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UPSC 2023 Maths Optional Paper 1 Q5c — Step-by-Step Solution

10 marks · Section B

Equilibrium of Forces in Three Dimensions · Dynamics & Statics · Read the full method →

Question

A cylinder of radius '$a$' touches a vertical wall along a generating line. Axis of the cylinder is fixed horizontally. A uniform flat beam of length '$l$' and weight '$W$' rests with its extremities in contact with the wall and the cylinder, making an angle of $45°$ with the vertical. If frictional forces are neglected, then show that

Also, find the reactions of the cylinder and wall.

Technique

2D vertical cross-section; place wall and cylinder explicitly; force balance + moments. Take moments about the bottom contact to eliminate $R_c$ and find $R_w=W/2$ cleanly. Then horizontal balance and the cylinder-constraint pin down $a/l$.

Solution

Setup

Work in the vertical 2D cross-section perpendicular to the cylinder’s axis. Wall = $x=0$. Cylinder centre $O=(a,0)$, touching wall at origin. Beam: top end on wall at $(0,h)$, bottom end on cylinder at $(l/\sqrt 2,h-l/\sqrt 2)$ (since beam is at $45°$ with vertical, length $l$).

Bottom of beam is on the cylinder: $(l/\sqrt 2-a)^2+(h-l/\sqrt 2)^2=a^2$.

Step 1 — Force balance

Forces on beam:

• Weight $W$ at midpoint, downward.
• Reaction $R_w$ at top: horizontal (perpendicular to wall, pointing $+x$).
• Reaction $R_c$ at bottom: along outward normal of cylinder at contact, in direction $((l/\sqrt 2-a)/a,(h-l/\sqrt 2)/a)$.

Vertical equilibrium: $R_c\cdot(h-l/\sqrt 2)/a=W$, so $R_c=\dfrac{Wa}{h-l/\sqrt 2}$.

Step 2 — Take moments about the bottom of the beam

About the bottom $(l/\sqrt 2,h-l/\sqrt 2)$:

• Weight $W$ acts at midpoint, at horizontal distance $-(l/\sqrt 2)/2=-l/(2\sqrt 2)$ from the bottom (toward wall). Moment of $W$ about bottom: $W\cdot l/(2\sqrt 2)$ (clockwise).
• Reaction $R_w$ at top, horizontal, perpendicular distance from bottom = vertical separation $=l/\sqrt 2$. Moment: $R_w\cdot l/\sqrt 2$ (counter-clockwise).
• Reaction $R_c$ at bottom: zero moment.

Set $\sum M=0$: $W\cdot l/(2\sqrt 2)=R_w\cdot l/\sqrt 2$, so

Step 3 — Horizontal equilibrium gives the geometry

$\sum F_x=0$: $R_w+R_c\cdot(l/\sqrt 2-a)/a=0$, so $R_c\cdot(a-l/\sqrt 2)/a=R_w=W/2$.

Using $R_c=Wa/(h-l/\sqrt 2)$:

So $h=2a-l\sqrt 2+l/\sqrt 2=2a-l/\sqrt 2$ (since $l\sqrt 2-l/\sqrt 2=l(2-1)/\sqrt 2=l/\sqrt 2$).

Step 4 — Apply the cylinder constraint

$(l/\sqrt 2-a)^2+(h-l/\sqrt 2)^2=a^2$. Substitute $h-l/\sqrt 2=2a-l\sqrt 2$:

Expand: $(l/\sqrt 2-a)^2=l^2/2-\sqrt 2\,al+a^2.$ $(2a-l\sqrt 2)^2=4a^2-4\sqrt 2\,al+2l^2.$

Sum: $l^2/2+2l^2-\sqrt 2 al-4\sqrt 2 al+a^2+4a^2=(5/2)l^2-5\sqrt 2\,al+5a^2$.

Set $=a^2$: $(5/2)l^2-5\sqrt 2\,al+4a^2=0$. Multiply by $2$: $5l^2-10\sqrt 2\,al+8a^2=0$.

Solve for $a/l$. Let $u=a/l$:

Take the larger root (physically the configuration where beam reaches farther down the cylinder):

Rationalise: $\dfrac{5\sqrt 2+\sqrt{10}}{8}=\dfrac{\sqrt 2(5+\sqrt 5)}{8}=\dfrac{5+\sqrt 5}{4\sqrt 2}=\dfrac{\sqrt 5+5}{4\sqrt 2}$.

Step 5 — Compute reactions

$R_w=W/2$ (already found).

$R_c=Wa/(h-l/\sqrt 2)=Wa/(2a-l\sqrt 2)$. With $a/l=(\sqrt 5+5)/(4\sqrt 2)$, $a=l(\sqrt 5+5)/(4\sqrt 2)$:

Rationalise: multiply by $(\sqrt 5-1)/(\sqrt 5-1)$:

$(\sqrt 5+5)(\sqrt 5-1)=5-\sqrt 5+5\sqrt 5-5=4\sqrt 5$.

Answer