Mechanics

Part (i).

Begin by drawing a clear diagram with all forces labelled. Place the painter at a general position, a distance $xa$ from the base of the ladder, measured along it. Working in multiples of $a$ ensures the factor cancels neatly in the moments equation.

The ladder makes an angle $\theta =\arctan 2$ with the horizontal, so $\tan \theta =2$, giving $\sin \theta =\frac{2}{\sqrt{5}}$ and $\cos \theta =\frac{1}{\sqrt{5}}$.

Let $R$ be the normal reaction from the wall (horizontal) and $N$ the normal reaction from the table top (vertical). Let $F$ be the friction force at the foot of the ladder.

Resolving horizontally: $F=R$.

Resolving vertically: $N=W+kW=(1+k)W$.

Taking moments about the base of the ladder:$$R\cdot 6a\sin \theta =W\cdot 3a\cos \theta +kW\cdot xa\cos \theta .$$Substituting and simplifying:$$R\cdot \frac{12a}{\sqrt{5}}=\frac{3Wa}{\sqrt{5}}+\frac{kWxa}{\sqrt{5}},$$$$12R=W(3+kx).$$So $R=\frac{(3+kx)W}{12}$ and therefore $F=\frac{(3+kx)W}{12}$.

For the ladder not to slip, we need $F\le \frac{1}{2}N$, that is:$$\frac{(3+kx)W}{12}\le \frac{(1+k)W}{2}.$$$$3+kx\le 6(1+k)=6+6k.$$$$kx\le 3+6k.$$$$x\le \frac{3}{k}+6.$$Since the painter is at most at the top of the ladder, $x\le 6$. The condition $x\le \frac{3}{k}+6$ holds for all $x\le 6$ whenever $\frac{3}{k}\ge 0$, which is always true for $k>0$. Hence the ladder never slips on the table, regardless of how high the painter climbs.

Part (ii).

Now $k=9$. The table has two pairs of legs: two near the wall and two away from it. Draw a fresh diagram including the table's weight $2W$ acting at its centre, and the reaction forces at each pair of legs.

Let $N_1$ be the total normal reaction from the two legs nearest the wall, and $N_2$ from the two legs furthest from the wall. The ladder pushes down on the table with force $(1+k)W=10W$ at the centre of the table top, and also pushes horizontally with force $R$ (away from the wall).

From Part (i) with $k=9$: $R=\frac{(3+9x)W}{12}=\frac{3(1+3x)W}{12}=\frac{(1+3x)W}{4}$.

Resolving vertically for the table:$$N_1+N_2=2W+10W=12W.$$Taking moments about the far edge (legs furthest from wall) for the table:

The table top has side $\frac{a}{2}$, so its centre is $\frac{a}{4}$ from each edge. The foot of the ladder is at the centre of the table top.$$N_1\cdot \frac{a}{2}=2W\cdot \frac{a}{4}+10W\cdot \frac{a}{4}-R\cdot a,$$$$\frac{N_{1}a}{2}=\frac{12Wa}{4}-\frac{(1+3x)Wa}{4}=\frac{(12-1-3x)Wa}{4}=\frac{(11-3x)Wa}{4},$$$$N_1=\frac{(11-3x)W}{2}.$$The table tilts (lifts off at the near legs) when $N_1=0$, giving $x=\frac{11}{3}a$, so the painter is $\frac{11}{3}a\approx 3.67a$ up the ladder.

For slipping, the total friction available is $\frac{1}{3}(N_1+N_2)=\frac{12W}{3}=4W$. The horizontal force the ladder exerts on the table equals $R$. The table slips when $R=4W$:$$\frac{(1+3x)W}{4}=4W⟹1+3x=16⟹x=5.$$The painter is $5a$ up the ladder.

Comparing: tilting occurs at $x=\frac{11}{3}a\approx 3.67a$, slipping at $x=5a$. Since $\frac{11}{3}<5$, tilting occurs first, when the painter is a distance $\frac{11}{3}a$ up the ladder.

Part (i): Relating $R_1$, $R_2$ and $\varphi$.

Place the rod inside the cylinder with its lower end at $A$ and upper end at $B$. The cylinder exerts normal reactions $R_1$ and $R_2$ at $A$ and $B$ respectively, directed radially inward (toward the rod's ends). Since each end is a point contact on the curved inner surface, the normal at each end points along the radius to that end. Let $\varphi$ be the angle between the rod and the radius at either end — by symmetry this is the same at both ends.

When the rod is on the point of slipping, friction is fully mobilised: $F_1=\mu R_1$ at $A$ and $F_2=\mu R_2$ at $B$, both directed along the rod (opposing the tendency to slide).

