Falling Ladder Problem

Problem

Given certain parameters, the problem is to find some rate of change regarding a falling ladder leaning against a wall.

Here is one manifestation:

Suppose a ladder $\ell$ meters long is leaning against a house. I have slathered grease all over the ground and am pulling the ladder away from the wall at a velocity $\vec{v_g}$ meters per second in such a way that the top of the ladder is sliding toward the ground. Assume friction is negligible. When the bottom of the ladder is $c$ meters away from the wall, at what velocity is the top of the ladder descending?

Solution

The crux of solving the problem is in finding a suitable model.

Define a coordinate system on the $xy$-plane where:

• The origin is the bottom of the wall of the house.

• $x$ is the horizontal displacement of the bottom of the ladder

• $y$ is the vertical displacement of the top of the ladder.

• $t$ is the amount of time elapsed since I have started my nefarious ladder-pulling scheme.

The assumption is that $x$ and $y$ are implicitly defined differentiable real functions of $t$.

We use the convention that:

• Horizontal movement to the right is positive

• Horizontal movement to the left is negative

• Vertical movement towards the sky is positive

• Vertical movement towards the ground is negative.

By this model, the given that the ladder is moving to the right at $\vec{v_g}$ is:

• $\dfrac {\mathrm dx} {\mathrm dt} = \vec {v_g}$

and the unknown is:

• $\left.{\dfrac {\mathrm dy} {\mathrm dt}}\right \vert_{x=c} = \ ?$

We make the simplifying assumption that the wall of the house is perpendicular to the ground.

Then, the position of the ladder can be given by the Pythagorean Theorem:

$x^2 + y^2 = \ell^2$

Differentiating both sides WRT $t$:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle D_t\left({x^2 + y^2}\right)\)	\(=\)	\(\displaystyle D_t\left({\ell^2}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	differentiate both sides WRT time
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle 2x\frac {\mathrm dx}{\mathrm dt} + 2y\frac {\mathrm dy}{\mathrm dt}\)	\(=\)	\(\displaystyle 0\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	By the chain rule, $\dfrac {\mathrm da}{\mathrm db} = \dfrac {\mathrm da}{\mathrm dc}\dfrac {\mathrm dc}{\mathrm db}$
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle x\frac {\mathrm dx}{\mathrm dt} + y\frac {\mathrm dy}{\mathrm dt}\)	\(=\)	\(\displaystyle 0\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	divide both sides by $2$

To solve for $\left.{y}\right \vert_{x=c}$, we can use the Pythagorean Theorem again:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle c^2 + y^2\)	\(=\)	\(\displaystyle \ell^2\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle y^2\)	\(=\)	\(\displaystyle \ell^2 - c^2\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle y\)	\(=\)	\(\displaystyle \sqrt{\ell^2- c^2}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	as length is a scalar, we ignore the negative solution
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle x \frac {\mathrm dx}{\mathrm dt} + y \frac {\mathrm dy}{\mathrm dt}\)	\(=\)	\(\displaystyle 0\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	from above
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle c \cdot \vec{v_g} + \sqrt{\ell^2- c^2} \cdot \frac {\mathrm dy}{\mathrm dt}\)	\(=\)	\(\displaystyle 0\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle \frac {\mathrm dy}{\mathrm dt}\)	\(=\)	\(\displaystyle \frac {-c \cdot \vec{v_g} }{\sqrt{\ell^2- c^2} }\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

$\blacksquare$

Sources

• For a video presentation of the contents of this page, visit the Khan Academy.