NBPhO 2026

Overview

The bar acts as a frictionless re-direction point: pulling the free end of the rope at speed $v$ shortens the segment between bar and weight at rate $v$. The weight itself does not move at speed $v$ — its motion is constrained to the floor and must be projected onto the rope direction.

Setting the origin at the foot of the vertical from the bar, with the weight at $(x,0)$ and the bar at $(0,H)$, the rope segment has length

$$L \;=\; \sqrt{x^{2}+H^{2}} \;=\; \frac{H}{\cos\alpha}, \qquad x \;=\; H\tan\alpha, \qquad \sin\alpha \;=\; \frac{x}{L}.$$

The single kinematic constraint is

$$\dot L \;=\; -v \qquad (\text{constant, hence } \ddot L = 0).$$

Two consequences set up the whole problem:

• The horizontal speed of the weight follows from the first derivative of the constraint.
• The horizontal acceleration follows from the second derivative — and crucially, $\ddot L = 0$ because $v$ is held constant. This is the fact most easily forgotten.

The natural dimensionless group is the Froude-like ratio

$$\mathrm{Fr}\;\equiv\;\frac{v^{2}}{gH},$$

which sets the lift-off angle in part (iii). $\mathrm{Fr}\ll 1$ is the slow-pull regime (weight comes in close to the bar before lifting); $\mathrm{Fr}\gg 1$ is the fast-pull regime (the weight is hoisted off almost as soon as the rope starts to shorten).

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Part (i) — Speed of the weight

Velocity projection along the rope

The rope is inextensible. Pulling its free end at speed $v$ means the bar-to-weight segment shortens at rate $v$, so the rate at which the weight approaches the bar, measured along the rope, equals $v$.

The weight is constrained to the floor, so its velocity is purely horizontal: $\vec u = -u\,\hat x$ (taking $u>0$ since the weight moves toward the bar). The unit vector along the rope, from the weight toward the bar, is

$$\hat e \;=\; (-\sin\alpha,\;+\cos\alpha).$$

Projecting $\vec u$ onto $\hat e$ must give the rate at which the bar-to-weight distance decreases:

$$\vec u\cdot\hat e \;=\; (-u)(-\sin\alpha) \;=\; u\sin\alpha \;=\; v.$$

Hence

$$\boxed{\,u \;=\; \frac{v}{\sin\alpha}.\,}$$

Consistency check from the geometric constraint

Differentiating $L^{2}=x^{2}+H^{2}$ once,

$$L\dot L \;=\; x\dot x \;\;\Longrightarrow\;\; \dot x \;=\; \frac{L}{x}\,\dot L \;=\; \frac{1}{\sin\alpha}\,(-v) \;=\; -\frac{v}{\sin\alpha},$$

reproducing $u=|\dot x|=v/\sin\alpha$. ✓

Physics remark

Notice that $u\to\infty$ as $\alpha\to 0$. Geometrically, the weight is sweeping toward the point directly under the bar, where its velocity vector becomes orthogonal to the rope and so cannot project onto it at all — to keep the rope shortening at rate $v$ the weight would have to move infinitely fast. In practice this divergence never materialises: at some earlier angle the weight lifts off, and the kinematic constraint changes character (the weight is no longer pinned to $y=0$).

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Part (ii) — Magnitude of the acceleration

Differentiating the constraint twice

Take the time derivative of $L\dot L = x\dot x$:

$$\dot L^{2} + L\ddot L \;=\; \dot x^{2} + x\ddot x.$$

Because $\dot L=-v$ is constant, $\ddot L=0$. The constraint reduces to

$$\dot L^{2} \;=\; \dot x^{2} + x\ddot x \;\;\Longrightarrow\;\; \ddot x \;=\; \frac{\dot L^{2}-\dot x^{2}}{x} \;=\; \frac{v^{2}-u^{2}}{x}.$$