Take moments about the midpoint $M$ of the rod. The weight $W$ acts through $M$ and contributes nothing. Each reaction $R_i$ acts perpendicularly to the radius, so at distance $a\sin \varphi$ from $M$; each friction $F_i$ acts along the rod, at distance $a\cos \varphi$ from $M$. Balancing:$$R_1\cdot a\sin \varphi =R_2\cdot a\sin \varphi +F_1\cdot a\cos \varphi +F_2\cdot a\cos \varphi .$$Substitute $F_i=\mu R_i$ and divide through by $a\cos \varphi$:$$(R_1-R_2)\tan \varphi =\mu (R_1+R_2).✓$$Part (ii): Finding $\tan \theta$.

Geometry first: each end of the rod lies on the cylinder, so the distance from the axis $O$ to each end is $r$. The midpoint $M$ lies at distance $a$ from each end along the rod, giving $O{M}^2=r^2-a^2$, so $\cos \varphi =a/r$ (the angle $\varphi$ satisfies $\sin \varphi =\sqrt{r^2-a^2}/r$).

Now resolve forces. Along the rod:$$(R_1-R_2)\cos \varphi +\mu (R_1+R_2)\sin \varphi =W\sin \theta .$$Perpendicular to the rod:$$(R_1+R_2)\sin \varphi -\mu (R_1-R_2)\cos \varphi =W\cos \theta .$$Divide these two equations to eliminate $W$:$$\tan \theta =\frac{(R_1-R_2)\cos \varphi +\mu (R_1+R_2)\sin \varphi }{(R_1+R_2)\sin \varphi -\mu (R_1-R_2)\cos \varphi }.$$From Part (i), $(R_1-R_2)\tan \varphi =\mu (R_1+R_2)$, so $(R_1-R_2)=\mu (R_1+R_2)/\tan \varphi$. Substitute and factor out $(R_1+R_2)/\tan \varphi$:$$\tan \theta =\frac{\mu \cos \varphi /\tan \varphi +\mu \sin \varphi }{\sin \varphi -\mu \cos \varphi /\tan \varphi }=\frac{\mu {\cos }^2\varphi +\mu {\sin }^2\varphi }{\sin \varphi \tan \varphi -\mu {\cos }^2\varphi }=\frac{\mu }{\sin \varphi \tan \varphi -\mu {\cos }^2\varphi }.$$Using $\cos \varphi =a/r$: ${\sin }^2\varphi =1-a^2/r^2$, and $\sin \varphi \tan \varphi ={\sin }^2\varphi /\cos \varphi =(r^2-a^2)/(ar)\cdot (r/a)\cdot (a/r)$... working more carefully:$$\sin \varphi \tan \varphi =\frac{{\sin }^2\varphi }{\cos \varphi }=\frac{(r^2-a^2)/r^2}{a/r}=\frac{r^2-a^2}{ar},$$$$\mu {\cos }^2\varphi =\frac{\mu a^2}{r^2}.$$So the denominator is $\frac{r^2-a^2}{ar}-\frac{\mu a^2}{r^2}$... multiply numerator and denominator of $\tan \theta$ by $a{r}^2$:$$\tan \theta =\frac{\mu r^2}{r^2-a^2-\mu a^2\cdot (r/a)\cdot (a/r)}=\frac{\mu r^2}{r^2-a^2(1+{\mu }^2)}.✓$$Deduction: $\lambda <\varphi$.

From Part (i), $\tan \varphi =\mu (R_1+R_2)/(R_1-R_2)$. Since the rod is inclined ($\theta \ne 0$) the reactions are unequal with $R_1>R_2$, so $(R_1+R_2)/(R_1-R_2)>1$, giving $\tan \varphi >\mu =\tan \lambda$. Since $\varphi ,\lambda \in (0,\pi /2)$, this means $\lambda <\varphi$.

Part (i). Consider the three-disc system as a whole. The system is in equilibrium under three external vertical forces: its total weight $3W$ downward, and two normal reactions $N_1$ from the floor (one under each lower disc). By symmetry, these reactions are equal. Resolving vertically for the whole system:$$2N_1=3W⟹N_1=\frac{3}{2}W.$$Part (ii). Let $R$ be the normal reaction between two discs in contact (acting perpendicular to the line joining their centres) and $F$ the frictional force between them (acting along the common tangent, i.e. along the line of contact).

First, take moments about the centre of one of the lower discs. The only forces with a moment about this point are the friction from the floor $f_1$ and the friction from the upper disc $F$. Each acts at a perpendicular distance equal to the radius $r$ from the centre. Since all three discs are identical, the geometry is symmetric, so $F=f_1$: all frictional forces are equal in magnitude. Call this common value $F$.

