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UPSC 2013 Maths Optional Paper 1 Q7c — Step-by-Step Solution

15 marks · Section B

Equilibrium of a system of particles · Dynamics & Statics · asked 8× in 14 yrs · Read the full method →

Question

Six equal rods $AB,BC,CD,DE,EF$ and $FA$ are each of weight $W$ and are freely jointed at their extremities so as to form a hexagon; the rod $AB$ is fixed in a horizontal position and the middle points of $AB$ and $DE$ are joined by a string. Find the tension in the string.

Technique

Force/torque equilibrium of individual rods, working “up” from $DE$ through $CD$ and $BC$; symmetry eliminates the left-half analysis.

Solution

Strategy. The freely-jointed hexagon, when the rods are equal and the structure has the hexagon shape, is taken to be a regular hexagon (standard convention in UPSC mechanics). Use symmetry + force/torque equilibrium of individual rods to back out the internal forces, ending with the string tension.

Step 1 — Geometry of the regular hexagon

Place the centre of the hexagon at the origin, with $AB$ horizontal at top:

Midpoint of $AB$: $(0,\,\ell\sqrt 3/2)$. Midpoint of $DE$: $(0,\,-\ell\sqrt 3/2)$. String length $\ell\sqrt 3$, vertical.

Step 2 — Equilibrium of rod $DE$

Forces on $DE$:

• Weight $W$ at midpoint, downward.
• Tension $T$ at midpoint, upward (toward midpoint of $AB$).
• Forces $\vec F_D=(F_{Dx},F_{Dy})$ at $D$ (from rod $CD$) and $\vec F_E=(-F_{Dx},F_{Dy})$ at $E$ (by left-right symmetry).

Vertical balance: $2F_{Dy}+T-W=0$, so

(Horizontal balance and torque about midpoint give no new info by symmetry.)

Step 3 — Equilibrium of rod $CD$

$C=(\ell,0),\;D=(\ell/2,-\ell\sqrt 3/2)$. Midpoint $M_{CD}=(3\ell/4,-\ell\sqrt 3/4)$.

Forces:

• Weight $W$ at $M_{CD}$, downward.
• $\vec F_C=(F_{Cx},F_{Cy})$ at $C$ (from rod $BC$).
• $-\vec F_D$ at $D$ (Newton’s third law, from rod $DE$).

Vertical balance: $F_{Cy}-F_{Dy}-W=0\;\Rightarrow\;F_{Cy}=F_{Dy}+W$. Horizontal: $F_{Cx}=F_{Dx}$.

Torque about $C$ (positions and forces):

• $-\vec F_D$ at $D$, lever $D-C=(-\ell/2,-\ell\sqrt 3/2)$: torque $(-\ell/2)(-F_{Dy})-(-\ell\sqrt 3/2)(-F_{Dx})=\dfrac{\ell}{2}F_{Dy}-\dfrac{\ell\sqrt 3}{2}F_{Dx}$.
• Weight $(0,-W)$ at $M_{CD}$, lever $M_{CD}-C=(-\ell/4,-\ell\sqrt 3/4)$: torque $(-\ell/4)(-W)-(-\ell\sqrt 3/4)(0)=\dfrac{W\ell}{4}$.

Sum $=0$:

Step 4 — Equilibrium of rod $BC$

$B=(\ell/2,\ell\sqrt 3/2),\;C=(\ell,0)$. Midpoint $M_{BC}=(3\ell/4,\ell\sqrt 3/4)$.

Forces:

• Weight $W$ at $M_{BC}$, downward.
• $\vec F_B$ at $B$ (from rod $AB$).
• $-\vec F_C=-(F_{Dx},F_{Dy}+W)$ at $C$.

Torque about $B$ (lever-arm bookkeeping):

• $-\vec F_C$ at $C$, lever $C-B=(\ell/2,-\ell\sqrt 3/2)$: torque $(\ell/2)(-(F_{Dy}+W))-(-\ell\sqrt 3/2)(-F_{Dx})=-\dfrac{\ell}{2}(F_{Dy}+W)-\dfrac{\ell\sqrt 3}{2}F_{Dx}$.
• Weight $(0,-W)$ at $M_{BC}$, lever $M_{BC}-B=(\ell/4,-\ell\sqrt 3/4)$: torque $(\ell/4)(-W)-(-\ell\sqrt 3/4)(0)=-\dfrac{W\ell}{4}$.

Sum $=0$:

Step 5 — Solve (II) and (III)

Add (II) and (III):

Better — solve directly. From (II): $F_{Dy}=\sqrt 3 F_{Dx}-W/2$. Substitute into (III):

Then $F_{Dy}=\sqrt 3\cdot(-W/(2\sqrt 3))-W/2=-W/2-W/2=-W$.

Step 6 — Tension from (I)

Answer