Substituting $u = v/\sin\alpha$ and $x=H\tan\alpha$:

$$\ddot x \;=\; \frac{v^{2}-v^{2}/\sin^{2}\alpha}{H\tan\alpha} \;=\; -\frac{v^{2}\cos^{2}\alpha/\sin^{2}\alpha}{H\sin\alpha/\cos\alpha} \;=\; -\frac{v^{2}\cos^{3}\alpha}{H\sin^{3}\alpha}.$$

The minus sign says the weight accelerates toward the bar, as it should. The magnitude is

$$\boxed{\,|a| \;=\; \frac{v^{2}\cos^{3}\alpha}{H\sin^{3}\alpha} \;=\; \frac{v^{2}}{H}\cot^{3}\alpha.\,}$$

Physics remark — why the acceleration exists at all

The free end of the rope is moving at constant speed, so one might naïvely expect a constant-speed motion of the weight too. It is the geometry that breaks the symmetry: as $\alpha$ shrinks, more and more of the weight’s horizontal motion is wasted across the rope rather than along it, so the weight has to speed up to keep $\dot L$ fixed. This same effect — the fact that a constant velocity along one variable forces an acceleration in another — is the kinematic content of related-rates problems generally and of the falling-ladder problem in particular.

Limiting checks

• $\alpha\to\pi/2$ (rope horizontal). Then $\cot\alpha\to 0$, so $|a|\to 0$. Correct: when the weight is far from the bar the rope is nearly horizontal, and the weight’s velocity is fully aligned with the rope; a small change in $\alpha$ produces no first-order change in speed.
• $\alpha\to 0$ (weight under the bar). Then $|a|\to\infty$ as $\cot^{3}\alpha$. The infinite-speed singularity from part (i) is accompanied by an even more violent acceleration singularity. The lift-off in part (iii) cuts both off.
• Dimensions. $[v^{2}/H] = \text{m s}^{-2}$ ✓.

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Part (iii) — Lift-off angle

Forces on the weight while it is still on the floor

Let $M$ denote the mass of the weight (the ratio $M$ will cancel — only $g$, $H$, $v$ enter the answer, so the term “heavy” is just shorthand for “rope-mass negligible”). Three forces act on it:

• gravity $-Mg\,\hat y$,
• floor normal $N\,\hat y$ with $N\ge 0$,
• rope tension $T\hat e \;=\; T(-\sin\alpha,\;+\cos\alpha)$, with $T\ge 0$ (rope can only pull).

Newton’s second law, with the weight constrained to $y=0$ (hence $\ddot y=0$) and accelerating purely horizontally as found above:

$$\begin{aligned} \text{horizontal:}\quad & -T\sin\alpha \;=\; M\ddot x \;=\; -\frac{Mv^{2}\cos^{3}\alpha}{H\sin^{3}\alpha},\\[2pt] \text{vertical:}\quad & T\cos\alpha + N - Mg \;=\; 0. \end{aligned}$$

The horizontal equation gives the rope tension while the weight is still on the floor:

$$T \;=\; \frac{M v^{2}\cos^{3}\alpha}{H\sin^{4}\alpha}.$$

The vertical equation then fixes the normal force,

$$N \;=\; Mg - T\cos\alpha \;=\; Mg - \frac{Mv^{2}\cos^{4}\alpha}{H\sin^{4}\alpha}.$$

The lift-off condition

The floor-contact constraint is unilateral: the floor can push up but not pull down. The weight is in contact precisely while $N\ge 0$. Lift-off is the angle $\alpha_{0}$ at which $N$ first reaches zero (any smaller $\alpha$ would require $N<0$, which the floor cannot supply).

