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There are two questions:

  1. A ladder, L[m] long, is leaning against a vertical wall. If the bottom of the ladder is pulled horizontally away from the wall at $v_h[\frac{m}{s}]$, how fast is the top of the ladder sliding down the wall when the bottom is the distance $x_0[m]$ from the wall

  2. Suppose a particular downward vertical speed is given as $v_d[\frac{m}{s}]$ . Determine how long it takes the top of the ladder to reach the speed by expressing the time in terms of $v_h, v_d, and L$

So for the first one I used related rates and set up my triangle such that


Given:

$L [m] \rightarrow$ Length of ladder

$V_h[\frac{m}{s}] \rightarrow$ Speed at which the ladder is sliding away from the wall

$x_0[m] \rightarrow$ Some distance the ladder is away from the wall.


Let us represent our rate at which the ladder is sliding down as

$\frac{\delta d}{\delta t} $ which we will later call $V_d[\frac{m}{s}]$


using the Pythagorean Theorem:

$$x_0^2 + d^2 = L^2\quad$$ where d is the point where the ladder is leaning against the wall (height of ladder)

$$2x_0 \frac{\delta x_0}{\delta t} + 2d\frac{\delta d}{\delta t} = 0\quad$$

$L^2$ goes to zero because it is a constant and never changes (ladder)

So solving for the rate at which the ladder is sliding down the wall when the bottom is a $x_0$ distance we isolate $\frac{\delta d}{\delta t}$ such that:

$$ \frac{\delta d}{\delta t} = - \frac{x_0}{d}\frac{\delta x_0}{\delta t}$$

Here we can replace the rate at which $x_0$ changes with our $V_h[m]$ which gives us

$$ \frac{\delta d}{\delta t} = - \frac{x_0}{d}V_h$$

We do not have a value for $d$, but can express it in terms of $x_0$ and $L$:

$$ d = \sqrt{-x_0^2+L^2} $$

So replacing $d$ with this equation we have

$$ \frac{\delta d}{\delta t} = - \frac{x_0}{\sqrt{-x_0^2+L^2}}V_h $$

Checking our units: $$ \frac{m}{s} = \frac{m}{m} \frac{m}{s} \implies \frac{m}{s} = \frac{m}{s} $$


This solves (1) I believe. Moving onto (2), is where I am confused. How do I find the time it takes the ladder to reach a downward velocity when we are given $V_h, V_d$, and $L$.

I know that the first answer is the velocity at which the ladder slides down, and the integral of velocity is position. So would I have to integrate which would give me the right hand side of (1) * t for some equation $y(t)$ ?

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You have the equation for a downward velocity in terms of $x$. Now the downward velocity is given. Knowing it, solve the equation for $x$. The time necessary for a bottom to reach this position would be $\frac{x}{V_h}$.

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  • $\begingroup$ Yes, here is what I got. $$ - \frac{ (-x^2 + L^2)^{3/2}}{V_h} = \frac{x}{V_d} $$ Is this correct, doing the algebra I was not sure if the answer was supposed to be in this format as it looks rough to look at. $\endgroup$ – Hawaiian Rolls Apr 10 '17 at 2:44
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Thank you @user58697 for hinting at how to solve it. By using the relation between time, velocity, and the x axis we can represent time as follows:

$$ t = \frac{x}{V_h} \rightarrow x = tV_h $$

We can use the relationship the following way. This is a simplified version and algebraically manipulated to solve for t

\begin{equation} t = \frac{V_dL}{V_h\sqrt{V_d^2 - V_h^2}} \end{equation}

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