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UPSC 2014 Maths Optional Paper 1 Q7c — Step-by-Step Solution

20 marks · Section B

Equilibrium of a system of particles · Dynamics & Statics · asked 8× in 13 yrs · Read the full method →

Question

A regular pentagon $ABCDE$, formed of equal heavy uniform bars jointed together, is suspended from the joint $A$, and is maintained in form by a light rod joining the middle points of $BC$ and $DE$. Find the stress in this rod.

Technique

Symmetry → analyse only right half; rod-by-rod equilibrium from bottom ($CD$) up through $BC$ and $AB$; combine vertical, horizontal, and torque equations.

Solution

Setup. Regular pentagon, bars of length $s$ and weight $W$ each, jointed at vertices. Suspended from $A$ (top). The light rod between midpoints $M_{BC}$ and $M_{DE}$ keeps the shape regular.

By 5-fold symmetry (well, mirror symmetry about the vertical through $A$), $A$ is on the vertical axis; $B,E$ are symmetric; $C,D$ are symmetric; the rod is horizontal.

Step 1 — Geometry

Interior angle of regular pentagon: $108°$. Each bar makes specific angles with the vertical:

• Bar $AB$ (from $A$): makes angle $54°$ with downward vertical.
• Bar $BC$: makes angle $18°$ with vertical (tilts inward as we go down).
• Bar $CD$: horizontal.

Unit direction vectors (taking $A$ at origin, downward $y$ negative): $\vec{AB}/s=(\cos 36°,-\sin 36°)$, $\vec{BC}/s=(-\cos 72°,-\sin 72°)$.

Step 2 — Equilibrium of bar $CD$ (bottom)

By symmetry, $CD$ is horizontal with $C$ on right, $D$ on left. Forces from $BC$ at $C$ (call it $-\vec R_C$, by Newton 3rd from force $\vec R_C$ on $BC$ at $C$) and symmetric force at $D$.

Vertical balance on $CD$: $-R_{Cy}-R_{Cy}-W=0\Rightarrow R_{Cy}=-W/2$ (force on $BC$ at $C$ from $CD$ has $y$-component $-W/2$).

Step 3 — Equilibrium of bar $BC$

Let $\vec R_B$ = force on $BC$ at $B$ from $AB$. Let $T$ = tension in the rod (positive = tension, pulling midpoints toward each other; force on $M_{BC}$ in $-x$ direction).

Vertical balance on $BC$: $R_{By}+R_{Cy}-W=0\Rightarrow R_{By}=3W/2$.

Horizontal balance: $R_{Bx}+R_{Cx}-T=0\Rightarrow R_{Cx}=T-R_{Bx}$.

Torque about $B$ for $BC$ (CCW positive). Lever arms in direction $\vec{BC}/s=(-\cos 72°,-\sin 72°)$, magnitudes $s/2$ (to midpoint) and $s$ (to $C$):

• Weight at midpoint: torque $(sW/2)\cos 72°$.
• $\vec R_C$ at $C$: torque $-sR_{Cy}\cos 72°+sR_{Cx}\sin 72°$.
• String at midpoint, force $(-T,0)$: torque $-(sT/2)\sin 72°$.

Sum $=0$, substitute $R_{Cy}=-W/2$:

Then $R_{Bx}=T-R_{Cx}=\dfrac{T}{2}+W\cot 72°$.

Step 4 — Equilibrium of bar $AB$, torque about $A$

Force on $AB$ at $B$ from $BC$ is $-\vec R_B=(-R_{Bx},-3W/2)$. Lever from $A$ is $\vec{AB}=s(\cos 36°,-\sin 36°)$.

• Weight $W$ at midpoint $\vec{AM_{AB}}=(s/2)\vec{AB}/s$: torque $-(sW/2)\cos 36°$.
• $-\vec R_B$ at $B$: torque $-s\cos 36°\cdot(3W/2)-s\sin 36°\cdot R_{Bx}=-(3sW/2)\cos 36°-sR_{Bx}\sin 36°$.

Sum $=0$:

Step 5 — Solve for $T$

Equate the two expressions for $R_{Bx}$:

The negative sign indicates the rod is in compression (it pushes the midpoints apart, opposing the pentagon’s tendency to splay outward under gravity).

Magnitude:

Answer