $AB$ has length

s_{1}=l-x .

Step 2 — Why the tangent at $B$ is horizontal (the key step).

The chain is perfectly flexible, so the tension at any point acts along the tangent to the chain there, and the direction of the chain cannot change abruptly at $B$ : a finite kink would require a finite concentrated force at $B$ , and the only forces there are the distributed weight and the distributed table reaction, both of which are finite per unit length. Hence the chain leaves the table smoothly, i.e. the tangent to the hanging arc at $B$ is horizontal (it coincides with the direction $BC$ of the part on the table).

Therefore $B$ is the lowest point (vertex) of the catenary formed by $AB$ , and the tension at $B$ is horizontal.

Step 3 — Standard catenary relations for the hanging arc.

Let $c$ be the parameter of the catenary, defined by $T_{0}=wc$ , where $T_{0}$ is the (horizontal) tension at the vertex $B$ . Take the directrix as the horizontal line at depth $c$ below $B$ , and measure $y$ upward from it; measure the arc $s$ from $B$ .

Resolving horizontally and vertically for the arc from the vertex $B$ to a general point $P$ of the hanging portion, where the tangent makes angle $\psi$ with the horizontal and the tension is $T$ :

T\cos\psi = T_{0}=wc,\qquad T\sin\psi = ws .

Hence $\tan\psi = \dfrac{s}{c}$ (intrinsic equation of the common catenary) and

T^{2}=w^{2}\bigl(c^{2}+s^{2}\bigr)\ \Longrightarrow\ T=w\sqrt{c^{2}+s^{2}} = wy,\qquad\text{i.e.}\qquad \boxed{y^{2}=s^{2}+c^{2}}

with $y=c\cosh(X/c)$ , $s=c\sinh(X/c)$ in Cartesians measured from the vertex.

At $B$ (the vertex): $s=0$ , $y=c$ — consistent, since $B$ is at height $c$ above the directrix. At $A$ : $A$ is $h$ above the table, i.e. $h$ above $B$ , so

y_{1}=c+h,\qquad s_{1}=l-x .

Substituting into $y^{2}=s^{2}+c^{2}$ :

(c+h)^{2}=(l-x)^{2}+c^{2}\quad\Longrightarrow\quad (l-x)^{2}=h^{2}+2ch. \tag{1}

Step 4 — Statics of the portion lying on the table.

Consider the straight portion $BC$ , of length $x$ and weight $wx$ . The forces on it are:

its weight $wx$ , vertically down;
the total normal reaction $R$ of the table, vertically up;
the total frictional force $F$ , horizontal, acting along the table towards $C$ (opposing the impending motion);
the pull of the hanging part at $B$ , which by Step 2 is horizontal, of magnitude $T_{0}=wc$ , directed towards $A$ ‘s side;
zero tension at the free end $C$ .

Resolving vertically: since the pull at $B$ is horizontal it contributes nothing vertically, so

R=wx

— the normal reaction is exactly the weight of the part on the table, no more and no less.

Resolving horizontally:

F=T_{0}=wc .

The chain is on the point of slipping, so friction is limiting: $F=\mu R$ . Hence

wc=\mu\,wx\quad\Longrightarrow\quad c=\mu x. \tag{2}

Step 5 — Eliminate $c$ and solve.

Put (2) into (1):

(l-x)^{2}=h^{2}+2\mu h x. \tag{3}

Expand:

x^{2}-2lx+l^{2}=h^{2}+2\mu hx\quad\Longrightarrow\quad x^{2}-2(l+\mu h)x+\bigl(l^{2}-h^{2}\bigr)=0 .

Therefore

x=(l+\mu h)\pm\sqrt{(l+\mu h)^{2}-l^{2}+h^{2}}=(l+\mu h)\pm\sqrt{(\mu^{2}+1)h^{2}+2\mu l h},

using $(l+\mu h)^{2}-l^{2}+h^{2}=2\mu lh+\mu^{2}h^{2}+h^{2}$ .

Step 6 — Rejecting the upper sign.

Equation (3) came from squaring $l-x=\sqrt{h^{2}+2\mu hx}$ , whose left side must be non-negative; so any admissible root must satisfy $x\le l$ . The upper-sign root

x=(l+\mu h)+\sqrt{(\mu^{2}+1)h^{2}+2\mu lh}>l

(strictly, since $\mu h\ge0$ and the surd is positive) violates this and is an artefact of squaring — it is rejected. The lower-sign root is admissible, because

x\le l\iff \mu h\le\sqrt{(\mu^{2}+1)h^{2}+2\mu lh}\iff \mu^{2}h^{2}\le \mu^{2}h^{2}+h^{2}+2\mu lh,

which is true; and $x>0\iff (l+\mu h)^{2}>(\mu^{2}+1)h^{2}+2\mu lh\iff l^{2}>h^{2}$ , true since the chain must be longer than the height of its fixed end.

Hence the length on the table is

x=(l+\mu h)-\sqrt{(\mu^{2}+1)h^{2}+2\mu l h}.\qquad\blacksquare

Answer

\boxed{\;x=(l+\mu h)-\sqrt{(\mu^{2}+1)h^{2}+2\mu l h}\;}