Now isolate one lower disc. The forces acting on it are:
- Weight $W$ downward,
- Normal from floor $N_1=\frac{3}{2}W$ upward,
- Friction from floor $F$ horizontal (directed inward, toward the centre),
- Normal from upper disc $R$ (directed along the line of centres, i.e. at angle $\theta$ from the vertical, pushing outward and downward),
- Friction from upper disc $F$ (perpendicular to $R$, directed along the line of centres' tangent, i.e. at angle $\theta$ from the horizontal).

The line of centres makes angle $\theta$ with the vertical (since $2\theta$ is the angle at the top vertex). The reaction $R$ acts along this line (outward from the upper disc centre), so its components are:$$R_x=R\sin \theta (\text{outward}),R_y=-R\cos \theta (\text{downward}).$$The friction $F$ from the upper disc acts perpendicular to $R$, directed so as to prevent slipping — it has components:$$F_x=-F\cos \theta (\text{inward}),F_y=-F\sin \theta (\text{downward, by geometry}).$$Resolving horizontally for the lower disc (taking outward as positive):$$R\sin \theta -F\cos \theta -F=0⟹R\sin \theta =F(1+\cos \theta ).$$Resolving vertically for the lower disc:$$N_1-W-R\cos \theta -F\sin \theta =0⟹\frac{3}{2}W-W=R\cos \theta +F\sin \theta ⟹\frac{W}{2}=R\cos \theta +F\sin \theta .$$From the horizontal equation, $R=\frac{F(1+\cos \theta )}{\sin \theta }$. Substituting into the vertical equation:$$\frac{W}{2}=\frac{F(1+\cos \theta )}{\sin \theta }\cdot \cos \theta +F\sin \theta =F\cdot \frac{(1+\cos \theta )\cos \theta +{\sin }^2\theta }{\sin \theta }=F\cdot \frac{\cos \theta +1}{\sin \theta }.$$Thus $F=\frac{W\sin \theta }{2(1+\cos \theta )}$, as required. Substituting back, $R=\frac{W}{2\sin \theta }\cdot \frac{(1+\cos \theta {)}^2}{(1+\cos \theta )}\cdot \frac{\sin \theta }{(1+\cos \theta )}$; more cleanly from $R\sin \theta =F(1+\cos \theta )$:$$R=\frac{F(1+\cos \theta )}{\sin \theta }=\frac{W\sin \theta }{2(1+\cos \theta )}\cdot \frac{1+\cos \theta }{\sin \theta }=\frac{W}{2}.$$Part (iii). For equilibrium to hold, friction must not exceed $\mu$ times the normal reaction at both contact types.

At a lower disc–floor contact: $F\le \mu N_1$, i.e. $F\le \frac{3\mu W}{2}$. Since $F=\frac{W\sin \theta }{2(1+\cos \theta )}$ and the half-angle formula gives $\frac{\sin \theta }{1+\cos \theta }=\tan (\theta /2)$, this becomes $\tan (\theta /2)\le 3\mu$, which is satisfiable for any $\mu >0$.

At a disc–disc contact: $F\le \mu R$, i.e.$$\frac{W\sin \theta }{2(1+\cos \theta )}\le \mu \cdot \frac{W}{2}⟹\frac{\sin \theta }{1+\cos \theta }\le \mu ⟹\tan \left(\frac{\theta }{2}\right)\le \mu .$$The discs cannot overlap, so there is a geometric constraint: the lower discs must be separated, meaning $\theta >0$. Also, the upper disc must rest on both lower discs, so $\theta <90^{\circ }$, giving $\tan (\theta /2)\in (0,\tan 45^{\circ })=(0,1)$.

For the lower discs not to overlap each other (their centres are $2r\sin \theta$ apart horizontally and both at height $r$), they are just touching when $\theta =30^{\circ }$. As $\theta \to 0^{+}$, $\tan (\theta /2)\to 0$; the infimum of $\tan (\theta /2)$ over all valid $\theta >0$ is $0^{+}$, but the discs must be strictly separated so the minimum of $\tan (\theta /2)$ is approached but not attained.

Actually, the binding constraint is that the two lower discs do not touch: their centre-to-centre distance is $2r\sin \theta$ (since each centre is at distance $2r$ from the upper disc's centre along lines at angle $\theta$ from the vertical), and this must be $\ge 2r$, so $\sin \theta \ge 1$... Re-examining: the triangle of centres is equilateral when all discs touch, giving $\theta =30^{\circ }$. For the lower discs not to be in contact with each other, we need $\theta >30^{\circ }$, so $\theta /2>15^{\circ }$ and $\tan (\theta /2)>\tan 15^{\circ }=2-\sqrt{3}$.

Hence for equilibrium we need $\mu \ge \tan (\theta /2)>2-\sqrt{3}$. If $\mu <2-\sqrt{3}$, then the inequality $\tan (\theta /2)\le \mu$ cannot be satisfied for any $\theta >30^{\circ }$, so no equilibrium is possible.