Setting $N=0$:

$$Mg \;=\; \frac{Mv^{2}\cos^{4}\alpha_{0}}{H\sin^{4}\alpha_{0}} \;\;\Longrightarrow\;\; \tan^{4}\alpha_{0} \;=\; \frac{v^{2}}{gH}.$$

Therefore

$$\boxed{\,\alpha_{0} \;=\; \arctan\!\left[\left(\frac{v^{2}}{gH}\right)^{1/4}\right].\,}$$

A cleaner derivation: tension cancels out

Both Newton equations are needed at the lift-off instant:

$$T\sin\alpha \;=\; M|a|, \qquad T\cos\alpha \;=\; Mg.$$

Dividing,

$$\tan\alpha_{0} \;=\; \frac{|a|}{g}.$$

This is a remarkably clean characterisation: lift-off occurs precisely when the (horizontal) acceleration of the weight has grown to $g\tan\alpha$. Substituting $|a| = v^{2}\cos^{3}\alpha_{0}/(H\sin^{3}\alpha_{0})$ recovers $\tan^{4}\alpha_{0}=v^{2}/(gH)$ in one line.

Limits of $\mathrm{Fr}\equiv v^{2}/(gH)$

• $\mathrm{Fr}=1$ (i.e. $v=\sqrt{gH}$). Then $\tan\alpha_{0}=1$, so $\alpha_{0}=45^{\circ}$. The pull speed is exactly the free-fall speed accumulated over height $H$.
• $\mathrm{Fr}\to 0$ (slow pull). $\alpha_{0}\to 0$: the rope must be drawn nearly vertical before the weight lifts. Physically reasonable — at low pull speed the inertial term that lifts the weight, $\sim v^{2}/H$, is small compared to $g$, so the geometric advantage of $\cos\alpha\approx 1$ is required to overcome gravity.
• $\mathrm{Fr}\to\infty$ (fast pull). $\alpha_{0}\to\pi/2$: lift-off almost immediately, while the rope is still nearly horizontal.

The fourth-root scaling, $\tan\alpha_{0}\propto\mathrm{Fr}^{1/4}$, is sluggish: increasing $v$ by a factor of $16$ only doubles $\tan\alpha_{0}$. Two factors of two come from the $\sin^{4}/\cos^{4}$ in the lift-off condition (one from $T\propto 1/\sin^{4}$, one from the vertical projection $\cos\alpha$).

Physics remark — why it isn’t simply ”$T = Mg$ when rope is vertical”

A common but wrong shortcut is: “the weight lifts off when the rope tension exceeds the weight’s mass.” That would imply lift-off requires the rope to be vertical, with no role for $v$ or $H$ at all. The error is forgetting that only the vertical component of $T$ supports the weight; while the rope is at an angle, only $T\cos\alpha$ is doing the lifting, and $T$ itself is fixed by the horizontal equation $T\sin\alpha = M|a|$. The two together produce the $\tan\alpha_{0}=|a|/g$ condition.

Sanity check — what happens just after lift-off?

Just after lift-off the floor is gone and the weight is a 2-D pendulum-like object on a rope of decreasing length, with initial velocity purely horizontal and equal to $u_{0} = v/\sin\alpha_{0}$. Plugging $\sin\alpha_{0}$ from the lift-off condition,

$$u_{0}^{2} \;=\; \frac{v^{2}}{\sin^{2}\alpha_{0}}\;=\;v^{2}\!\left(1+\cot^{2}\alpha_{0}\right) \;=\;v^{2}\!\left(1+\sqrt{\frac{gH}{v^{2}}}\right) \;=\;v^{2}+v\sqrt{gH}.$$

So the lift-off speed is order $\sqrt{gH}$ when $v\ll\sqrt{gH}$ and order $v$ when $v\gg\sqrt{gH}$ — consistent with the two regimes identified above. (The subsequent motion is beyond the scope of the problem.)

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Summary of results

Part	Quantity	Result
(i)	speed of the weight	$\displaystyle u \;=\; \frac{v}{\sin\alpha}$
(ii)	magnitude of acceleration	$\displaystyle \lvert a\rvert \;=\; \frac{v^{2}\cos^{3}\alpha}{H\sin^{3}\alpha} \;=\; \frac{v^{2}}{H}\cot^{3}\alpha$
(iii)	lift-off angle	$\displaystyle \alpha_{0} \;=\; \arctan\!\left[\left(\frac{v^{2}}{gH}\right)^{1/4}\right]$, equivalently $\tan^{4}\alpha_{0}=v^{2}/(gH)$